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

Question

Six fair dice are rolled. The probability that the product of the numbers appearing on top faces is prime, is (A) (1/2)((1/6))4 (B) ((1/2))6 (C) (1/64) (D) (1/2)((1/6))5

Tardigrade Admin · Accepted Answer

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 =6C5((1/6))5((3/6)) =6((1/6))5((1/2)) =(1/2)((1/6))4

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

$\frac{1}{2} (\frac{1}{6})^{4}$

$(\frac{1}{2})^{6}$

$\frac{1}{6 ^{4}}$

$\frac{1}{2} (\frac{1}{6})^{5}$

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 $=^{6} C_{5} (\frac{1}{6})^{5} (\frac{3}{6})$
$= 6 (\frac{1}{6})^{5} (\frac{1}{2})$
$= \frac{1}{2} (\frac{1}{6})^{4}$

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

21​(61​)4

(21​)6

641​

21​(61​)5

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 =6C5​(61​)5(63​) =6(61​)5(21​) =21​(61​)4

$\frac{1}{2} (\frac{1}{6})^{4}$

$(\frac{1}{2})^{6}$

$\frac{1}{6 ^{4}}$

$\frac{1}{2} (\frac{1}{6})^{5}$

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 $=^{6} C_{5} (\frac{1}{6})^{5} (\frac{3}{6})$
$= 6 (\frac{1}{6})^{5} (\frac{1}{2})$
$= \frac{1}{2} (\frac{1}{6})^{4}$