Probability
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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

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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.