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 p be the probability of success and q = 1 − p. In a binomial distribution with n = 3, \(P(X = r) = {}^{3}C_{r}\,p^r q^{3-r}\):
\(P(X = 2) = {}^{3}C_{2}\,p^2q = 3p^2q\)
\(P(X = 3) = p^3\)
Given \(P(X = 2) = 9P(X = 3)\):
\(3p^2q = 9p^3\)
Dividing by \(3p^2\) (p ≠ 0):
\(q = 3p\)
\(1 - p = 3p\)
\( \displaystyle p = \frac{1}{4} \)