# Easy Probability Solved QuestionAptitude Discussion

 Q. Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.
 ✖ A. 5/12 ✔ B. 11/36 ✖ C. 5/36 ✖ D. 13/36

Solution:
Option(B) is correct

Let $X$ be required events and $S$ be the sample space,

then $X$ ={(2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2),(6,2),(3,4),(6,4),(3,6)}

$n(X)=11$, $n(S)=36$

Hence, required probability

$=\dfrac{n(X)}{n(S)}$

$=\dfrac{11}{36}$

## (3) Comment(s)

Anubhav Singh
()

There are 11 favourable cases as (6,6) will be repeated.

Poonam Pipaliya
()

why can't we take (2,1)

Ritika
()

Since second dice contains multiple of 3 only. In your case it is 1, which is not a multiple of 3.