Question

Six fair dice are rolled. The probability that the product of the numbers appearing on top faces is prime, is

• A. $\frac{1}{2}\left(\frac{1}{6}\right)^{4}$
• B. $\left(\frac{1}{2}\right)^{6}$
• C. $\frac{1}{6^{4}}$
• D. $\frac{1}{2}\left(\frac{1}{6}\right)^{5}$

Answer 1

Answer: (A)

This can be possible if five of the dice have $1$ on top faces and one die has $2,3$ or $5.$
Hence, required probability $=^{6}C_{5}\left(\frac{1}{6}\right)^{5}\left(\frac{3}{6}\right)$
$=6\left(\frac{1}{6}\right)^{5}\left(\frac{1}{2}\right)$
$=\frac{1}{2}\left(\frac{1}{6}\right)^{4}$