Question

Two dice are tossed $6$ times, then the probability that $7$ will show an exactly four of tosses is

• A. $\frac{225}{18442}$
• B. $\frac{116}{20003}$
• C. $\frac{125}{15552}$
• D. None of these

Answer 1

Answer: (C)

$\frac{125}{15552}$

Two dice will show $7$ if they show : $\{1,6\},\{2,5\},\{6,1\}\{5,2\},\{4,3\},\{3,4\}$

the probability of success $\left(p\right)=6/(6\times 6)=1/6$

the probability of failure $\left(q\right)=1-1/6=5/6$

Probability of showing 7 exactly 4 times $={}^6{C}_4\times p^4\times q^2=15\times (1/6{)}^4\times (5/6{)}^2=\frac{125}{15552}$