Question

Ben rolls $5$ fair $12$ sided dice. the $12$ faces of each die are numbered from $1$ to $12$.

What is the probability that exactly two of the dice show an even number?

Answer 1

What is the probability that exactly two of the dice show an even number:

$12$ sided dice has $6$ even number and $6$ odd numbers.

To show exactly two even numbers, it must also show $4$ odd numbers.

Probability of getting even number is $\frac{6}{12}=0.5$.

Probability of getting odd number is $\frac{6}{12}=0.5$.

Apply the formula $\mathrm{P}(\mathrm{exactly}2\mathrm{even}\mathrm{numbers})={}^5\mathrm{C}_2\times \mathrm{P}(\mathrm{getting}\mathrm{even}\mathrm{number})\times \mathrm{P}(\mathrm{getting}\mathrm{odd}\mathrm{number})$.

Here, ${}^{5}C_2$ is number of ways $2$ numbers can be chosen from $5$.

Substitute the known values;

$\mathrm{P}(\mathrm{exactly}2\mathrm{even}\mathrm{numbers})={}^5\mathrm{C}_2\times \frac{1}{2}\times \frac{1}{2}$

Apply combination formula ${}^{n}C_r=\frac{n!}{\left(n-r\right)!\times r!}$.

$$\begin{gathered} \mathrm{P}(\mathrm{exactly}2\mathrm{even}\mathrm{numbers})=\frac{5!}{\left(5-2\right)!\times 2!}\times \frac{1}{2}\times \frac{1}{2} \\ \mathrm{P}(\mathrm{exactly}2\mathrm{even}\mathrm{numbers})=\frac{5!}{3!\times 2!}\times \frac{1}{2}\times \frac{1}{2} \\ \mathrm{P}(\mathrm{exactly}2\mathrm{even}\mathrm{numbers})=\frac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 2}\times \frac{1}{2}\times \frac{1}{2} \\ \mathrm{P}(\mathrm{exactly}2\mathrm{even}\mathrm{numbers})=\frac{5}{2} \end{gathered}$$

Hence, the probability that exactly two of the dice show an even number is $\frac{5}{2}$.