Question

Two dice are tossed 6 times. Then the probability that 7 will show an exactly four of the tosses is:

• A. $\frac{225}{18442}$
• B. $\frac{116}{20003}$
• C. $\frac{125}{15552}$
• D. $\frac{117}{17442}$

Answer 1

Answer: (C)

$\frac{125}{15552}$

Answer 2

This is the question of binomial distribution because dice tossed 6 times we know, $P(x) = {^nC_r} p^r \, q^{n-r}$ where p = prob of success q = prob of failure n = no. of toss Now, Here n = 6, p = prob of getting 7 = $\frac{1}{6}$ and q = 1 - p = 1 - $\frac{1}{6} = \frac{5}{6}$ , r = 4 $\therefore$ Required prob = ${^6C_4} \left( \frac{1}{6}\right)^4 \left( \frac{5}{6} \right)^2 = \frac{125}{15552}$