Question

If in $6$ trials, $X$ is a binomial variate which follows the relation $9P(X=4)=P(X=2)$, then what is the probability of success ?

• A. $3/4$
• B. $1/4$
• C. $3/8$
• D. $1/8$

Answer 1

The correct option is A $3/4$

$9{\times }^6{C}_4{P}^4{(1-p)}^2{=}^6{C}_4{P}^4{(1-p)}^2$

$\Rightarrow 9P^2={(1-p)}^2$

$\Rightarrow \frac{P}{1-p}=3\Rightarrow P=3-3p$

$=4p=3\Rightarrow p=3/4$