Question

Two dice are thrown simultaneously. What is the probability of obtaining a multiple of 2 on one of them and a multiple of 3 on the other 

(A) 5/36

(B) 11/36

(C) 1/6

(D) 1/3

Answer 1

Answer is (B) 11/36

Favourable cases for one are three, that is, 2, 4 and 6 and for other are two, that is, 3, 6. Hence,

required probability = [((3 x 2)/36) 2 - 1/36] = 11/36

{As same way happen when dice changes numbers among themselves}