If the probability of two successes is 9 times the probability of 3 successes in 3 trials of a binomial distribution, then the probability of success in each trial is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1/4
Let probability of success in each trial be $p$ and failure be $q=1-p$
Given number of trials $n=3$
Probability of exactly $2$ successes
$P(2)=\frac{3!}{2!1!}p^2q=3p^2q$
Probability of exactly $3$ successes
$P(3)=p^3$
Given $P(2)=9P(3)$
$3p^2q=9p^3$
Divide by $3p^2$
$q=3p$
$1-p=3p$
$1=4p$
$p=\frac{1}{4}$
The probability of success in each trial is $\frac{1}{4}$.