Probability

P(happening) + P’(not happening) = 1

Given: P = 0.85

∴ P’ = 1 – 0.85 = 0.15

Sample space, S = 6

(i) Number of event of getting a prime number, n(E) = 3

(ii)

(iii)

(iv)

(v)

(vi)

Sample space = 36

(i) n(E) = 5

(ii) n(E) = 6

(iii) n(E) = 3

(iv)

(v)

(vi)

(vii)

(viii) Number of event with sum 9 or 11 = 6

∴ Number of events of not getting a sum of either 9 or 11 = 36 – 6 = 30

(ix)

(x)

(xi)

(xii)

(xiii)

Sample space, n(S) = 8

(i)

(ii)

(iii)

(iv)

Ordinary year has 365 days (52 weeks + 1 day)

That 1 day may be any day of the 7 days of a week

Therefore, total number of possible outcomes, n(S) = 7

And event of getting 53 Sunday is that 1 day is Sunday. Thus, n(E) = 1

A leap year has 366 days (52 weeks + 2 days)

That two days may be (sun, mon) (mon, tue) (tue, wed) (wed, thurs) (thurs, fri) (fri, sat) (sat, sun)

∴ n(S) = 7

Number of events for the occurrence of 53 Tuesdays and 53 Mondays, n(E) = 1

Given: A and B throw a pair of dice. A throws 9.

To find: B’s chance of throwing a higher number.

Solution:

Sample space of throwing a pair of dice is:








Sample space, n(S) = 36

let E be the number of events of getting sum higher than 9 i.e getting 10,11 and 12.

The pairs are:



then n(E) = 6

So,

Possible outcomes, n(S) = 36

Number of events of getting sum greater than 10, n(E) = 3

Total number of outcomes, n(S) = 52

(i) n(E) = 2

(ii) n(E) = 26 + 2 = 28

(iii) n(E) = 2

(iv) n(E) = 12

(v) n(E) = 36

(vi) n(E) = 16

(vii) n(E) = 44

(viii) n(E) = 24

(ix) n(E) = 48

(x) n(E) = 4

(xi) n(E) = 13

(xii) n(E) = 26

(xiii) n(E) = 1

(xiv) n(E) = 4

(xv) n(E) = 1

(xvi) n(E) = 4

(xvii) n(E) = 13

(xviii) n(E) = 26

(xix) n(E) = 44

Total number of possible outcomes, n(S) = 50

Number of events of getting a prime number, N(E) = 15

Probability,

Total number of possible outcomes, n(S) = 10 + 8 = 18

Number of events of getting white ball, N(E) = 8

Probability,

Total number of possible outcomes, n(S) = 12 {3 red balls, 5 black balls and 4 white balls)

(i) n(E) = 4

(ii) n(E) = 3

(iii) n(E) = 5

(iv) n(E) = 9

Sample space, n(S) = 15

Number of events of getting numbers multiple of 4, n(E) = 3

Sample space, n(S) = 18

Number of events of getting balls not black, n(E) = 10

Sample space, n(S) = 12

Number of events of getting white ball, n(E) = 5

Sample space, n(S) = 20

Number of events of getting a multiples of 3 or 7 on the drawn ticket, n(E) = 8 {3,6,9,12,15,18,7,14}

Sample space, n(S) = 35

Number of events of getting a prize, n(E) = 10

We not that, P(E) + P’(E) = 1

Here, P(E) = 0.3

P’(E) = 1 – 0.3 = 0.7

Sample space, n(S) = 15

(i) n(E) = 7

(ii) n(E) = 10

(iii) n(E) = 5

Here, n(S) = 15

(i) n(E) = 6

(ii) n(E) = 4

(iii) number of events of getting black balls, n(E) = 5

∴ probability of getting black balls, P(E) =

Hence, probability of getting black balls = 1 – P(E) = 1 –

(iv) number of events of getting a red or white ball, n(E) = 10

Sample space, n(S) = 36

(i) N(E) = 5

(ii) n(E) = 5

(iii) n(E) = 3

(iv) n(E) = 6

(v) n(E) = 6

(vi) The maximum sum is 12. So, getting a sum of number appearing on the top of the two dice as 13 is an impossible event.

∴ Probability, P(E) = 0

(vii) n(E) = 36

(viii) n(E) = 2

(ix) n(E) = 8

Number of possible outcomes, n(S) = 52

(i) n(E) = 2

(ii) n(E) = 12

(iii) n(E) = 6

(iv) n(E) = 2

(v) n(E) = 6

(vi) n(E) = 13

Total number of possible outcomes, n(S) = 5

(i) Number of events of drawing a queen, n(E) = 1

(ii) If a king is drawn first and put aside, total number of remaining cards, n(S) = 4

a) Probability that second card is ace,

b) Since there was only a single king and when it is put aside, then n(E) = 0

Total number of possible outcomes, n(S) = 8

(i) Number of possible outcomes of drawing a red ball, n(E) = 3

(ii) Number of possible outcomes of drawing a black ball, n(E) = 5

Total number of possible outcomes, n(S) = 89

(i) Number of favorable outcomes of getting a two digit number = 81

(ii) Number of favorable outcomes of getting a perfect square = 8

The possible number of outcomes, n(S) = 12

(i) Number of favorable outcomes, n(E) = 1

(ii) Number of favorable outcomes, n(E) = 6

(iii) Number of favorable outcomes, n(E) = 4

(iv) Number of favorable outcomes, n(E) = 6

Total number of days to visit the shop, n(S) = 36

(i) Number of favorable outcomes, n(E) = 6

(ii) n(E’) = 1 – n(E) =

(iii) Number of favorable outcomes, n(E) = 5

Total number of possible outcomes, n(S) = 34

(i) Number of favorable outcomes, n(E) = 18

(ii) Number of favorable outcomes, n(E) = 16

Number of possible outcomes, n(S) = 2

Probability of getting a head, P(E’) = � = probability of getting a tail

Since both probabilities are equal for both of the events, therefore both of the players have equal chances to win the game and thus this is a fair way to make this decision that who will play first.

Total number of possible outcomes = 10

Average of the numbers =

∴ Number of favorable outcomes, n(E) = 3

Total number of possible outcomes, n(S) = 36

Number of favorable outcomes, n(E) = 6

Let the number of green marbles be x and the number of white marbles be y

Total number of possible outcomes, n(S) = x + y + 10

Probability of green marbles = 1/4

⇒ x + y + 10 = 4x

⇒ 3x – y – 10 = 0 -------------(i)

Probability of white marbles =

⇒ x – 2y + 10= 0 -------------(ii)

Solving eq. (i) and (ii), we get

x = 6 and y = 8

Thus, total number of marbles in the jar = x + y + 10 = 6 + 8 + 10 = 24

Total number of possible outcomes, n(S) = 30

Number of favorable outcomes of selecting a card divisible by 3, n(E) = 10

Number of favorable outcomes of not selecting a card divisible by 3 = 1 – P(E)

Total number of possible outcomes, n(S) = 20

(i) Number of favorable outcomes, n(E) = 13

(ii) Number of events of getting a black ball, n(E) = 7

Probability of not getting a black ball = 1 – P(E)

(iii) Number of favorable outcomes, n(E) = 5

Total number of possible outcomes, n(S) = 25

Number of events of getting a prime number, n(E) = 9

Number of events of not getting a prime number = 1 – P(E)

Total number of possible outcomes, n(S) = 18

(i) Number of favorable outcomes, n(E) = 14

(ii) Number of events of getting a black ball, n(E) = 4

Probability of not getting a black ball = 1 – P(E)

(iii) Number of favorable outcomes, n(E) = 8

Total number of possible outcomes, n(S) = 35

(i) Number of favorable outcomes, n(E) = 11

(ii) Number of favorable outcomes, n(E) = 5

(iii) Number of favorable outcomes, n(E) = 16

Total number of possible outcomes, n(S) = 52 – 2 – 2 – 2 – 2 = 44

(i) Number of favorable outcomes, n(E) = 2

(ii) Number of favorable outcomes, n(E) = 26 – 8 = 18

(iii) Number of favorable outcomes, n(E) = 2

(iv) Number of favorable outcomes, n(E) = 6

(i) Since the bag contains lemon flavored candies only thus there is no chance of getting orange flavored candy.

∴ P(E) = 0

(ii) The bag contains lemon flavored candies only thus this is a sure event of getting lemon flavored candy

∴ P(E) = 1

Let E be the event of 2 students having the same birthday and _E’ be the event of not getting the same birthday

Given: P(E) = 0.992

We know that, P(E) + P(E’) = 1

∴ P(E’) = 1 – P(E)

= 1 – 0.992

= 0.008

Total number of possible outcomes, n(S) = 8

(i) Number of favorable outcomes, n(E) = 3

(ii) Probability of not drawing a red ball = 1 – P(E)

Total number of possible outcomes, n(S) = 5 + 8 + 4 = 17

(i) Number of favorable outcomes, n(E) = 5

(ii) Number of favorable outcomes, n(E) = 8

(iii) Number of events of drawing a green marble = 4

Probability of not drawing green marble = 1 – P(E)

(i) Total number of possible outcomes, n(S) = 20

Number of favorable outcomes, n(E) = 4

(ii) Total number of possible outcomes, n(S) = 19

Number of favorable outcomes, n(E) = 15

Total number of possible outcomes, n(S) = 90

(i) Number of favorable outcomes, n(E) = 81

(ii) Number of favorable outcomes, n(E) = 9

(iii) Number of favorable outcomes, n(E) = 18

Total number of possible outcomes, n(S) = 144

(i) Number of favorable outcomes, n(E) = 124

(ii) Probability of not buying a defective pen = 1 – P(E)

Total number of possible outcomes, n(S) = 144

Number of favorable outcomes, n(E) = 132

Total number of possible outcomes, n(S) = 5

(i) Number of favorable outcomes, n(E) = 1

(ii) If queen is drawn and put aside, total number of remaining cards, n(S) = 4

a) Number of favorable outcomes, n(E) = 1

b) Total number of events of getting a queen = 0 as there is no queen

∴ P(E) = 0

Total number of possible outcomes, n(S) = 4

Number of favorable outcomes, n(E) = 3

Total number of possible outcomes, n(S) = 36

Number of favorable outcomes = n(E)

We know that,

Total number of possible outcomes, n(S) = 48

(i) Number of favorable outcomes, n(E) = 10

(ii) Number of favorable outcomes, n(E) = 4

Given: number of red balls = 6

Let number of blue balls be x

Total number of possible outcomes, n(S) = 6 + x

Number of favorable outcomes = n(E)

P(blue ball) = 2P(red ball)

⇒ x = 12

∴ number of blue balls = 12

Total number of possible outcomes, n(S) = 20

(i) Number of favorable outcomes, n(E) = 3

(ii) Number of favorable outcomes, n(E) = 3

Total number of possible outcomes, n(S) = 52 – 3 = 49

(i) Number of favorable outcomes, n(E) = 13

(ii) Number of favorable outcomes, n(E) = 4 – 1 = 3

(iii) Number of favorable outcomes, n(E) = 13 – 3 = 10

Total number of possible outcomes, n(S) = 36

(i) Number of favorable outcomes, n(E) = 25

(ii) Number of favorable outcomes, n(E) = 11

(iii) Number of favorable outcomes, n(E) = 1

(i) 1

(ii) 0

(iii) 0 and 1

(iv) Equal

(v) 1

(vi) 1

(i) Incorrect statement

If 2 coins are tossed at the same time = 4 possible outcomes (HH, HT, TH, TT}

Therefore probability of each outcome = 1/4

(ii) Correct statement

When a dice is thrown once, event of occurrence of 3 odd numbers and 3 even numbers are possible out of 6 outcomes. Therefore, P(even number) = P(odd number) =

Total number of possible outcomes, n(S) = 100 + 200 + 50 = 350

(i) Number of favorable outcomes, n(E) = 50

(ii) Number of favorable outcomes, n(E) = 100 + 50 = 150

(iii) Number of favorable outcomes, n(E) = 100

Total number of possible outcomes, n(S) = 50

Number of favorable outcomes, n(E) = 4

Total number of possible outcomes, n(S) = 18

Number of favorable outcomes, n(E) = 10

Total number of possible outcomes, n(S) = 12

(i) Number of favorable outcomes, n(E) = 3

(ii) Number of favorable outcomes, n(E) = 9

Total number of possible outcomes, n(S) = 30

(i) Number of favorable outcomes of numbers divisible by 3, n(E) = 10

Probability of getting a number not divisible by 3 = 1 – P(E)

(ii) Number of favorable outcomes, n(E) = 6

(iii) Number of favorable outcomes of getting a perfect square, n(E) = 5

Probability of not getting a perfect square = 1 – P(E)

Total number of possible outcomes, n(S) = 36

(i) Number of favorable outcomes, n(E) = 11 {(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5) (4,5), (6,5)}

(ii) Number of favorable outcomes, n(E) = 11 – 1 = 10

Total number of possible outcomes, n(S) = 100 + 50 + 20 + 10 = 180

(i) Number of favorable outcomes, n(E) = 100

(ii) Number of favorable outcomes, n(E) = 20 + 10 = 30

(iii) Number of favorable outcomes, n(E) = 100 + 50 + 20 = 170

(iv) Number of favorable outcomes, n(E) = 50 + 20 = 70

Total number of possible outcomes, n(S) = 52 – 6 = 46

(i) Number of favorable outcomes, n(E) = 12 – 6 = 6

(ii) Number of favorable outcomes, n(E) = 26

(iii) Number of favorable outcomes, n(E) = 26 – 6 = 20

(iv) Number of favorable outcomes, n(E) = 4 – 2 = 2

Total number of possible outcomes, n(S) = 50

(i) Number of favorable outcomes, n(E) = 25

(ii) Number of favorable outcomes, n(E) = 4

(iii) Number of favorable outcomes, n(E) = 10

(iv) Number of favorable outcomes, n(E) = 4

Total number of possible outcomes, n(S) = 49

(i) Number of favorable outcomes, n(E) = 25

(ii) Number of favorable outcomes, n(E) = 9

(iii) Number of favorable outcomes, n(E) = 6

(iv) Number of favorable outcomes, n(E) = 1

Total number of possible outcomes, n(S) = 52 – 8 = 44

(i) Number of favorable outcomes, n(E) = 2

(ii) Number of favorable outcomes, n(E) = 26 – 4 = 22

Total number of possible outcomes, n(S) = 52 – 12 = 40

(i) Since all jacks,

Number of favorable outcomes, n(E) =

Total number of possible outcomes, n(S) = 52 – 2 – 2 = 48

(i) Number of favorable outcomes, n(E) = 4

(ii) Number of favorable outcomes, n(E) = 24

(iii) Number of favorable outcomes, n(E) = 12 – 4 = 8

(iv) Number of favorable outcomes, n(E) = 2

Total number of possible outcomes, n(S) = 20

(i) Number of favorable outcomes, n(E) = 10 + 3 = 13 {(2, 4, 6, 8, 10, 12, 14, 16, 18, 20) (3, 9, 15)}

(ii) Number of favorable outcomes, n(E) = 8

Total number of possible outcomes, n(S) = 44

(i) Number of favorable outcomes, n(E) = 20

(ii) Number of favorable outcomes, n(E) = (24 – 9) + (20 – 11) = 24

(iii) Number of favorable outcomes, n(E) = 9

(iv) Number of favorable outcomes, n(E) = 9

Total number of possible outcomes, n(S) = 52 – 6 = 46

(i) Number of favorable outcomes, n(E) = 26 – 6 = 20

(ii) Number of favorable outcomes, n(E) = 12 – 6 = 6

(iii) Number of favorable outcomes, n(E) = 13 – 3 = 10

Total numbers of elementary events are = 9

Let E be the event of getting an odd number

The favorable outcomes are: 1, 3, 5, 7, 9

Numbers of favorable events= 5

P (odd number) = P (E) = 5/9

The total number of elementary events are: 4,5,6 …..20 = 20 – 3= 17

Let E be the event of getting an even number on the random draw

Then total favorable outcomes are: 4, 6, 8, 10, 12, 14, 16, 18, 20

Total number of favorable outcomes = 9

P (even number) = P (E) = 9/17

Total numbers of elementary events are = 9

Let E be the event of getting an even number

Favorable outcomes are: 2, 4, 6, 8

Numbers of favorable outcomes are: 4

P (even number) = P (E) = 4/9

Total numbers of elementary events are: 52

Let E be the event of getting a non –face card

The total number of favorable outcomes are ; 52 – 12 = 40: being four cards each of king , queen, jack being a face card , deleted from the favorable outcomes number

P (non-face card) = P (E) = 40/52 = 20/ 26 = 10/ 13

Total numbers of elementary events are = 9

Let E be the event of getting a multiple of 3

Favorable outcomes are: 3, 6, 9

Numbers of favorable outcomes are: 3

P (multiple of 3) = P (E) = 3/9 = 1/3

Total numbers of elementary events are: 5+8 + 7 = 20

Let E be the event of getting a green ball or a white ball

Total numbers of favorable outcomes are: 8+ 7 = 15

P (white or green ball) = P (E) = 15/20 = 3/4

Total numbers of elementary events are: 2 x 2 x 2 = 8

Let E be the event of getting at least two heads

Favorable outcomes are: THH, HHT, HHH, HTH

Numbers of favorable outcomes are: 4

P (at least two heads) = P (E) = 4/8 = 1/2

Total numbers of elementary events are: 6 being a dice having six faces

Let E be the event of getting a prime number

Total favorable outcomes are: 1, 2, 3, 5

Total number of favorable events = 4

P (prime number) = P (E) = 4/6 = 2/3

Total numbers of elementary events are: 6 being a dice having six faces

Let E be the event of getting a prime number

The favorable outcomes are: 3, 4, 5,

Total number of favorable outcomes are = 3

P (number between 2 and 6) = P (E) = 3/6 =1/2

Total numbers of elementary events are: 6

Let E be the event of getting multiple of 3

Favorable events are: 3, 6

Numbers of favorable events are: 2

P (getting multiple of 3) = P (E) = 2/6 = 1/3

Total numbers of elementary events are: 7

Let E be the event of getting < 2

Favorable outcomes are: -1, 0, 1

Numbers of favorable outcomes are: 3

We know,

P (< 2) = P (E) = 3/7

Total numbers of elementary events are: 6 being a dice having six faces

Let E be the event of getting a prime number

The favorable events are: 1, 3, 5

Number of favorable events = 3

P (odd number) = P (E) = 3/6 =1/2

E denotes the complement of an event E then denotes the affirmation aspect and the probability of any event`s occurring or non-occurring , together is always 1

So P(E) + P( ) = 1

Probability of guessing the correct answer is = P (E) = x/12

Probability of guessing the incorrect answer is = P (not E) = 2/3

P (E) + P (not E) = 1

x/12 + 2/3 = 1

x+8 = 12

x = 4

Total numbers of elementary events are: 3+ 4 + 2= 9

Let E be the event of getting not an orange marble

Numbers of Favorable outcomes are: 3 + 4 = 7

P (not an orange marble) = P (E) = 7/9

Total numbers of elementary events are 52

Let E be the event of getting an ace

The numbers of favorable events are: 4

P (an ace) = P (E) = 4/52 = 1/13

Total numbers of elementary events are 2+ 2 = 4

Let E be the event of getting at least two heads

The favorable outcomes are: HH, HT, TH

Total numbers of favorable outcomes are 3

P (ATLEAST one head) = P (E) = 3/4

Note: at least one head means the outcome with one head or more will be considered favorable and the outcome with no head at all will be ruled out.

Total numbers of elementary events are: 10

Let E be the event of getting selected number being the average

Favorable outcome is = 70 /10 = 7

Average = total of observations/ numbers of observation

Total of observations = 3+ 5+ 5+ 7+ 7+ 7+ 9+ 9 + 9+ 9 = 70

Numbers of observation = 7

Numbers of favorable outcomes are: 3

P ( selecting average number) = P (E) = 3/10

Total numbers of elementary events are: 3 x 3 = 9

Let E be the event of (xy < 9)

Favorable outcomes are: 1× 1 =1, 1×4 = 4, 2×1 =2, 2×4= 8, 3×1 =3

Numbers of favorable outcomes are: 5

P (xy < 9) = 5/9

Total numbers of elementary events are = 20

Let E be the event of drawing a multiple of 3

The favorable outcomes are: 3, 6, 9, 12, 15, 18

Total numbers of favorable outcomes are = 6

P (multiple of 3) = P (E) = 6/20 =3/10

Total numbers of elementary events are: 6

Let E be the event of getting a number greater than 2

Favorable outcomes are: 3, 4, 5, 6

Numbers of favorable outcomes are: 4

P (number>2) = P (E) = 4/6 =2/3

Total numbers of elementary events are 52

Let E be the event of drawing a black queen

Total numbers of favorable outcomes are 2

P (black queen) = P (E) = 2/52 = 1/26

Total numbers of elementary events are: 52

Let E be the event of getting the dropped card as an ace

Favorable outcomes are: 4

P (ace) = P(E) = 4/52 = 1/13

Total number of elementary events are : 6

Let E be the event of getting a number less than 3

Favorable events are: 1, 2

Total numbers of favorable events are =2

P (number less than 3) = P (E) = 2/6 =1/3

Total numbers of elementary events are 2+ 2 = 4

Let E be the event of getting one head

Favorable outcome are: TH , HT

Numbers of favorable outcomes are= 2

P (exactly one head) = P (E) = 2/4= 1/2

Total numbers of elementary events are: 7

Let E be the event of having 53 Sundays

Note- a non-leap year has 365 days or 52 weeks and 1 day. This one day could be any day amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Out of these favorable event is one; Sunday.

Number of favorable outcome is = 1

P (Sunday) = P (E) = 1/7

Total numbers of elementary events are 6

Let E be the event of getting number greater than 4

The favorable outcomes are: 5, 6

Number of favorable outcomes are = 2

P (number greater than 4) = P (E) = 2/6 = 1/3

Total numbers of elementary events are: 25

Let E be the event of getting a prime number

Favorable outcomes are: 2, 3, 5, 7, 11, 13, 17, 19, 23

Numbers of favorable outcomes are = 9

P (prime number) = P (E) = 9/25

Total number of elementary events are = 7

Let E be the event of getting a multiple of 4

Favorable outcome are: 4, 8

Numbers of favorable outcome are= 2

P (multiple of 4) = P (E) = 2/7

Total numbers of elementary events are: 6 x 6 = 36

Let E be the event of getting a perfect square

Favorable outcomes are: 4, 9

4 and 9 can be rolled in combinations of: (2,2), (1,3), (3,1) , (3,6), (4,5), (5,4), (6,3)

Numbers of favorable outcomes are = 7

P (perfect square) = P (E) = 7/36

Total number of elementary events are= 26

Let E be the event of choosing a consonant

The favorable outcomes are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z

Numbers of favorable outcomes are = 21

P (getting a consonant) = P (E) = 21/26

Probability of an event cannot be negative or more than one so -1.5 can never be probability of any event.

P (E) = 0.05 given

P (E) + P (not E) = 1

P (not E) = 1 – 0.05 = 0.95

Probability of the occurrence of an event cannot be greater than 1.

1.6 Which is greater than 1 is the correct answer.

Probability of a certain event is always 1.

Probability of an impossible event is always zero (0)

Total numbers of elementary events are: 7

Let E be the event of having 53 Sundays

Note- a non-leap year has 365 days or 52 weeks and 1 day. This one day could be any day amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Out of these favorable event is one; Sunday.

Number of favorable outcome is = 1

P (Sunday) = P (E) = 1/7

Total numbers of elementary events are: 100

Let E be the event of winning a prize

Numbers of favorable outcomes are = 5

P (prize winning) = P (E) = 5/100 = 1/20

Total numbers of elementary event are: 50

Let E be the event of getting a multiple of 3 or 5

Favorable outcomes are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 5, 10, 20, 25, 35, 40, 50

Numbers of favorable outcomes are: 23

P (multiple of 3 or 5) = P (E) = 23/50

Total numbers of elementary events are: 10 x 9= 90

Being numbers are of non-replacing nature and order matters

Let E be the event of a/b as an integer

Favorable events are:

2/1, 3/1 , 4/1, 4/2, 5/1 , 6/1, 6/2 , 6/3 , 7/1 , 8/4 , 8/2 , 8/1, 9/3 , 9/1 , 10/5 , 10/2 , 10/1

Numbers of favorable outcomes are: 17/90

P (a/b is an integer) = P (E) = 17/90

Total numbers of elementary events are: 6 x 6 = 36

Let E be the event of getting of showing different faces on both the dice

Numbers of favorable events on first dice are: 6

Numbers of favorable events on second dice are: 5

Total numbers of favorable events are = 6 × 5 = 30

P (each dice showing different face) = 30/36 = 10/12 = 5/6

Total numbers of elementary events are: 7

Let E be the event of having exactly 52 Mondays

Note- a leap year has 366 days or 52 weeks and 2 day. Two days could be any days amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday.

Favorable outcomes are: Tuesday, Wednesday, Thursday, Friday, Saturday,

Number of favorable outcome is = 5

P (52 Mondays) = P (E) = 5/7

Total numbers of elementary events are: 12

Let E be the event of having month as March or October

Numbers of favorable events are = 2

P (March or October) = P(E) = 2/12 = 1/6

Total numbers of elementary events are: 6

Let E be the event of selecting a vowel

Favorable outcomes are: O, I, E

Numbers of favorable events are = 3

P (vowel) = P (E) = 3/6 = 1/2

Total numbers of elementary event are: 90

Let E be the event of choosing a multiple of 3

Favorable outcomes are: 3× 4 =12 , 3×5= 15, 3×6= 18, 3×7= 21,….. 3×33= 99

Numbers of favorable outcomes are= 30

P( multiple of 3) = P (E) = 30/90 = 1/3

Total numbers of elementary events are: 6x 6= 36

Let E be the event of getting same number on both dice together

Favorable outcomes are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

Numbers of favorable outcomes are: 6

P( both same numbers) = P (E) = 6/36 = 1/6

Total numbers of elementary events are: 6

Let E be the event of getting a prime number

Favorable events are: 2, 3, 5

Numbers of favorable events are= 3

P (prime number) = P (E) = 3/6 = 1/2

Total numbers of elementary events are: 6

Let E be the event of getting an even number

Favorable events are : 2, 4, 6

Numbers of favorable events are: 3

P (even number on dice) = P (E) = 3/6 = 1/2

Total numbers of elementary events are: 90

Let E be the event of drawing prime number less than 23

Favorable outcomes are: 2, 3, 5, 7, 11, 13, 17, 19

Numbers of favorable outcomes are: 8

P (prime number less than 23) = P (E) = 8/90 = 4/45

Total numbers of elementary events are: 15

Let E be the event of getting a multiple of 4

Favorable outcomes are: 4, 8, 12

Numbers of favorable outcomes are = 3

P (multiple of 4) = P (E) = 3/15 = 1/5

Total numbers of elementary events are: 2×2×2 = 8

(Each child could be either a girl or a boy)

Let E be the event of having at least one boy

(Note- at least one boy means there could be one or two or three boys in the family. The possibility of all the three girls is not considerable.)

P (having all girls) = P (not E) = 1/8

P (HAVING at least one boy) = 1- P (not E) = 1- 1/8 = 7/8

Total numbers of elementary events are: 25

Let E be the event of getting number divisible by both 2 and 3

Favorable outcomes are: { 6, 12, 18, 24}

Numbers of favorable outcomes are: 4

P (number divisible by 2 and 3) = P (E) = 4/25

Total numbers of elementary events are 2×2 =4

Let E be the event of at least one head

Favorable events are: having one head or both the heads. The possibility with all tails is not considerable.

P (both tails) = 1/4

P (having at least one head) = P (E) = 1 – P (both tails) = 1 – 1/4 = 3/4

Total numbers of elementary events are: 6 x 6 = 36

Let E be the event of getting an even number on both the dice

Favorable outcomes are: (2,2) , (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)

Numbers of favorable outcomes are = 9

P (getting even number on both dice) = P (E) = 9/36 = 1/4

Total numbers of elementary events are: 30

Let E be the event of selecting a prime number

Favorable outcomes are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Numbers of favorable outcomes are: 10

P (prime number) = P (E) = 10/30 = 1/3

Total numbers of elementary events are: 52

Let E be the event of drawing card other than ace

Favorable numbers of events are: 52 – 4= 48

P (not drawing ace) = P (E) = 48/52 = 24/26 = 12/13