Probability

Given, Total number of defective ballpens = 15

and total number of good ballpens = 135

Therefore, total number of ball pens = number of defective ballpens + number of good ball pens

So, total number of ball pens = 15 + 135 = 150

Now, let E be the event that the ballpen picked is defective and F be the event that the ballpen picked is good.

Therefore, the number of favourable outcomes of event E = 15

and , the number of favourable outcomes of event F = 135

Total number of favourable outcomes = 15 + 135 = 150

Probability that the ball pen selected is good = Probability of event F occuring = P(F)

P(F) = = = 0.9

So, the probability that the ballpen selected is good is 0.9.

Number of green balls in the box = 5

Number of yellow balls in the box = 8

Number of brown balls in the box = 7

Total number of balls including green, yellow and brown in the box = 5 + 8 + 7 = 20

Now, let the event that a green ball is taken from the box be G.

the event that a yellow ball is taken from the box be Y.

the event that a green ball is taken from the box be B.

Number of favourable outcomes of event G = 5

Number of favourable outcomes of event Y = 8

Number of favourable outcomes of event B = 7

Total number of favourable outcomes = 5 + 8 + 7

= 20

(i) Probability of a yellow ball to be taken from the box = Probability of event Y occurring = P(Y)

P(Y) =

(ii) Probability of a brown ball to be taken from the box = Probability of event B occurring = P(B)

P(B) =

(iii) Probability of the taken ball is nether green nor brown = 1 – Probability of ball taken is either green or brown = 1- P(G or B)

P(G or B)’ = 1-P(G or B)

=

=

(iv) Probability of the ball taken is not brown = 1-Probability of the ball taken is brown = 1-P(B)

P(B’) = 1- P(B)

=

=

Let the total number of orange candies in the bag be T.

As the bags contain only orange flavoured candies so the total number of candies in the bag

= T

Let the event that an orange candy is taken out be O.

Number of favourable outcomes of the event O = T

Also, total number of favourable outcomes = T

(i) Probability that the orange flavoured candy is taken out = Probability of event occurring O = P(O)

P(O) = = 1

(ii) Probability that a lemon flavoured candy is taken out = Probability that an orange flavoured candy is not taken out = P(O’)

P(O’) = 1 – P(O) = 1 – 1 = 0

Given, total number of cards in the box is 100 with each card marked with numbers from 1 to 100.

Therefore, we know that from 1 to 100 , number of single digit numbers = 1 to 9 = 9

Also, number of two-digit numbers from 1 to 100 = 10 to 99 = 90

Only three digit number from 1 to 100 is 100.

So, number of three-digit numbers from 1 to 100 = 1

Numbers divisible by 8 from 1 to 100 are 8,16,24,32,40,48,56,64,72,80,88,96 = 12

Numbers which are multiple of 9 are 9,18,27,36,45,54,63,72,81,90,99 = 11

Numbers which are multiple of 5 are 5,10,15,20,……….100 = 20

Now, let the event that a single digit number is drawn from the box be A.

Event that card bearing a two-digit number is drawn from the box be B.

Event that a card bearing a three-digit number is drawn from the box be C.

Event that a card bearing a number divisible by 8 is drawn from the box be D.

Event that a card bearing a number which is multiple of 9 is drawn from the box be E.

Event that a card bearing a number which is multiple of 5 is drawn from the box be F

Number of favourable outcomes of event A = 9

Number of favourable outcomes of event B = 90

Number of favourable outcomes of event C = 1

Number of favourable outcomes of event D = 12

Number of favourable outcomes of event E = 11

Number of favourable outcomes of event F = 20

Total number of favourable outcomes = 100

(i) Probability of the card bearing a single digit number = Probability of occurrence of event A = P(A)

P(A) = 0.09

(ii) Probability of the card bearing a two-digit number = Probability of occurrence of event B = P(B)

P(B) = 0.9

(iii) Probability of the card bearing a three-digit number = Probability of occurrence of event C = P(C)

P(C) = 0.01

(iv) Probability of the card bearing a number divisible by 8 = Probability of occurrence of event D = P(D)

P(D) = 0.12

(v) Probability of the card bearing a number which is multiple 9 = Probability of occurrence of event E = P(E)

P(E) = 0.11

(vi) Probability of the card bearing a number which is multiple 5 = Probability of occurrence of event F = P(F)

P(F) = 0.2

Total number of trousers = 100

Number of trousers which are good = 73

Number of trousers with minor defects = 12

Number of trousers with major defects = 15

Now, let the event that a trouser is acceptable to Kanu be E.

Event that a trouser is acceptable to Radha be F.

Number of favourable outcomes of event E = 73

Number of favourable outcomes of event F = 15

Total number of favourable outcomes = 100

(i)Probability that a trouser is acceptable = Probability of occurrence of event E = P(E)

P(E) = = 0.73

(ii)Probability that a trouser is not acceptable = Probability of occurrence of event F = P(F)

P(F) = = 0.15

The above chart shows the marks obtained by 50 students from total marks 100.

Total marks = 100

Total number of students = 50

Number of students scoring between 0-34 = 8

Number of students scoring between 35-50 = 9

Number of students scoring between 51-70 = 14

Number of students scoring between 71-90 = 11

Number of students scoring between 91-100 = 8

Now, Let the event that students getting marks below 34 be A.

Event that students getting marks between 71-90 be B.

Event that students getting marks more than 70 be C.

Event that students getting marks less than or equal to 50 be D.

Event that students getting marks above 90 be E.

Number of favourable outcomes of event A = 8

Number of favourable outcomes of event B = 11

Number of favourable outcomes of event C = 11 + 8 = 19

Number of favourable outcomes of event D = 8 + 9 = 17

Number of favourable outcomes of event E = 8

(i) Probability that a student get marks below 34 = Probability of occurrence of event A = P(A)

P(A) = = 0.16

(ii) Probability that a student get marks between 71-90 = Probability of occurrence of event B = P(B)

P(B) = = 0.22

(iii) Probability that a student get marks more than 70 = Probability of occurrence of event C = P(C)

P(C) = = 0.38

(iv) Probability that a student get marks less than or equal to 50 = Probability of occurrence of event D = P(D)

P(D) = = 0.34

(v) Probability that a student get marks above 90 = Probability of occurrence of event E = P(E)

P(E) = = 0.16

If two dice are rolled together then we get 36 outcomes as

(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)

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

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

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

Favourable outcomes of event A i.e, getting the same number on both sides are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

Number of favourable outcomes of event A = 6

Favourable outcomes of event B i.e, the sum of the integers on two dice is more than 4 but less than 8 are (1,4), (1,5), (1,6), (2,3), (2,4),(2,5), (3,2),

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

Number of favourable outcomes of event B = 15

Favourable outcomes of event C i.e, the product of numbers on two dice is divisible by 2 are

(1,2),(1,4),(1,6)

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

(3,2),(3,4)(3,6)

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

(5,2),(5,4)(5,6)

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

Number of favourable outcomes of event C = 27

Favourable outcomes of event D i.e, the sum of numbers on two dice is greater than 12 is 0 as the greatest value of the sum of numbers on two dice is 6 + 6 = 12.

Number of favourable outcomes of event D = 0

Total number of favourable outcomes = 36

(i) Probability of occurrence of event A = P(A)

P(A) = = 0.16667

(ii) Probability of occurrence of event B = P(B)

P(B) = = 0.41667

(iii) Probability of occurrence of event C = P(C)

P(C) = = 0.75

(iv) Probability of occurrence of event D = P(D)

P(D) = = 0

If a coin is tossed three times their will be 8 outcomes.

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

Now, favourable outcomes of event A i.e, getting at least two heads are HHH,HHT,HTH,THH

Number of favourable outcomes of event A = 4

Now, favourable outcomes of event B i.e, getting exactly two heads are HHT,HTH,THH

Number of favourable outcomes of event B = 3

Now, favourable outcomes of event C i.e, getting at most one head are TTH,HTT,THT,TTT

Number of favourable outcomes of event C = 4

Now, favourable outcomes of event D i.e, getting more heads than tails are HHH,HHT,HTH,THH

Number of favourable outcomes of event D = 4

Total number of outcomes = 8

(i) Probability of occurrence of event A = P(A)

P(A) = = 0.5

(ii) Probability of occurrence of event B = P(B)

P(B) = = 0.375

(iii) Probability of occurrence of event C = P(C)

P(C) = = 0.5

(iv) Probability of occurrence of event D = P(D)

P(D) = = 0.5

Total numbers are from 1 to 12 i.e, 12 numbers.

Now, let the event that the arrow will point at 7 be A.

Event that arrow will point at a number greater than 5 be B.

Event that arrow will point at an odd number be C.

Event that arrow will point at an even number be D.

Event that arrow will point at a number less than 5 be E.

Number of favourable outcomes of event A = 1

Favourable outcomes of event B i.e., a number greater than 5 are 6, 7, 8 ,9 10, 11, 12.

Number of favourable outcomes of event B = 7

Favourable outcomes of event C i.e., an odd number are 1,3,5,7,9,11.

Number of favourable outcomes of event C = 6

Favourable outcomes of event D i.e., an even number are 2,4,5,8,10,12.

Number of favourable outcomes of event D = 6

Favourable outcomes of event E i.e., a number less than 5 are 1,2 ,3,4.

Number of favourable outcomes of event E = 4

Total number of favourable outcomes = 12

(i) Probability that the arrow will point at 7 = Probability of occurrence of event A = P(A)

P(A) = = 0.0833

(ii) Probability that arrow will point at a number greater than 5 = Probability of occurrence of event B = P(B)

P(B) = = 0.58333

(iii) Probability that arrow will point at an odd number = Probability of occurrence of event C = P(C)

P(C) = = 0.5

(iv) Probability that arrow will point at an even number = Probability of occurrence of event D = P(D)

P(D) = = 0.5

(v) Probability that arrow will point at a number less than 5 = Probability of occurrence of event E = P(E)

P(E) = = 0.3333

We know that P(A) + P(A’) = 1

If an event is certain then it will surely occur.

So, the probability of certain occurrence of event is always 1.

If an event is impossible that means it will not occur in any case.

So, the probability of impossible event is always 0.

The minimum probability of an event is 0 that is the event is not occurring.

P(A) 0

So, the probability of an event is greater than or equal to 0 .

The maximum probability of an event is 1 that is the event is certainly occurring.

P(A) 1

So, the probability of an event is less than or equal to 1 .

We know that P(A) + P(A’) = 1

P(A) = 0.35

P(A’) = 1-P(A) = 1- 0.35 = 0.65

We know that P(E) + P(E’) = 1

P(E) = 0.35

P(E’) = 1-P(E) = 1- 0.47 = 0.53

Assuming the paper in your hand is of total 100 marks and to get 101 out of 100 is an impossible event.

Therefore, Probability of getting 101 marks out of 100 is 0.

We know that The Sun rises in the East. It is a fact.

So, it is an impossible event for the sun to rise in the west.

Therefore, Probability of the event "The Sun rises in West" is 0

We know that the sum of probabilities of all the elementary events of an experiment is 1.