Question

Apoorve throws two dice at once and computes the product of the numbers appearing on the dice. Peehu throws one dice and square the number that appears on it. Who has the better chance of getting the number $36$? Why?

Answer 1

Apoorv throws two dice once.
So, total number of outcomes, $n\left(S\right)=36$.
Number of outcomes for getting product $36$.
$n\left(E_1\right)=1(6\times 6)$
$\therefore ProbabilityofApoorv=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{1}{36}$
Also, Peehu throw one die.
So, the total number of outcomes $n\left(S\right)=6$
Number of outcomes for getting square of a number is $36$.
$n\left(E_2\right)=1(\because 6^2=36)$
$\therefore ProbabilityforPeehu=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{1}{6}=\frac{6}{36}$
Hence, Peehu has better chance of getting the number $36$.