Question

A cubical die is thrown $6$ times. The probability that $2$ and $4$ will turn up exactly $3$ times each is

• A. $\frac{1}{6^6}$
• B. $\frac{6!}{6^6}$
• C. $\frac{6!}{3!3!}\times \frac{1}{6^6}$
• D. $1$

Answer 1

Answer: (C)

$\frac{6!}{3!3!}\times \frac{1}{6^6}$

Number of ways of occering $2$ and $4$ exactly three times.

$=6C_3(\frac{1}{6}\ast \frac{1}{6}\ast \frac{1}{6})(\frac{1}{6}\ast \frac{1}{6}\ast 16)=6C_3\frac{1}{6^6}$