Probability

Ten cards are numbered, hence the sample space is

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n(S) = 10

Let event A be the even number card drawn, then

A = {2, 4, 6, 8, 10}, n(A)=5

Now probability that an even card is drawn is

Let event B be the cards greater than 3, then

B = {4, 5, 6, 7, 8, 9, 10}, n(B) = 7

So the probability of A∩B

So, A∩B = {4, 6, 8, 10}, n(A∩B)=4

So the required probability will become

Hence the probability that an even card is drawn out of cards that are more than 3 is

Let b and g represents the boy and the girl child respectively.

Now if a family has two children, the sample space will be

S = {(b, b), (b, g), (g, b), (g, g)}, n(S)=4

Let A be the event that both children are girls, the

A= {(g, g)}, n(A)=1

So the probability that both children are girls is

If the youngest is a girl

Let B be the event that the youngest child is a girl

Then B = {(b, g), (g, g)}, n(B) = 2

And the corresponding probability becomes

The sample space for the both girls is girl child and the youngest is a girl will become

(A∩B) = {(g, g)}, n(A∩B)=1

And the corresponding probability becomes

So the conditional probability that both are girls given that the youngest is a girl is

At least one is a girl

Let C be the event that at least one is a girl

Then C = {(b, g), (g, b), (g, g)}, n(C) = 3

And the corresponding probability becomes

The sample space for the both girls is girl child and at least one is a girl will become

(A∩C) = {(g, g)}, n(A∩C)=1

And the corresponding probability becomes

So the conditional probability that both are girls given that the youngest is a girl is

Let S be the sample space of throwing two dice and the two numbers are different, then the sample space is

S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

n(S) = 30

Let B be the event that the sum if the numbers on the dice is 4, then the sample space becomes

B = {(1, 3), (3, 1)}, n(B) = 2

So the probability of the event ‘the sum of numbers on the dice is 4’ is

If h and t represents heads and tails respectively, then when a coin is tossed three times, the sample space becomes

S = {(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)},n(S)=8

Let E be the event of head occurring on the third toss, then the favorable outcomes will be

E = {(HHH, HTH, THH, TTH)}, n(E) = 4

So the corresponding probability will be

Let F be the event that head occurs on first two toss, then the favorable outcomes will be

F = {(HHH,HHT)}, n(F) = 2

So the corresponding probability will be

And the favorable outcome for getting head on all the three toss will be

(E∩F)={(HHH)}, n(E∩F)=1

And the corresponding probability becomes

So if head occurs on first two tosses, the probability of getting head on third tosses

When a die is thrown three times the number of all favorable outcomes are n(S) = 216

Let E be the event that 4 occurs on the third toss, then the favorable outcomes will be

E ={(1, 1, 4),(1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4)

(2, 1, 4),(2, 2, 4), (2, 3,4), (2, 4, 4), (2, 5, 4), (2, 6, 4)

(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6,4)

(4, 1 ,4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4)

(5, 1, 4), (5, 2,4), (5, 3, 4),(5,4, 4), (5, 5, 4), (5, 6, 4)

(6, 1, 4),(6, 2, 4), (6, 3, 4), (6,4, 4),(6, 5, 4), (6, 6, 4)}

n(E) = 36

So the probability that 4 appears on the third toss is

Let F be the event that 6 and 5 appear on first two tosses respectively, and then the favorable outcomes will be

F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 6)}

n(F) = 6

So the probability that 6 and 5 appear on first two tosses respectively will be

So the favorable outcomes that 6 and 5 appear respectively on first two tosses and 4 appears on the third toss will be

(E∩F) = {(6, 5, 4)}, n(E∩F) = 1

And the corresponding probability will be

So the conditional probability that if it is given that 6 and 5 appear respectively on first two tosses and 4 appears on the third toss will be

Given: P(B) = 0.5, P(A∩B) = 0.32

To find: P(A|B)

We know that,

By substituting the values we get

is the required value

Given: P(A) = 0.4, P(B)=0.3, P(B|A) = 0.5

To find: P(A|B) and P(A∩B)

We know that,

By substituting the values we get

is the required value

Similarly,

By substituting the values we get

is the required value

Given:

To find: P(A|B) and P(B|A)

We know that,

By substituting the values we get

We also know that,

By substituting the values we get

is the required value

Similarly,

By substituting the values we get

is the required valu

Let b and g represents the boy and the girl child respectively.

Now if a family has two children, the sample space will be

S = {(b, b), (b, g), (g, b), (g, g)}, n(S)=4

(i) Let A be the event that both children are males, then

A= {(b, b)}, n(A)=1

So the probability that both children are males is

Let B be the event that at least one of the children is male

Then B = {(b, b), (b, g), (g, b)}, n(B) = 3

And the corresponding probability becomes

The sample space for the at least male and both being male will become

(A∩B) = {(b, b)}, n(A∩B)=1

And the corresponding probability becomes

So the conditional probability that both are males given that the at least one is male is

(ii) Let P be the event that both children are females, then

P= {(g, g)}, n(P)=1

So the probability that both children are females is

Let Q be the event that the elder child is a female

Then Q = {(g, b), (g, g)}, n(Q) = 2

And the corresponding probability becomes

The sample space for the elder child being female and both child being female will become

(P∩Q) = {(g, g)}, n(P∩Q)=1

And the corresponding probability becomes

So the conditional probability that if the elder child is female then both children are females is

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is a king. There are four kings in the pack. Hence, the probability of the first card is a king is

Let B be the event that second card is also king without replacement. Then there are 3 kings left in the pack as the cards are not replaced. Therefore, the probability of the second card is also king is

Then the probability of getting two kings without replacement is

The probability that both of them are kings is

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is ace (or, kings). There are four aces (or, kings) in the pack. Hence, the probability of the first card is ace (or, kings) is

Let B be the event that second card is also ace (or, kings) without replacement. Then there are 3 aces (or, kings) left in the pack as the cards are not replaced. Therefore, the probability of the second card is also ace (or, kings) is

Let C be the event that third card is also ace (or, kings) without replacement. Then there are 2 aces (or, kings) left in the pack as the cards are not replaced. Therefore, the probability of the third card is also ace (or, kings) is

Let D be the event that fourth card is also ace (or, kings) without replacement. Then there are 1 ace (or, kings) left in the pack as the cards are not replaced. Therefore, the probability of the fourth card is also ace (or, kings) is

Then the probability all are aces (or, kings) is

The probability that all are aces (or, kings) is

Bag contains 5 red balls and 7 white balls. So the total number of all favorable cases is n(S) = 5+7=12

Let A represents first ball as white ball, and B be second ball as white ball.

Then the probability of drawing two white balls without replacement is

P(2 white balls without replacement)

(as there are 7 white balls in first draw out of 12 balls, and 6 white balls in second draw out of 11 balls as the balls are not replaced)

Hence the required probability is

There are 25 tickets numbered from 1 to 25, so the sample space is

S = {1, 2, 3, …., 25}, n(S) = 25

Number of even numbered tickets from 1 to 25 is

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24} = 12

Let A represents first ticket with even number and B represents second ticket with even number

Then the probability of both tickets being even number without replacement is

P(both tickets showing even number without replacement)

(as there are 12 even numbered tickets out of 25 tickets in first draw, and 11 even numbered tickets out of 24 tickets in second draw as the tickets are not replaced)

Hence the required probability is

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is card of spade. There are 13 spade cards in the pack. Hence, the probability of the first card is spade is

Let B be the event that second card is also card of spade without replacement. Then there are 12 spade cards left in the pack as the cards are not replaced. Therefore, the probability of the second card is also spade card is

Let C be the event that third card is also spade card without replacement. Then there are 11 spade cards left in the pack as the cards are not replaced. Therefore, the probability of the third card is also a spade card is

Then the probability all three are spade cards without replacement is

(as there are 13 spade cards in the pack of 52 in first draw, 12 spade cards in the pack of 51cards in the second draw as the cards are not replaced and 11spadecards in the pack of 50 cards in the third draw as the cards are not replaced.)

The required probability is

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is a king. There are four kings in the pack. Hence, the probability of the first card is a king is

Let B be the event that second card is also king without replacement. Then there are 3 kings left in the pack as the cards are not replaced. Therefore, the probability of the second card is also king is

Then the probability of getting two kings without replacement is

(as there are 4 kings out of 52 cards in first draw, and 3 kings out of 51 cards in the second draw as the cards are not replaced)

The probability that both of them are kings is

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is a king. There are four kings in the pack. Hence, the probability of the first card is a king is

Let B be the event that second card is an ace without replacement. Then there are 4 aces in the pack as the cards are not replaced. Therefore, the probability of the second card is an ace is

Then the probability of getting first is a king and the second is an ace without replacement is

(as there are 4 kings out of 52 cards in first draw, and 4 aces out of 51 cards in the second draw as the cards are not replaced)

The probability that first is a king and the second is an ace without replacement is

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is a heart. There are 13 hearts in the pack. Hence, the probability of the first card is a heart is

Let B be the event that second card is red without replacement. Then there are 26 red cards in the pack but as the cards are not replaced now there are 25 red cards as one heart which is red in color is already drawn out. Therefore, the probability of the second card is a red is

Then the probability of getting first is a heart and the second is a red card without replacement is

The probability that first is a heart and the second is red card without replacement is

There are 20 tickets numbered from 1 to 25, so the sample space is

S = {1, 2, 3, …., 20}, n(S) = 20

Number of even numbered tickets from 1 to 20 is

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20} = 10

Number of odd numbered tickets from 1 to 20 is

{1, 3, 5, 7, 9, 11, 13, 15, 17, 19} = 10

Let A represents first ticket with even number and B represents second ticket with odd number

Then the probability of first ticket having even number and second ticket having odd number without replacement is

(as there are 10 even numbered tickets out of 20 tickets in first draw, and 10 odd numbered tickets out of 19 tickets in second draw as the tickets are not replaced)

Hence the required probability is

There are 3 white, 4 red and 5 black balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 3+4+5=12

Let A be the event of not getting a black ball in the first draw, this means getting another color (red or white) ball out of 12 balls in the first draw. Hence the probability becomes

(as there are 3 white + 4 red = 7 balls)

Let B represents the event of not getting a black ball in the second draw, this means getting another color (red or white) ball out of 11 balls in the second draw as the balls are not replaced. Hence the probability becomes

(as there are 3 white + 4 red = 7 balls and one ball is already drawn in first draw so now there are total of 6 red and white balls)

Then the probability of at least one black ball without replacement

Hence the required probability is

There are 5 white, 7 red and 3 black balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 5+7+3=15

Let A be the event of getting a red ball in the first draw. Hence the probability becomes

(as there are 7 red balls out of 15 balls)

Let B represents the event of getting a red ball in the second draw. Hence the probability becomes

(as there are 7 red balls and one red ball is already drawn in first draw so now there are total of 6 red balls)

Let C represents the event of getting a red ball in the third draw. Hence the probability becomes

(as there are 7 red balls and two red balls are already drawn in first and second draw so now there are total of 5 red balls)

Then the probability of none being red balls without replacement

Hence the required probability is

There are 52 cards in a deck. Let A be the event that first card drawn is a heart. There are 13 hearts in the pack. Hence, the probability of the first card is a heart is

Let B be the event that second card is a diamond without replacement. Then there are 13 diamond cards in the pack but as the cards are not replaced now there are 51 cards in total in the deck. Therefore, the probability of the second card is a red is

Then the probability of getting first is a heart and the second is a diamond card without replacement is

The probability that first is a heart and the second is diamond card without replacement is

There are 10 black and 5 white balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 10+5=15

Let A be the event of getting a black ball in the first draw. Hence the probability becomes

(as there are 10 black balls out of 15 balls)

Let B represents the event of getting a black ball in the second draw. Hence the probability becomes

(as there are 10 black balls and one black ball is already drawn in first draw so now there are total of 9 black balls)

Then the probability of all being black balls without replacement

Hence the required probability is

There are 52 cards in a deck. Let A be the event that first card drawn is a king. There are 4 hearts in the pack. Hence, the probability of the first card is a king is

Let B be the event that second card is also a king without replacement. Then there are 3 king cards out of 51 cards in the pack as the cards are not replaced. Therefore, the probability of the second card is a red is

Let C be the event that third card is an ace card without replacement. Then there are 4 ace cards out of 50 cards in the pack as the cards are not replaced. Therefore, the probability of the third card is an ace card is

Then the probabilities of getting first two cards are kings and third card drawn is an ace without replacement is

The probability that first two cards are kings and third card drawn is an ace without replacement is

Number of oranges the box contains = 15

Number of good oranges = 12

Number of bad oranges = 3

Probability that box is approved for sale

= Probability that first orange is good × probability that second orange is good, given first is good × probability that third orange is good, given first two are good

Let A represents a good orange

Then P(A)=P(getting first orange as good)

And

P(A│A)=P(getting second orange good, given first is good)

(as now there are 11 good oranges left out of 14 total oranges as one good orange is already drawn in first draw and are not replaced)

And also,

P(AA│A)=P(getting third orange good, given first two are good)

(as now there are 10 good oranges left out of 13 total oranges as two good orange is already drawn in first draw and are not replaced)

So the probability that box is approved for sale is

Therefore probability that box is approved for sale is

There are 4 white, 7 black and 5 red balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 4+7+5=16

Let A be the event of getting a white ball in the first draw. Hence the probability becomes

(as there are 4 white balls out of 16 balls)

Let B represents the event of getting a black ball in the second draw. Hence the probability becomes

(as there are 7 black balls out of 15 balls as balls are not replaced back in the bag)

Let C represents the event of getting a red ball in the third draw. Hence the probability becomes

(as there are 5 red balls out of 14 balls as balls are not replaced back in the bag)

Then the probability that the balls drawn are white, black and red respectively without replacement

Hence the required probability is

We have, =

From the given data, P (A ∩ B) =

And P (B) =

Hence, =

= (answer)

We have, =

And =

And also P (A ∩ B) = P (B ∩ A)

Hence = (answer)

And (answer)

We have, =

Therefore =

{multiply numerator and denominator by 2 to convert into the whole number}

= 0.64 (answer)

We have, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

From equation, =

P (B ∩ A) = × P (A) = 0.6× 0.4

= 0.24

Hence, = =

= 0.3 (answer)

And P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

= 0.4+0.8-0.24

= 0.96 (answer)

We have, P (A) = , P (B)= and P(A ∪ B) =

We have, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

Therefore, P (A ∩ B) = P (A)+ P (B) - P (A ∪ B)

= )

= {Make denominator same by taking LCM)

= (Answer)

Now, = = )

= (answer)

And = = = (answer)

We have P (A) = 6/11, P (B) = 5/11 and P (A ∪ B) = 7/11

We have, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

Therefore, P (A ∩ B) = P (A)+ P (B) - P (A ∪ B)

=

= { 11 is common in denominator }

= (answer)

Now, = =

= (answer)

And = =

(answer)

We have, P (A)= 7/13, P (B)=9/13 and P(A ∩ B)=4/13

Now, =

= = (answer)1

We have, P(A)=1/2,P (B)=1/3 and P(A ∩ B) = 1/4

Now, =

=

= (answer)

And = =

(answer)

We have, 2P (A) = P (B) =

2P (A) =

P (A) = =

Now, we need P (A ∩ B)

We have, =

P (A ∩ B) =

=

=

Now, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

=

(taking lcm as 26 is common factor)

= (answer)

We have, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B)

i. From above formula,

P (A ∩ B)= P (A)+ P (B) - P (A ∪ B)

=

= (answer)

ii. =

(answer)

iii. = = (answer)

When a coin is tossed three times, we have following outcomes

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total outcomes =8

From above outcomes,

A= Heads on third toss = {HHH, HTH, THH, TTH} =4

Therefore probability of occurrence of event A = P(A) =

B= heads on first two tosses = {HHH, HHT} =2

Therefore probability of occurrence of event B = P(B) =

Also we want P (A ∩ B) = probability of occurrence of both events A and B

=heads on first two tosses and heads on third toss

=heads on all tosses = {HHH} = 1(Occurrence Of both A and B events

Therefore, P (A ∩ B) =

Hence, =

==(answer)

When a coin is tossed three times, we have following outcomes

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total outcomes =8

A= at least two heads ={HHH, HHT, THH, HTH} = 4

Therefore probability of occurrence of event A = P(A) =

B= at most two heads ={HHT, HTH, HHT, THT, TTH, TTT, HTT} =7

Therefore probability of occurrence of event B = P(B) =

Now Also we want P (A ∩ B) = probability of occurrence of both events A and B

= occurrence of atleast and atmost two heads ={HHT, HTH, THH} =3

Hence, probability of occurrence of both events A and B = P (A ∩ B)=

Therefore, =

= (answer)

When a coin is tossed three times, we have following outcomes

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total outcomes =8

A= atmost two tails ={HHH, HHT, HTH, THH, HTT, THT, TTH, } = 7

Therefore probability of occurrence of event A = P(A) =

B= at least one tail ={HHT, HTH, HHT, THT, TTH, TTT, HTT} =7

Therefore probability of occurrence of event B = P(B) =

Now Also we want P (A ∩ B) = probability of occurrence of both events A and B

= occurrence of atleast two tail and atleast one tail ={HHT, HTH, THH, HTT, THT, TTH,} =6

Hence, probability of occurrence of both events A and B = P (A ∩ B)=

Therefore, = = (answer)

When two coins are tossed, total outcomes are

{HH, HT, TH, TT} =4

A=Tail appears on one coin= {HT, TH} =2

Probability of A = P(A) =

B= One coin shows head ={HT, TH} =2

Probability of B =P(B) =

(A ∩ B) = one coin shows head and another shows tail = {HT, TH} =2

P (A ∩ B) =

Hence, =

= (answer)

When two coins are tossed, total outcomes are

{HH, HT, TH, TT} =4

A=No tail occurs = {HH} =1

Probability f A = P(A) =

B= No head occurs={TT} =1

Probability of B =P(B) =

(A ∩ B) = No tail and no head occurs = not possible =0

P (A ∩ B) =

Hence, =

= (answer)

When a die is thrown 3 times, total possible outcomes are, 6³ = 216

A= 4 appears on third toss =

{(1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4),(1,6,4)

(2,1,4),(2,2,4),(2,3,4),(2,4,4),(2,5,4),(2,6,4)

(3,1,4),(3,2,4),(3,3,4),(3,4,4),(3,5,4),(3,6,4)

(4,1,4), (4,2,4)(4,3,4),(4,4,4),(4,5,4),(4,6,4)

(5,1,4),(5,2,4),(5,3,4)(5,4,4),(5,5,4),(5,6,4)

(6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4),(6,6,4)}

=36

Hence P(A) =

B= 6,5 appears respectively on first two tosses

{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}

= 6

Hence P(B) =

(A ∩ B) = 6,5 occurs on first two toss and 4 occur on third toss = (6,5,4) = 1

P (A ∩ B) =

Hence =

= = (answer)

= (answer)

Total possible outcomes =

{FMS, FSM, MFS, SMF, SFM, MSF} =6

A= Son on end = {FMS, MFS, SMF, SFM} =4

Therefore P(A) =

B= father in middle = {MFS, SFM} = 2

Therefore P(B) =

(A ∩ B) = son on one end and father in middle

= {MFS, SFM} =2

P(A ∩ B) =

Hence = (answer)

. = (answer)

When a dice is thrown 2 times, total outcomes are

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

= 36 possible outcomes

A= sum of numbers is 6

={(1,5),(2,4),(3,3),(4,2),(5,1)} = 5

P(A) =

B= 4 has appeared ateast once

= {(1,4),(2,4),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,4),(6,4)} =11

P(B)=

(A ∩ B) = sum of two number is 6 and also 4 has occurred ateast once

(A ∩ B) ={ (2,4),(4,2)} = 2

P(A ∩ B) =

Conditional probability = = (answer)

We know total possible outcomes when two dice are thrown = 36

A= sum is 8

{(2,6),(3,5),(4,4),(5,3),(6,2)} = 5

P(A) =

B =second die exhibit 4

{(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}= 6

P(B) =

(A ∩ B) = sum of two number is 8 and also second die exhibit 4

=(4, 4) =1

P(A ∩ B) =

Therefore, = (answer)

We know total possible outcomes when two dice are thrown = 36

A= sum is 7

{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1l} = 6

P(A) =

B =second die exhibit odd number

{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3),(5,5),(6,1),(6,3),(6,5)}= 18

P(B) =

(A ∩ B) = sum of two number is 7 and also second die exhibit odd number

={(2,5),(4,3),(6,1)} =3

P(A ∩ B) =

Therefore, =

= (answer)

We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting 7 as sum are {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6

Therefore, P(A) = 6/36 =

B= No of outcomes for second die exhibiting prime no =

{(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(4,2),(4,3),(4,5),(5,2),(5,3),(5,5),(6,2),(6,3),

(6,5)} = 18 {since 2,3,5 are prime nos )

Hence P(B) =

(A B) = outcomes of getting 7 as a sum with a second die showing prime no

= {(2,5), (5,2),(4,3)} = 3

Hence P(A B) = 3/36 =

Therefore, = = (answer)

Total outcomes when a die is rolled = {1,2,3,4,5,6} = 6

A = outcome is odd number = {1,3,5} = 3

Hence P(A) =

B= Outcome is prime number = { 2,3,5) = 3

Hence P(B) =

(A B) = outcome is odd number and also prime number = {3,5} =2

Hence P(A B) =

We require, =

(answer)

We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting sum of 8 or more are

{(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)} = 15

Therefore, P(A) =

B= No of outcomes for first die showing 4 =

{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)} = 6

Hence P(B) =

(A B) = outcomes of getting sum as 8 or more with a first die showing 4

= {(4,4),(4,5),(4,6)} = 3

Hence P(A B) =

Therefore, = (answer)

We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting sum of 8 are

{(2,6),(3,5),(4,4),(5,3),(6,2)} = 5

Therefore, P(A) =

B= No of outcomes for at least one die does not show 5

= {(1,1),(1,2),(1,3),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,6),(3,1),(3,2),(3,3),(3,4),(3,6),(4,1),(4,2),

(4,3),(4,4),(4,6),(6,1),(6,2),(6,3),(6,4),(6,6) } = 25

Hence P(B) =

(A B) = outcomes of getting sum as 8 with at least one die not showing 5

= {(2,6),(4,4),(6,2)= 3

Hence P(A B) =

Therefore, =

(answer)

Given: Two numbers are selected at random from integers 1 to 9.

Let A be the event when both the numbers are odd.

A = {(3, 1), (5, 1), (7, 1), (9, 1), (3, 5), (3, 7), (3, 9), (5, 3), (5, 7), (5, 9), (7, 3), (7, 5), (7, 9), (9, 3), (9, 5), (9, 7)}

Let B be the event when sum of both numbers is even.

B = {(1, 3(, (1, 5), (2, 4), (1, 7), (2, 6), (3, 5), (1, 9), (2, 8), (3, 7), (4, 6), (7, 5), (8, 4), (9, 3), (8, 6), (9, 5), (9, 7)}

A B = {(1, 3), (1, 5), (1, 7), (3, 5), (1, 9), (3, 7), (7, 5), (9, 3), (9, 5), (9, 7)}

We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting sum of 8 are

{(2,6),(3,5),(4,4),(5,3),(6,2)} = 5

Therefore, P(A) =

B= No of outcomes for at least 5 has appeared once =

= {(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5)} =11

Hence P(B) =

(A B) = outcomes of getting sum as 8 with at least one die showing 5

= {(3,5),(5,3)= 2

Hence P(A B)

Therefore, =

(answer)

Given: Two dice are thrown.

Let A be the event when the sum of the numbers = 7

Therefore, A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Let B be the event when first dice shows 6

Therefore, B = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

A B = {(6, 1)}

We know, when a pair of dice is thrown, total possible outcomes are = 36

E=No of outcomes for getting sum greater than or equal to 10

{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)} = 6

Therefore, P(E) =

Case 1:

F= No of outcomes for 5 appears on first die =

= {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)} =6

Hence P(B) =

(E F ) = outcomes of getting a sum greater than or equal to 10 and first die showing 5

= {(5,6),(5,5)= 2

Hence P(E F) =

Therefore, = (answer)

Case 2:

F = 5 appears on atleast one die =

{(1,5),(2,5),(3,5),(4,5),( 5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5)} = 11

P(F) =

(E F ) = outcomes of getting a sum greater than or equal to 10 and at least die showing 5

= {(5,6),(5,5),(6,5)}= 3

Hence P(E F) =

Therefore, = (answer)

Let M = Mathematics , C= computer science

By given data,

P(M) = and P(M C) =

We need to find, (Answer)

Let S= Shirt, T= Trouser

P(S) = 0.2 , P(T) = 0.3 and = 0.4

We need to find P(S T) and

We know, From given data, 0.4 = P (S T) / 0.3

P (S T) = = 0.12 (Answer).

Now we want,

= (answer)

Given: Total Number of students = 1000

Number of Girls = 430

% of girls in Class XII = 10%

Let A be the event of Student chosen studies in Class XII

And Let B be the event that the student chosen is a girl

Now,

Given: Total Number of Cards = 10

Let A be the event of drawing a number more than 3.

Let B be the event of drawing an even number.

Let B= boy G= girl

And let us consider, in a sample space, the first child is elder and second child is younger.

Total possible outcome = {BB, BG, GB, GG} = 4

Let A= be the event that both the children are girls = 1

Therefore P(A) =

Case 1.

Let B = event that youngest is girl = {BG, GG} =2

{Since we have considered second is younger in a sample space}

Therefore P(B) =

And (A ∩ B) = both are girls and younger is also girl = (GG) = 1

Therefore , P (A ∩ B) =

We require

=

= (answer)

Case 2.

Let B = event that at least one is girl = {BG,GB GG} =3

{Since we have considered second is younger in a sample space}

Therefore P(B) =

And (A ∩ B) = both are girls and atlas one is girl = (GG) = 1

Therefore , P (A ∩ B) =

We require

=

= (answer)

It is given that the coin is tossed thrice, so the sample space will be,

S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

For independent event, P(A)*P(B)=P(A∩B)

i. A=the first throw results in head, B=the last throw results in tail

A={HHH,HHT,HTH,HTT}

B={HHT,HTT,THT,TTT}

A∩B={HHT,HTT}

Therefore A and B are independent events.

ii. A=the number of heads is odd, B=the number of tails is odd

A={HTT,THT,TTH}

B={HTH,THH,HHT}

A∩B={}=Ø

A and B are not independent

iii. A=the number of heads is two, B=the last throw results in head

A={HHT,HTH,THT}

B={HHH,HTH,THH,TTH}

A∩B={HTH}

Therefore A and B are not independent events.

We are throwing two dice so,

Sample space S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sample space contains 36 elements,

A= Number four appears on first die

A={(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}

B=Number 5 on second die

B={(1,5), (2,5), (3,5),(4,5),(5,5),(6,5)}

P(A∩B)={(4,5)}

Therefore A and B are independent events

(i) In a pack of 52 cards there are 4 king,4 queen,4 jack

A=the card drawn is a king or queen

B=the card drawn is a queen or jack

A∩B=Card is a queen

Therefore A and B are not independent events

(ii) In a pack of 52 cards there are 4 king,26 black

A=the card drawn is a black

B=the card drawn is a king

A∩B=Card is a black king

Therefore A and B are independent events

(iii) In a pack of 52 cards there are 13 spade,4 ace

A=the card drawn is a spade

B=the card drawn is an ace

A∩B=Card is an ace from spade

Therefore A and B are independent events

Sample space S={(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)}

A=first toss is head

A={HHH,HTH,HHT,HTT}

B=second toss is head

B={HHH,HHT,THT,THH}

A∩B={HHH,HHT}

Therefore A and B are independent events.

(ii) Sample space S={(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)}

B=second toss is head

P(B)={HHH,HHT,THT,THH}

C= exactly two heads are tossed in a row.

P(C)={HHT,THH}

A∩B={HHT,THH}

Therefore B and C are not independent events.

(iii) Sample space S={(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)}

C= exactly two heads are tossed in a row.

C={HHT,THH}

A=first toss is head

A={HHH,HTH,HHT,HTT}

A∩B={HHT}

Therefore A and C are independent events.

P (A ∪ B)=P(A)+P(B)–P(A∩B)

Therefore A and B are independent events.

For independent event,

P(A)*P(B)=0.3*0.6=0.18

ii.

iii.

iv.

=[(1–P(A)][1–P(B)] =[1–0.3][1–0.6]

=0.7*0.4=0.28

iv. P (A ∪ B)

P (A ∪ B)=P(A)+P(B)–P(A∩B)

=0.3+0.6–0.18=0.72

v. P (A / B)

vi. P (B / A)

We know that,

P (A ∪ B)=P(A)+P(B)–P(A∩B)

P(not B)=1–P(B)

=1–0.65=0.35

Given that A and B are independent events,

Therefore,

P (A ∪ B)=P(A)+P(B)–P(A∩B)

=P(A)+P(B)–[P(A)*P(B)]

P (A ∪ B)=P(A)[1–P(B)]+P(B)

0.85=P(A)*0.65+0.35

P(A)*0.65=0.50

Let , denote the complements of A, B respectively.

Given that, P(A∩) )

P(A)P()

P(A)[1–P(B)]

P(A)=1/6[1–P(B)]

P(A'∩B)

P(A')P(B)

[1–P(A)]P(B) )

[1–1/6{1–P(B)}]P(B) )

[{6–6P(B)–1}/{6–6P(B)}]P(B) )

15[5–6P(B)]P(B)=2[6–6P(B)]

15[5–6P(B)]P(B)=12[1–P(B)]

5[5–6P(B)]P(B)=4[1–P(B)]

25P(B)–30[P(B)]²=4–4P(B)

–30[P(B)]²+25P(B)+4P(B)–4=0

30[P(B)]²–29P(B)+4=0

30a²–29a+4=0 where P(B)=a

30a²–24a–5a+4=0

6a(5a–4)–1(5a–4)=0

(6a–1)(5a–4)=0

6a–1=0

6a=1

a

P(B)

5a–4=0

5a=4

a=

P(B)

Therefore, P(B),

P(A∩B)=

6P(A)²–5P(A)+1=0

6P(A)²–3P(A)–2P(A)+1=0

[2P(A)–1][3P(A)–1]=0

2P(A)–1=0

2P(A)=1

3P(A)–1=0

3P(A)=1

Given that A and B are independent events,

We know that,

P (A ∪ B)=P(A)+P(B)–P(A∩B)

Therefore,

P (A ∪ B)=P(A)+P(B)–P(A∩B)

=P(A)+P(B)–[P(A)*P(B)]

0.6=0.2+P(B)–0.2P(B)

0.6–0.2=P(B)[1–0.2]

0.4=P(B)*0.8

We are throwing two dice so,

Sample space S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sample space contains 36 elements,

Let ,

A= getting a number greater than 3 on first toss

A={(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

B= getting a number greater than 3 on second toss

B={(1,4),(1,5),(1,6), (2,4),(2,5),(2,6), (3,4),(3,5),(3,6), (4,4),(4,5),(4,6), (5,4),(5,5),(5,6), (6,4),(6,5),(6,6)}

P(A∩B)=P(getting a number on each toss)

Given that A and B are independent events,

Given that P(A can solve problem)=

P(B can solve the problem)=

P(none can solve the problem)=

Therefore, required probability=

Given an unbiased die is tossed twice so the sample space contain 36 elements.

Let A be the probability of getting 4,5,6 on first toss

A={(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Let B be the probability of getting 1,2,3,4 on the second toss

B={1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),

(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4),(5,1),(5,2),(5,3),(5,4)

(6,1),(6,2),(6,3),(6,4)}

P(getting 4,5,6 on the first toss and 1,2,3 or 4 on the second toss)

Given, bag contains 3 red and 2 black balls.

Probability of getting red ball=

Probability of getting black ball=

(i) two red balls

P(getting two getting red balls)=P(A)*P(A)

(ii) two black balls

P(getting two black balls)=P(B)*P(B)

(iii) first red and second black ball.

P( getting first red and second black ball)=P(A)*P(B)

In a pack of 52 cards, there are 4 king,4 queen and 4 jack

Let A be drawn card is a king

Let B be drawn card is a queen

Let C be drawn card is a jack

P(cards drawn are king, queen, jack )=

=P(A)P(B)P(C)+P(A)P(C)P(B)+P(B)P(A)P(C)+P(B)P(C)P(A)+P(C)P(A)P(B)+P(C)P(B)P(A)

=6*P(A)P(B)P(C)

Given,

An article manufactured by a company consists of two parts X and Y.

Part X has 9 out of 100 defective

That is, Part X has 91 out of 100 non defective

Part Y has 5 out of 100 defective

That is, Part Y has 95 out of 100 non defective

Consider,

X=a non defective part of X

Y=a non defective part of Y

P(assembled product will not be defective)

P(neither X defective nor Y defective)

=P(X∩Y)

=P(X)P(Y)

Given,

Probability that A hits a target is 1/3

P(A)=

The probability that B hits it, is 2/5

P(b)=

P(target will be hit)=1–P(target will not be hit)

=1–P(neither A nor B hit the target)

Given that an anti–aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it.

Let

A=hitting the plane at first shot

P(A)=0.4

B=hitting the plane at second shot

P(B)=0.3

C=hitting the plane at third shot

P(C)=0.2

D=hitting the plane at fourth shot

P(D)=0.1

P(gun hits the plane)=1–P(gun does not hit the plane)

=1–P(none of the four shots hit the plane)

=1–[(1–0.4)(1–0.3)(1–0.2)(1–0.1)]

=1–(0.6)(0.7)(0.8)(0.9)

=1–0.3024=0.6976

(i) The odds against certain event are 5 to 2

odds in favour of another event are 6 to5

Find the probability that (i) at least one of the events will occur

=1–P(none of the events will occur)

(ii) none of the events will occur.

P(none of the events will occur)

The sample space has 216 outcomes.

A be getting an odd number in a throw of die

P(getting an odd number at least once)=1–P(getting no odd number)

Given an urn contains 10 black balls and 8 red balls

Probability of getting red ball=

Probability of getting black ball=

(i) both balls are red

P(getting two getting red balls)=P(R)*P(R)

(ii) first balls is black and second is red

P(first balls is black and second is red)=P(B)*P(R)

(iv) one of them is black and other is red.

P(one of them is black and other is red)=P(B)P(R)+P(R)P(B)

Given an urn contains 4 red and 7 black balls

Probability of getting red ball=

Probability of getting black ball=

(i) two red balls

P(getting two getting red balls)=P(R)*P(R)

(ii) two black balls

P(getting two blue balls)=P(B)*P(B)

(iii) first red and second black ball.

P( getting one red and one black ball)=P(R)*P(B)+P(B)*P(R)

Given that A coming in time and B coming in time are independent.

Let A denote A coming in time

B denote B coming in time

denotes complementary event of A

denotes complementary event of B

P(only one coming in time)

We are throwing two dice so,

Sample space S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sample space contains 36 elements,

n(S)=36

let E be an event of getting sum 4

E={(1,3),(3,1),(2,2)}

n(E)=3

F be the event of getting 9 or more

F ={(3,6),(6,3),(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)}

n(F)=3

G be the event getting a total divisible by 5

G= {(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)}

n(G)=7

So E and F are not independent

So E and G are not independent

So F and G are not independent

Therefore no pair is independent

If the events are said to be independent, if the occurrence or non occurrence of one does not affect the probability of the occurrence or non occurrence of other.

_i. p₁p₂

p₁p₂=P(A)P(B)

Both events A and B will occur

_ii. (1–p₁) p₂

(1–p₁) p₂=[1–P(A)]P(B)

Event A does not occur but event B occur.

iii. 1–(1–p1)(1–p2)

1–(1–p1)(1–p2)=[1–(1–P(A))(1–P(B))]

At least one of the event will occur

iv. p₁+p₂ = 2p₁p₂

p₁+p₂ = 2p₁p₂

P(A)+P(B)=2P(A)P(B)

P(A)+P(B)–2P(A)P(B)=0

P(A)–P(A)P(B)+P(B)–P(A)P(B)=0

P(A)[1–P(B)]+P(B)[1–P(A)]=0

Exactly one of A and B occurs

Given:

⇒ Bag A contains 6 black balls and 3 white balls

⇒ Bag B contains 5 black balls and 4 white balls

It is told that one ball is drawn is drawn from is each bag.

We need to find the probability that the balls are of the same colour.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag A)

⇒

⇒

⇒

⇒ P(W₁) = P(drawing white ball from bag A)

⇒

⇒

⇒

⇒ P(B₂) = P(drawing black ball from bag B)

⇒

⇒

⇒

⇒ P(W₂) = P(drawing white ball from bag B)

⇒

⇒

⇒

We need to find the probability of drawing the same colour balls from two bags

⇒ P(S) = P(drawing two balls of same colours) = P(drawing black balls from each bag) + (P(drawing white balls from each bag)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probability is .

Given:

⇒ Bag A contains 3 red balls and 5 black balls

⇒ Bag B contains 6 red balls and 4 black balls

It is told that one ball is drawn is drawn from is each bag.

We need to find the probability that one ball is red and other is black.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag A)

⇒

⇒

⇒

⇒ P(R₁) = P(drawing Red ball from bag A)

⇒

⇒

⇒

⇒ P(B₂) = P(drawing black ball from bag B)

⇒

⇒

⇒

⇒ P(R₂) = P(drawing Red ball from bag B)

⇒

⇒

⇒

We need to find the probability of drawing the different colour balls from two bags

⇒ P(S) = P(drawing one red ball and one Black ball)

⇒ P(S) = P(drawing black balls from bag A and red ball from bag B) + P(drawing black balls from bag B and red ball from bag A)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probability is .

Given:

⇒ Bag contains 10 black balls and 8 red balls

It is told that two balls are drawn from bag with replacement.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag)

⇒

⇒

⇒

⇒ P(R₁) = P(drawing red ball from bag)

⇒

⇒

⇒

We need to find:

(i) P(D_rr) = P(both balls drawn are red)

(ii) P(D_br) = P(first drawn is black and next is red)

(iii) P(S_rd) = P(one ball is red and other is black)

⇒ P(D_rr) = P(both balls drawn are red)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ P(D_br) = P(first drawn is black and next is red)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ P(S_rb) = P(one ball is red and other is black)

⇒ P(S_rb) = P(first drawn is red and next is black) + P(first drawn black and next is red)

⇒ P(S_rb) = P(D_rb) + P(D_br)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒

∴ The required probabilities are .

Given that two cards are drawn from a well-shuffled deck of 52 cards.

We know that there will 4 aces present in a deck.

It is told that two cards successively without replacement.

Let us find the probability of drawing the cards.

⇒ P(A₁) = P(Drawing ace from 52 cards deck)

⇒

⇒

⇒ P(O₁) = P(Drawing cards other than ace from 52 cards deck)

⇒

⇒

⇒ P(A₂) = P(Drawing ace from remaining 51 cards deck)

⇒

⇒

⇒ P(O₂) = P(Drawing a card other than ace from remaining 51 cards deck)

⇒

⇒

We need to find the probability of drawing exactly one ace

⇒ P(D_A) = P(Drawing exactly 1 ace in the drawn two cards)

⇒ P(D_A) = P(Drawing Ace first and others next) + (P(Drawing Other cards first and ace next)

Since drawing cards are independent their probabilities multiply each other,

⇒ P(D_A) = (P(A₁)P(O₂)) + (P(O₁)P(A₂))

⇒

⇒

⇒

∴ The required probability is .

Given:

⇒ A speaks truth in 75% of cases

⇒ B speaks truth in 80% of cases

Now,

⇒ P(T_A) = P(A speaking truth) = 0.75

⇒ P(N_A) = P(A not speaking truth) = 1-0.75

⇒ P(N_A) = 0.25

⇒ P(T_B) = P(B speaking truth) = 0.80

⇒ P(N_B) = P(B not speaking truth) = 1-0.80

⇒ P(N_B) = 0.20

We need to find the probability for case in which A and B contradict each other for narrating an incident.

This happens only when A not telling truth while B is telling truth and vice-versa.

⇒ P(C_AB) = P(A and B contradict each other)

⇒ P(C_AB) = P(A tells truth and B doesn’t) + P(B tells truth and A doesn’t)

Since the speaking of A and B are independent events their probabilities will multiply each other.

⇒ P(C_AB ) = (P(T_A)P(N_B)) + (P(N_A)P(T_B))

⇒ P(C_AB) = (0.75×0.20) + (0.25×0.80)

⇒ P(C_AB) = 0.15 + 0.20

⇒ P(C_AB) = 0.35

∴ The required probability is 0.35.

Given that,

⇒ P(K_S) = P(Kamal’s selection)

⇒

⇒ P(K_N) = P(Not selecting Kamal)

⇒

⇒

⇒ P(V_S) = P(Monika’s selection)

⇒

⇒ P(V_N) = P(Not selecting Monika)

⇒

⇒

We need to find:

i. Both of them will be selected

ii. None of them will be selected

iii. At least one of them will be selected

iv. Only one of them will be selected.

⇒ P(S_both) = P(Both of them are selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_none) = P(None of them are selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_atone) = P(Selecting at least one of them)

⇒ P(S_atone) = P(selecting only Kamal) + P(selecting only Monika) + P(Selecting both)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒

⇒ P(S_one) = P(Only one of them is selected)

⇒ P(S_one) = P(only Kamal is selected) + P(only Monika is selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

∴ The required probabilities are .

Given:

⇒ Bag contains 3 white, 4 red and 5 black balls.

It is told that two balls are drawn from bag without replacement.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag on first draw)

⇒

⇒

⇒

⇒ P(W₁) = P(drawing white ball from bag on first draw)

⇒

⇒

⇒

⇒ P(B₂) = P(drawing black ball from bag on second draw)

⇒

⇒

⇒ P(W₂) = P(drawing white ball from bag on second draw)

⇒

⇒

We need to find:

⇒ P(S_WB) = P(one ball is White and other is black)

⇒ P(S_WB) = P(first drawn is White and next is black) + P(first drawn black and next is white)

⇒ P(S_WB) = P(D_WB) + P(D_BW)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒

⇒

∴ The required probability is .

Given:

Bag contains 8 red and 6 green balls

It is told that three balls are drawn without replacement.

We need to find the probability that at least two balls drawn are green.

The possible cases of drawing balls are as shown below:

i. Green,Green,Red

ii. Green,Red,Green

iii. Red,Green,Green

iv. Green,Green,Green

⇒ P(D_at2greens) = P(GGR) + P(GRG) + P(RGG) + P(GGG)

Since drawing balls are independent events so, the probabilities multiply each other.

⇒

⇒

⇒

⇒

⇒

∴ The required probability is .

Given that,

⇒ P(A_S) = P(Arun’s selection)

⇒

⇒ P(A_N) = P(Arun’s rejection)

⇒

⇒

⇒ P(T_N) = P(Tarun’s rejection)

⇒

⇒ P(T_S) = P(Tarun’s selection)

⇒

⇒

We need to find the probability that at least one of them will be selected

⇒ P(S_atone) = P(Selecting at least one of them)

⇒ P(S_atone) = P(selecting only Arun) + P(selecting only Tarun) + P(Selecting both)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒

∴ The required probability is .

Given that A and B toss a coin until one of them gets a head to win the game.

Let us find the probability of getting the head.

⇒ P(A_H) = P(A getting a head on tossing a coin)

⇒

⇒ P(A_N) = P(A not getting head on tossing a coin)

⇒

⇒ P(B_H) = P(B getting a head on tossing a coin)

⇒

⇒ P(B_N) = P(B not getting head on tossing a coin)

⇒

It is told that A starts the game.

We need to find the probability that B wins the games.

B wins the game only when A losses in1^st,3^rd,5^th,…… tosses.

This can be shown as follows:

⇒ P(W_B) = P(B wins the game)

⇒ P(W_B) = P(A_NB_H) + P(A_NB_NA_NB_H) + P(A_NB_NA_NB_NA_NB_H) + …………………

Since tossing a coin by each person is an independent event, the probabilities multiply each other.

⇒ P(W_B) = (P(A_N)P(B_H)) + (P(A_N)P(B_N)P(A_N)P(B_H)) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

⇒

⇒

⇒

⇒

∴ The required probability is .

Given that two cards are drawn from a well-shuffled deck of 52 cards.

We know that there will 26 Red and 26 Black cards present in a deck.

It is told that two cards successively without replacement.

Let us find the probability of drawing the cards.

⇒ P(R₁) = P(Drawing Red card from 52 cards deck)

⇒

⇒

⇒ P(B₁) = P(Drawing Black card from 52 cards deck)

⇒

⇒

⇒ P(R₂) = P(Drawing Red card from remaining 51 cards deck)

⇒

⇒

⇒ P(B₂) = P(Drawing Black card from remaining 51 cards deck)

⇒

⇒

We need to find the probability of drawing exactly one Red and one black card

⇒ P(D_RB) = P(Drawing exactly one red and one black card)

⇒ P(D_RB) = P(Drawing Red first and Black next) + (P(Drawing Black first and red next)

Since drawing cards are independent their probabilities multiply each other,

⇒ P(D_RB) = (P(R₁)P(B₂)) + (P(B₁)P(R₂))

⇒

⇒

⇒

∴ The required probability is .

Given that tickets are numbered from 1 to 10.

It is told that two tickets are drawn one after other in random.

We need to find the probability that number on ticket is a multiple of 5 and other number is multiple of 4.

Let us find the individual probabilities first

⇒ P(T₅₁) = P(Number on ticket is multiple of 5 in 1^st draw)

⇒

⇒

⇒

⇒ P(T₄₁) = P(Number on ticket is multiple of 4 in 1^st draw)

⇒

⇒

⇒

⇒ P(T₅₂) = P(Number on ticket is multiple of 5 in 2^nd draw)

⇒

⇒

⇒ P(T₄₂) = P(Number on ticket is multiple of 4 in 2^nd draw)

⇒

⇒

⇒ P(D₅₄) = P(Drawing one ticket which is multiple of 5 and other is multiple of 4)

⇒ P(D₅₄) = P(Drawing 5 multiple ticket first and 4 multiple ticket next) + (P(Drawing 4 multiple ticket first and 5 multiple ticket next)

Since drawing tickets are independent their probabilities multiply each other,

⇒ P(D₅₄) = (P(T₅₁)P(T₄₂)) + (P(T₄₁)P(T₅₂))

⇒

⇒

⇒

∴ The required probability is .

Given:

⇒ Husband tells lies in 30% of cases

⇒ Wife tells lies in 35% of cases

Now,

⇒ P(N_H) = P(Husband telling lies) = 0.30

⇒ P(T_H) = P(Husband telling truth) = 1-0.30

⇒ P(T_H) = 0.70

⇒ P(N_W) = P(Wife telling lies) = 0.35

⇒ P(T_W) = P(Wife telling truth) = 1-0.35

⇒ P(T_W) = 0.65

We need to find the probability for case in which husband and wife contradict each other for narrating a fact.

This happens only when husband not telling truth while Wife is telling truth and vice-versa.

⇒ P(C_HW) = P(Husband and Wife contradict each other)

⇒ P(C_HW) = P(Husband tells truth and Wife doesn’t) + P(Wife tells truth and Husband doesn’t)

Since the speaking of husband and Wife are independent events their probabilities will multiply each other.

⇒ P(C_HW ) = (P(T_H)P(N_W)) + (P(N_H)P(T_W))

⇒ P(C_HW) = (0.70×0.35) + (0.30×0.65)

⇒ P(C_HW) = 0.245 + 0.195

⇒ P(C_HW) = 0.44

∴ The required probability is 0.44.

Given that,

⇒ P(H_S) = P(Husband’s selection)

⇒

⇒ P(H_N) = P(Not selecting Husband)

⇒

⇒

⇒ P(W_S) = P(wife’s selection)

⇒

⇒ P(W_N) = P(Not selecting Wife)

⇒

⇒

We need to find:

i. Both of them will be selected

ii. Only one of them will be selected

iii. None of them will be selected

⇒ P(S_both) = P(Both of them are selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_one) = P(Only one of them is selected)

⇒ P(S_one) = P(only Husband is selected) + P(only Wife is selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_none) = P(None of them are selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

∴ The required probabilities are .

Given:

Bag contains 7 white, 5 black and 4 red balls

It is told that four balls are drawn without replacement.

We need to find the probability that at least three balls drawn are Black.

The possible cases of drawing balls are as shown below:

i. Black, Black, Black, Others

ii. Black, Black, Others, Black

iii. Black, Others, Black, Black

iv. Others, Black, Black, Black

v. Black, Black, Black, Black

⇒ P(D_at3blacks) = P(BBBO) + P(BBOB) + P(BOBB) + P(OBBB) + P(BBBB)

Since drawing balls are independent events so, the probabilities multiply each other.

⇒

⇒

⇒

⇒

⇒

∴ The required probability is .

Given:

A speaks truth three times out of four

B speaks truth four times out of five

C speaks truth five times out of six

⇒ P(T_A) = P(A speaks truth)

⇒

⇒ P(T_B) = P(B speaks truth)

⇒

⇒ P(T_C) = P(C speaks truth)

⇒

⇒ P(N_A) = P(A speaks lies)

⇒

⇒

⇒ P(N_B) = P(B speaks lies)

⇒

⇒

⇒ P(N_C) = P(C speaks lies)

⇒

⇒ .

It is told that the occurrence is will be reported by majority of witness. This is only possible when at least two of the witnesses speaks truth.

⇒ P(M) = P(T_AT_BN_C) + P(T_AN_BT_C) + P(N_AT_BT_c) + P(T_AT_BT_C)

Since speaking by a person is independent the probabilities will multiply each other.

⇒ P(M) = (P(T_A)P(T_B)P(N_C)) + (P(T_A)P(N_B)P(T_C)) + (P(N_A)P(T_B)P(T_c)) + (P(T_A)P(T_B)P(T_C))

⇒

⇒

⇒ .

∴ The required probability is .

Given:

⇒ Bag A contains 4 white balls and 2 black balls

⇒ Bag B contains 3 white balls and 5 black balls

It is told that one ball is drawn is drawn from is each bag.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag A)

⇒

⇒

⇒

⇒ P(W₁) = P(drawing white ball from bag A)

⇒

⇒

⇒

⇒ P(B₂) = P(drawing black ball from bag B)

⇒

⇒

⇒

⇒ P(W₂) = P(drawing white ball from bag B)

⇒

⇒

⇒

We need to find:

i. P(D_WW) = P(Both are white)

ii. P(D_BB) = P(Both are black)

iii. P(D_WB) = P(One drawn is white and other is black)

⇒ P(D_WW) = P(Both are White)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

P(D_BB) = P(Both balls are black)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

P(D_WB) = P(One ball drawn is white and other is black)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probabilities are .

Given:

Bag contains 4 white, 7 black and 5 red balls

It is told that four balls are drawn with replacement.

We need to find the probability that at least two balls drawn are white.

Let us find the individual probabilities to draw a ball

⇒ P(W) = P(drawing a white ball)

⇒

⇒

⇒ P(O) = P(drawing balls other than white)

⇒

⇒

The possible cases of drawing balls are as shown below:

i. Two whites and two others – 6 cases to draw

ii. Three whites and 1 others – 4 cases to draw

iii. Four whites – 1 case to draw

Since drawing balls are independent events so, the probabilities multiply each other.

⇒

⇒

⇒

⇒

∴ The required probability is .

Given that three cards are drawn from a well-shuffled deck of 52 cards with replacement.

We know that there will 4 kings, 4 queens, 4 jacks present in a deck.

Let us find the probability of drawing the cards.

⇒ P(K) = P(Drawing King from 52 cards deck)

⇒

⇒

⇒ P(Q) = P(Drawing Queen from 52 cards deck)

⇒

⇒

⇒ P(J) = P(Drawing Jack from 52 cards deck)

⇒

⇒

We need to find the probability of drawing one king, one queen, one jack

There will be 6 cases or sequences for drawing these three cards

⇒ P(D_KQJ) = P(Drawing King, queen and jack)

Since drawing cards are independent their probabilities multiply each other,

⇒

⇒

∴ The required probability is .

Given:

⇒ Bag A contains 4 red balls and 5 black balls

⇒ Bag B contains 3 red balls and 7 black balls

It is told that one ball is drawn is drawn from is each bag.

We need to find the probability that the balls are of same colour.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag A)

⇒

⇒

⇒

⇒ P(R₁) = P(drawing red ball from bag A)

⇒

⇒

⇒

⇒ P(B₂) = P(drawing black ball from bag B)

⇒

⇒

⇒

⇒ P(R₂) = P(drawing red ball from bag B)

⇒

⇒

⇒

We need to find the probability of drawing the different colour balls from two bags

⇒ P(D_RB) = P(drawing one red ball and one Black ball)

⇒ P(D_RB) = P(drawing black balls from bag A and red ball from bag B) + P(drawing black balls from bag B and red ball from bag A)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

We need to find the probability of drawing the same colour balls from two bags

⇒ P(S) = P(drawing two balls of same colours) = P(drawing black balls from each bag) + (P(drawing white balls from each bag)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probabilities are .

Given:

A hits a target 3 times out of 6 shots

B hits a target 2 times out of 6 shots

C hits a target 4 times out of 4 shots

⇒ P(T_A) = P(A hits target)

⇒

⇒ P(T_B) = P(B hits target)

⇒

⇒ P(T_C) = P(C hits target)

⇒

⇒ P(N_A) = P(A doesn’t hits target)

⇒

⇒

⇒ P(N_B) = P(B doesn’t hits target)

⇒

⇒

⇒ P(N_C) = P(C doesn’t hits target)

⇒

⇒ .

It is told that the target is to be hit by at least two shots. This is only possible when at least two of the Persons hits the target.

⇒ P(M) = P(T_AT_BN_C) + P(T_AN_BT_C) + P(N_AT_BT_c) + P(T_AT_BT_C)

Since hitting by a person is independent the probabilities will multiply each other.

⇒ P(M) = (P(T_A)P(T_B)P(N_C)) + (P(T_A)P(N_B)P(T_C)) + (P(N_A)P(T_B)P(T_c)) + (P(T_A)P(T_B)P(T_C))

⇒

⇒

⇒ .

⇒

∴ The required probability is .

Given that,

⇒ P(A_P) = P(A passing an examination)

⇒

⇒ P(A_N) = P(A Not passing an examination)

⇒

⇒

⇒ P(B_P) = P(B passing an examination)

⇒

⇒ P(B_N) = P(B Not passing an examination)

⇒

⇒

We need to find probability that:

i. Only A passing the examination

ii. Only one of them passing the examination

⇒ P(S_A) = P(Only A passing the examination)

This happens only in the case B must fail

Since passing examination is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_one) = P(Only one of them passed the examination)

⇒ P(S_one) = P(only A passed the examination) + P(only B passed the examination)

Since passing examination is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒

∴ The required probabilities are .

Given:

⇒ Urn A contains 4 red balls and 3 black balls

⇒ Urn B contains 5 red balls and 4 black balls

⇒ Urn C contains 4 red balls and 4 black balls

It is told that one ball is drawn is drawn from is each urn.

We need to find the probability that two balls are red and other ball is black.

Let us find the Probability of drawing each colour ball from the Urn.

⇒ P(B₁) = P(drawing black ball from Urn A)

⇒

⇒

⇒

⇒ P(R₁) = P(drawing Red ball from urn A)

⇒

⇒

⇒

⇒ P(B₂) = P(drawing black ball from Urn B)

⇒

⇒

⇒

⇒ P(R₂) = P(drawing Red ball from urn B)

⇒

⇒

⇒

⇒ P(B₃) = P(drawing black ball from Urn C)

⇒

⇒

⇒

⇒ P(R₃) = P(drawing Red ball from urn C)

⇒

⇒

⇒

We need to find the probability of 2 red balls and 1 black ball from three bags

⇒ P(S) = P(drawing two red ball and one Black ball)

⇒ P(S) = P(drawing black balls from bag A red ball from bag B and Red ball from bag C) + P(drawing black balls from bag B red ball from bag A and Red ball from bag C) + P(drawing black balls from bag C red ball from bag A and Red ball from bag B)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒

⇒ .

∴ The required probability is .

Given:

⇒ P(M_A) = P(getting A in mathematics)

⇒ P(M_A) = 0.2

⇒ P(M_N) = P(not getting A in mathematics)

⇒ P(M_N) = 1-0.2

⇒ P(M_N) = 0.8

⇒ P(P_A) = P(getting A in physics)

⇒ P(P_A) = 0.3

⇒ P(P_N) = P(not getting A in physics)

⇒ P(P_N) = 1-0.7

⇒ P(P_N) = 0.3

⇒ P(C_A) = P(getting A in Chemistry)

⇒ P(C_A) = 0.5

⇒ P(C_N) = P(not getting A in chemistry)

⇒ P(C_N) = 1-0.5

⇒ P(C_N) = 0.5

We need to find the probability that:

i. X gets A in all subjects

ii. X gets A in no subjects

iii. X gets A in two subjects

⇒ P(X_all) = P(getting A in all subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_all) = (P(M_A)P(P_A)P(C_A))

⇒ P(X_all) = 0.2×0.3×0.5

⇒ P(X_all) = 0.03

⇒ P(X_none) = P(getting A in no subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_none) = (P(M_N)P(P_N)P(C_N))

⇒ P(X_none) = 0.8×0.7×0.5

⇒ P(X_none) = 0.28

⇒ P(X_two) = P(getting A in any two subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_two) = (P(M_A)P(P_A)P(C_N)) + (P(M_A)P(P_N)P(C_A)) + (P(M_N)P(P_A)P(C_A))

⇒ P(X_two) = (0.2×0.3×0.5) + (0.2×0.7×0.5) + (0.8×0.3×0.5)

⇒ P(X_two) = 0.03 + 0.07 + 0.12

⇒ P(X_two) = 0.22

∴ The required probabilities are 0.03, 0.28, 0.22.

Given that A and B throws two dice.

The first who throw 9 awarded a prize.

The possibilities of getting 9 on throwing two dice are:

⇒ P(S₉) = P(getting sum 9)

⇒

⇒

⇒ P(S_N) = P(not getting sum 9)

⇒

⇒

Let us assume A starts the game, A wins the game only when he gets 9 while throwing dice in 1^st,3^rd,5^th,…… times

Here the probability of getting sum 9 on throwing a dice is same for both the players A and B

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(A_wins) = P(S₉) + P(S_N)P(S_N)P(S₉) + P(S_N)P(S_N)P(S_N)P(S_N)P(S₉) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

⇒ P(B_wins) = 1-P(A_wins)

⇒

⇒

⇒

⇒ P(A_wins):P(B_wins) = 9:8

∴ Thus proved

Given that A, B and C toss a coin until one of them gets a head to win the game.

Let us find the probability of getting the head.

⇒ P(A_H) = P(A getting a head on tossing a coin)

⇒

⇒ P(A_N) = P(A not getting head on tossing a coin)

⇒

⇒ P(B_H) = P(B getting a head on tossing a coin)

⇒

⇒ P(B_N) = P(B not getting head on tossing a coin)

⇒

⇒ P(C_H) = P(C getting a head on tossing a coin)

⇒

⇒ P(C_N) = P(C not getting head on tossing a coin)

⇒

It is told that A starts the game.

A tosses in1^st,4^th,7^th,…… tosses.

This can be shown as follows:

⇒ P(W_A) = P(A wins the game)

⇒ P(W_A) = P(A_H) + P(A_NB_NC_NA_H) + P(A_NB_NC_NA_NB_NC_NA_H) + …………………

Since tossing a coin by each person is an independent event, the probabilities multiply each other.

⇒ P(W_A) = (P(A_H)) + (P(A_N)P(B_N)P(C_N)P(A_H)) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

⇒

⇒

⇒

⇒

B tosses in2^nd,5^th,8^th,…… tosses.

This can be shown as follows:

⇒ P(W_B) = P(B wins the game)

⇒ P(W_B) = P(A_NB_H) + P(A_NB_NC_NA_NB_H) + P(A_NB_NC_NA_NB_NC_NA_NB_H) + …………………

Since tossing a coin by each person is an independent event, the probabilities multiply each other.

⇒ P(W_B) = (P(A_N)P(B_H)) + (P(A_N)P(B_N)P(C_N)P(A_N)P(B_H)) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

⇒

⇒

⇒

⇒

⇒ P(W_C) = P(winning of C)

⇒ P(W_C) = 1-P(W_A)-P(W_B)

⇒

⇒

∴ The chances of winning of A,B and C are .

Given that A,B and C throws a die.

The first who throw 6 wins the game.

⇒ P(S₆) = P(getting 6)

⇒

⇒ P(S_N) = P(not getting 6)

⇒

⇒

Let us assume A starts the game, A wins the game only when he gets 6 while throwing dice in 1^st,4^th,7^th,…… times

Here the probability of getting sum 6 on throwing a dice is same for the players A, B and C

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(A_wins) = P(S₉) + P(S_N)P(S_N)P(S_N)P(S₉) + P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S₉) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

B wins the game only when he gets 6 while throwing dice in 2^nd,5^th,8^th,…… times and others doesn’t get 6.

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(B_wins) = (P(S_N)P(S₉)) + (P(S_N)P(S_N)P(S_N)P(S_N)P(S₉)) + (P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S_N)P(S₉)) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

⇒ P(C_wins) = 1-P(A_wins)-P(B_wins)

⇒

⇒

∴ The probabilities of winning of A, B and C is .

Given that A and B throws two dice.

The first who throw 10 awarded a prize.

The possibilities of getting 10 on throwing two dice are:

⇒ P(S₁₀) = P(getting sum 10)

⇒

⇒

⇒ P(S_N) = P(not getting sum 10)

⇒

⇒

Let us assume A starts the game, A wins the game only when he gets 10 while throwing dice in 1^st,3^rd,5^th,…… times

Here the probability of getting sum 10 on throwing a dice is same for both the players A and B

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(A_wins) = P(S₁₀) + P(S_N)P(S_N)P(S₁₀) + P(S_N)P(S_N)P(S_N)P(S_N)P(S₁₀) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

⇒ P(B_wins) = 1-P(A_wins)

⇒

⇒

⇒

⇒ P(A_wins):P(B_wins) = 12:11

∴ Thus proved

Given:

⇒ Bag A contains 3 red balls and 5 black balls

⇒ Bag B contains 2 red balls and 3 black balls

It is told that one ball is drawn from bag A and two balls from bag B.

We need to find the probability that one ball is red and other two are black.

Let us find the Probability of drawing each colour ball from the bag.

⇒ P(B₁) = P(drawing black ball from bag A)

⇒

⇒

⇒

⇒ P(R₁) = P(drawing Red ball from bag A)

⇒

⇒

⇒

⇒ P(B₂₁) = P(drawing black ball from bag B in 1^st draw)

⇒

⇒

⇒

⇒ P(R₂₁) = P(drawing Red ball from bag B in 1^st draw)

⇒

⇒

⇒

⇒ P(B₂₂) = P(drawing black ball from bag B in 2^nd draw after drawing red ball)

⇒

⇒

⇒

⇒ P(R₂₂) = P(drawing Red ball from bag B in 2^nd draw after drawing Black ball)

⇒

⇒

⇒

⇒ P(B₂₂₁) = P(drawing black ball from bag B in 2^nd draw after drawing black ball)

⇒

⇒

⇒

We need to find the probability of drawing a red and two black balls from two bags

⇒ P(S) = P(drawing one red ball and two Black balls)

⇒ P(S) = P(drawing red ball from bag A and black balls from bag B) + P(drawing black ball from bag A and red and black balls from bag B)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probability is .

Given that,

⇒ P(F_S) = P(Fatima’s selection)

⇒

⇒ P(F_N) = P(Not selecting Fatima)

⇒

⇒

⇒ P(J_S) = P(John’s selection)

⇒

⇒ P(J_N) = P(Not selecting John)

⇒

⇒

We need to find:

i. Both of them will be selected

ii. Only one of them will be selected

iii. None of them will be selected

⇒ P(S_both) = P(Both of them are selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_one) = P(Only one of them is selected)

⇒ P(S_one) = P(only Fatima is selected) + P(only John is selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

⇒ P(S_none) = P(None of them are selected)

Since selection of each person is an independent event their probabilities multiply each other

⇒

⇒

⇒

∴ The required probabilities are .

Given:

Bag contains 3 blue and 5 red marbles

It is told that two marbles are drawn with replacement

Let us find the probability of drawing each marble from bag

⇒ P(B) = P(Drawing a Blue Marble)

⇒

⇒

⇒ P(R) = P(Drawing a Red Marble)

⇒

⇒

We need to find the probability that the marbles drawn:

i. Blue followed by red

ii. Blue and red in any order

iii. Of the same colour

⇒ P(S_BR) = P(drawing Blue marble followed by Red)

Since drawing a marble is an independent event, the probabilities multiply each other.

⇒

⇒

⇒

⇒ P(S_any) = P(drawing Blue and red marble in any order)

⇒ P(S_any) = P(drawing Blue marble followed by red) + P(drawing Red marble followed by Blue)

Since drawing a marble is an independent event, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

⇒ P(S) = P(drawing two marbles of same colour)

⇒ P(S) = P(drawing black balls from each bag) + (P(drawing white balls from each bag)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probabilities are .

Given:

Urn contains 7 red and 4 blue balls

It is told that two balls are drawn with replacement

Let us find the probability of drawing each marble from bag

⇒ P(B) = P(Drawing a Blue Ball)

⇒

⇒

⇒

⇒ P(R) = P(Drawing a Red Ball)

⇒

⇒

⇒

We need to find the probability that the balls drawn:

i. Both in red colour

ii. Both in blue colour

iii. One red and one blue

⇒ P(S_RR) = P(drawing Two red colour balls)

Since drawing a ball is an independent event, the probabilities multiply each other.

⇒

⇒

⇒

⇒ P(S_BB) = P(drawing two blue colour balls)

Since drawing a ball is independent for each bag, the probabilities multiply each other.

⇒

⇒

⇒

⇒ P(S_any) = P(drawing Blue and red ball in any order)

⇒ P(S_any) = P(drawing Blue ball followed by red) + P(drawing Red ball followed by Blue)

Since drawing a ball is an independent event, the probabilities multiply each other.

⇒

⇒

⇒

⇒ .

∴ The required probabilities are .

Given that two cards are drawn from the deck with replacement.

We know that there will four suits in a deck and each suit contains 13 cards namely Spades, Hearts, Diamonds, Clubs.

Also, there will total of 4 aces and 2 red queens and 2 black queens

Let us find the probability required:

⇒ P(D_same) = P(For selecting a card from a single suit out of 52 cards)

⇒

⇒

This probability will be same for all the suits.

⇒

⇒ P(D_ace) = P(Drawing an ace)

⇒

⇒

⇒ P(D_redQ) = P(Drawing a red queen)

⇒

⇒

We need to find the probability of getting:

i. Both cards from same deck

ii. First an ace and second a red queen

⇒ P(S_same) = P(getting both cards from same deck)

We may get two cards any of the four decks. So, each deck’s probability is taken into consideration.

⇒ P(S_same) = P(both cards from spade) + P(both cards from hearts) + P(both cards from diamond) + P(both cards from club)

Since drawing a card is an independent event, their probabilities multiply each other.

⇒ P(S_same) = (P(D_S)P(D_S)) + (P(D_H)P(D_H)) + (P(D_D)P(D_D)) + (P(D_C)P(D_C))

⇒

⇒

⇒

⇒

⇒ P(S_AR) = P(getting an ace first and followed red queen)

Since drawing a card is an independent event, their probabilities multiply each other.

⇒ P(S_AR) = (P(D_A)P(D_redQ))

⇒

⇒

∴ The required probabilities are .

Given:

Section A contains 40 students

Section B contains 60 students

It is told that you and your friend are among 100 students

We need to find:

i. You both enter the same section

ii. You both enter the different section

⇒ P(S_same) = P(You and your friend enter same section)

This happens only in the case if you and your friend belongs to either section A or section B.

⇒ P(S_same) = P(Both belongs to section A) + P(Both belongs to section B)

⇒

⇒

⇒

⇒

⇒

⇒

⇒ P(S_diff) = P(You and your friend enter different section)

This happens only in the case if you belongs to A and your friend belongs to section B.

⇒ P(S_diff) = 1-P(Both belongs to same section)

⇒ P(S_diff) = 1-P(S_same)

⇒

⇒

∴ The required probabilities are .

Given that teams A and B scored same number of goals.

It is asked captains of A and B to throw a die.

The first who throw 6 awarded a prize.

⇒ P(S₆) = P(getting 6)

⇒

⇒ P(S_N) = P(not getting 6)

⇒

⇒

It is given A starts the game, A wins the game only when he gets 6 while throwing die in 1^st,3^rd,5^th,…… times

Here the probability of getting 6 on throwing a die is same for both the players A and B

Since throwing a die is an independent event, their probabilities multiply each other

⇒ P(A_wins) = P(S₆) + P(S_N)P(S_N)P(S₆) + P(S_N)P(S_N)P(S_N)P(S_N)P(S₆) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

⇒ P(B_wins) = 1-P(A_wins)

⇒

⇒

Since the probabilities if winnings of A and B are not equal, the decision of the referee is not fair

Given that A and B throw a pair of die alternatively.

It is told that A wins if he gets sum of 7 and B wins if he gets sum of 10.

Let us find the probability of getting sum 7 and 10.

The possibilities of getting 7 on throwing a pair of die:

{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6 cases

The possibilities of getting 10 on throwing a pair of die:

{(4,6),(5,5),(6,4)} = 3 cases

⇒ P(S₇) = P(getting sum 7)

⇒

⇒

⇒

⇒ P(S_N7) = P(not getting sum 7)

⇒ P(S_N7) = 1-P(S₇)

⇒

⇒

⇒ P(S₁₀) = P(getting sum 10)

⇒

⇒

⇒

⇒ P(S_N10) = P(not getting sum 10)

⇒ P(S_N10) = 1-P(S₁₀)

⇒

⇒

It is told that A starts the game.

We need to find the probability that B wins the games.

B wins the game only when A losses in1^st,3^rd,5^th,…… throws.

This can be shown as follows:

⇒ P(B_WINS) = P(B wins the game)

⇒ P(B_Wins) = P(S_N7S₁₀) + P(S_N7S_N10S_N7S₁₀) + P(S_N7S_N10S_N7S_N10S_N7S₁₀) + …………………

Since throwing a pair of die by each person is an independent event, the probabilities multiply each other.

⇒ P(B_wins) = (P(S_N7)P(S₁₀)) + (P(S_N7)P(S_N10)P(S_N7)P(S₁₀)) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

⇒

⇒

⇒

∴ The required probability is .

Given:

Bag A contains 5 white and 6 black balls.

Bag B contains 4 white and 3 black balls.

A ball is transferred from bag A to bag B, and then a ball is drawn from bag B.

There are two mutually exclusive ways to draw a black ball from bag B –

a. A white ball is transferred from bag A to bag B, and then, a black ball is drawn from bag B

b. A black ball is transferred from bag A to bag B, and then, a black ball is drawn from bag B

Let E₁ be the event that white ball is drawn from bag A and E₂ be the event that black ball is drawn from bag A.

Now, we have

We also have

Let E₃ denote the event that black ball is drawn from bag B.

Hence, we have

We also have

Using the theorem of total probability, we get

P(E₃) = P(E₁)P(E₃|E₁) + P(E₂)P(E₃|E₂)

Thus, the probability of the drawn ball being black is.

Given:

First purse contains 2 silver and 4 copper coins.

Second purse contains 4 silver and 3 copper coins.

A coin is pulled a random from one of the two purses.

There are two mutually exclusive ways to pull a silver coin from one of the two purses –

a. The first purse is selected, and then, a silver coin is pulled from the first purse

b. The second purse is selected, and then, a silver coin is pulled from the second purse

Let E₁ be the event that the first purse is selected and E₂ be the event that the second purse is selected.

Since there are only two purses and each purse has an equal probability of being selected, we have

Let E₃ denote the event that a silver coin is pulled.

Hence, we have

We also have

Using the theorem of total probability, we get

P(E₃) = P(E₁)P(E₃|E₁) + P(E₂)P(E₃|E₂)

Thus, the probability of pulling a silver coin is.

Given:

The bag I contains 4 yellow and 5 red balls.

Bag II contains 6 yellow and 3 red balls.

A ball is transferred from bag I to bag II and then a ball is drawn from bag II.

There are two mutually exclusive ways to draw a yellow ball from bag II –

a. A yellow ball is transferred from the bag I to bag II, and then, a yellow ball is drawn from bag II

b. A red ball is transferred from the bag I to bag II, and then, a yellow ball is drawn from bag II

Let E₁ be the event that yellow ball is drawn from the bag I and E₂ be the event that red ball is drawn from the bag I.

Now, we have

We also have

Let E₃ denote the event that yellow ball is drawn from bag II.

Hence, we have

We also have

Using the theorem of total probability, we get

P(E₃) = P(E₁)P(E₃|E₁) + P(E₂)P(E₃|E₂)