Question

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

Answer 1

Step 1 - Probability Apoorv getting the number $36$

Apoorv throws two dice at once.

The total number of outcomes $=36$

Number of favorable outcomes for getting the product $36=1(\because 6\times 6=36)$

$\therefore$ Probability for Apoorv getting the number $\mathbf{=}\frac{\mathbf{f}\mathbf{a}\mathbf{v}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{a}\mathbf{b}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}\mathbf{}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}}$

$=\frac{1}{36}$

Step 2- Probability Peehu getting the number $36$

Peehu throws one die,

Hence, the total number of outcomes $=6$

Number of favorable outcomes for getting square $36=1(\because 6^2=36)$

$\therefore$ Probability for Peehu getting the number $\mathbf{=}\frac{\mathbf{f}\mathbf{a}\mathbf{v}\mathbf{o}\mathbf{r}\mathbf{a}\mathbf{b}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}\mathbf{}}$

$=\frac{1}{6}$

Since, $\frac{1}{6}>\frac{1}{36}$

Hence, Peehu has a better chance of getting the number $36$