Question

A pair of dice is thrown simultaneously. Find the probability of getting
$\left(i\right)$ a multiple of $3$ on both dice.
$\left(ii\right)$ Sum of the numbers on two dice is always less than $7$.
$\left(iii\right)$ An odd number on the first die and a prime number on the other.

• A. $\left(i\right)\frac{1}{5}\left(ii\right)\frac{7}{13}\left(iii\right)\frac{1}{3}$
• B. $\left(i\right)\frac{1}{7}\left(ii\right)\frac{11}{18}\left(iii\right)\frac{1}{7}$
• C. $\left(i\right)\frac{1}{9}\left(ii\right)\frac{1}{12}\left(iii\right)\frac{1}{4}$
• D. None of these

Answer 1

The correct option is D None of these

Pair of dices can be thrown in $6\times 6$$=36$ ways Then $n\left(s\right)=36$

(i)The event of getting multiple of $3$ is four.

Then provability=$\frac{n\left(A\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$

(ii)The event of sum of the numbers on two dice is always less than $7$ is $15$.

Then provability=$\frac{n\left(A\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$

(iii)The event of odd number on the first die and a prime number on the other are $9$.

Then provability=$\frac{n\left(A\right)}{n\left(S\right)}=\frac{9}{36}=\frac{9}{4}$