Question

Let the random variables $X$ follow $B(6,p)$ If $16P(X=4)=P(X=2)$, then what is the value of $p$:

• (A) $\frac{1}{3}$
• (B) $\frac{1}{4}$
• (C) $\frac{1}{5}$
• (D) $\frac{1}{6}$

Answer 1

The Correct Option is C

Explanation:
Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}$Where, n is number of observations, p is the probability of success.Formula:Calculate combination:$\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{(n-k)!k!}$Factorial:$n!=1\times 2\times 3\times \dots \times (n-1)\times n$Given: $16P(X=4)=P(X=2)$
To find the value of $p$
Calculating probabilities of random variable X,
$P(X=4)=\left(\begin{array}{l}6\\ 4\end{array}\right)p^4(1-p{)}^2$$P(X=2)=\left(\begin{array}{l}6\\ 2\end{array}\right)p^2(1-p{)}^4$
Given,
$16P(X=4)=P(X=2)$$\Rightarrow 16\left(\begin{array}{l}6\\ 4\end{array}\right)p^4(1-p{)}^2=\left(\begin{array}{l}6\\ 2\end{array}\right)p^2(1-p{)}^4$$\Rightarrow 16\times \frac{6!}{2!4!}{p}^4(1-p{)}^2=\frac{6!}{2!4!}{p}^2(1-p{)}^4$$\Rightarrow 16p^2=(1-p{)}^2$$\Rightarrow \pm 4p=1-p$$\Rightarrow p=\frac{1}{5}$ or $p=-\frac{1}{3}$So, $p=\frac{1}{5}$.Hence, the correct option is (C).