Question

Two dice were rolled simultaneously. Find the probability of getting:
a) prime number on both the dice
b) a doublet
c) 3 as a product
d) different number on both the dice.​

Answer 1

Answer:

let the n(s) be the total number of outcomes

n(s) = 36

(a) Let P(A) be the probability of getting prime number on both dice

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

n(A) = 9

P(A) = n(A) / n(s) = 9/36 = 1/4

(b) Let P(B) be the probability of getting doublet

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

n(B) = 6

P(B) = n(B) / n(s) = 6/36 = 1/6

(c) Let P(C) be the probability of getting 3 as product

C = {(1,3), (3,1)}

n(C) = 2

P(C) = n(C) / n(s) = 2/36 = 1/18

(d) Let P(D) be the probability of getting different number on the both the dice

D = n(s) - n(B) [ because except doublet all the outcomes are different]

= 36 - 6 = 30

P(D) = n(D) / n(s) = 30/36 = 5/6