Q. A pair of fair dice is thrown independently 4 times. The probability of getting a sum of exactly 7 twice is
A) $\frac{5}{81}$
B) $\frac{25}{243}$
C) $\frac{25}{216}$
D) $\frac{125}{648}$
Explanation
(c)
Let $E =$Event of getting sum of 7 in two dice
$= \{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$
Now, $P(E) = \frac{6}{36} = \frac{1}{6}$ (say)
$\Rightarrow \ p = \frac{1}{6}$
$\therefore\ q = 1 - p = \frac{5}{6}$
Required probability$= \ ^{4}C_{2}\left( \frac{1}{6} \right)^{2}\left( \frac{5}{6} \right)^{2}$
$= 6 \times \frac{5^{2}}{6^{4}} = \frac{25}{216}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.