Hiremee Probability Questions with Answers

n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2

Let E = event of getting 1 club and 1 diamond.

We know that there are 13 clubs and 13 diamonds in the total 52 cards.

Hence, n(E) = Number of ways of drawing one club from 13 and one diamond from 13

= 13C1 × 13C1 = 13/102

P(E) = n(E)/n(S)=(13C1×13C1)/52C2
=

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36

Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ... etc

Let E = the event that the total is a prime number
= {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4),
(4,1), (4,3), (5,2), (5,6), (6,1), (6,5)}

Hence, n(E) = 15

P(E) = n(E)/n(S)=15/36=5/12

Total number of balls = 4 + 5 + 6 = 15

Let S be the sample space.
n(S) = Total number of ways of drawing 3 balls out of 15 = 15C3

Let E = Event of drawing 3 balls, all of them are yellow.
n(E) = Number of ways of drawing 3 balls, all of them are yellow
= Number of ways of drawing 3 balls from the total 5 = 5C3
(∵ there are 5 yellow balls in the total balls)

P(E) = n(E)/n(S) = 2/91

Total Number of letters in the word ASSASSINATION, n(S) = 13

Total number of consonants in the word ASSASSINATION = 7

P(consonant) = 7/13

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)

Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36

Let E = the event of getting two numbers whose product is even

= {(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)}

Hence, n(E) = 27

P(E) = n(E)/n(S)=27/36=3/4

Total number of balls = (8 + 7 + 6) = 21.

Let E= event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7.

P(E) =n(E)/n(S)= 7/21 =1/3 .

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

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) = n(E)/n(S) = 7/8 .

Let S be the sample space.

Then, n(S) = 52C2 = (52 x 51)/(2 x 1) = 1326.

Let E = event of getting 2 kings out of 4.

n(E) = 4C2 = 6.

P(E) = n(E)/n(S) = 6/1326 = 1/221 .

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/ n(S)= 9/20 .