Question

A coin that comes up heads with probability $p$ and tails with probability $q$ independently on each flip is tossed $8$ times. Suppose that the probability of occurrence of three heads and five tails is equal to $\frac{1}{25}$ times the probability of occurrence of five heads and three tails. Let $p=\frac{m}{n}$ where $m$ and $n$ are relatively prime. Then the value of $m+n$ is

• A. $15$
• B. $5$
• C. $11$
• D. $14$

Answer 1

The correct option is C $11$

$p+q=1$
Given : ${}^8{C}_3{p}^3(1-p{)}^5=\frac{1}{25}{}^8{C}_5{p}^5(1-p{)}^3$
$\Rightarrow 25(1-p{)}^2=p^2$
$\Rightarrow \frac{p}{1-p}=\pm 5$
$\Rightarrow p=\frac{5}{6},\frac{5}{4}$
$\Rightarrow p=\frac{5}{6}=\frac{m}{n}(\because 0<p<1)$
$\therefore m+n=11$