Question

If in a binomial distribution n=4, P(X=0)=\(\frac{16}{81}\), then P(X = 4) equals :

• A. \(\frac{1}{16}\)
• B. \(\frac{1}{81}\)
• C. \(\frac{1}{27}\)
• D. \(\frac{1}{8}\)

Answer 1

The Correct Option is B

To solve the problem, we begin by recalling the basic formula for the probability of exactly \(k\) successes in a binomial distribution, which is:

\(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Given \(n=4\), \(P(X=0)=\frac{16}{81}\), we use the formula:

\(P(X=0) = \binom{4}{0} p^0 (1-p)^4 = (1-p)^4 = \frac{16}{81}\)

Taking the fourth root from both sides to solve for \((1-p)\):

\(1-p = \sqrt[4]{\frac{16}{81}} = \frac{2}{3}\)

Thus, \(p = 1-\frac{2}{3} = \frac{1}{3}\)

We need to find \(P(X=4)\):

\(P(X=4) = \binom{4}{4} p^4 (1-p)^0 = p^4 = \left(\frac{1}{3}\right)^4 = \frac{1}{81}\)

Therefore, the probability \(P(X=4)\) is \(\frac{1}{81}\).