Question

In a binomial distribution, if \(n=4\) and \( P(X=0) = \frac{16{81} \), then \( P(X=4) = \)}

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

Hint:

For a binomial distribution B(n,p): The probability mass function is \( P(X=k) = {}^n C_k p^k q^{n-k} \), where \( q=1-p \). \( P(X=0) = {}^n C_0 p^0 q^n = q^n \). \( P(X=n) = {}^n C_n p^n q^0 = p^n \). Use the given information to find \(p\) and \(q\), then calculate the required probability.

Answer 1

The Correct Option is D

Step 1: Recall the formula for binomial distribution.
\( P(X=k) = {}^n C_k \cdot p^k \cdot q^{n-k} \), where \(p\) is the probability of success, \(q=1-p\) is the probability of failure, and \(n\) is the number of trials.

Step 2: Use the given information for \( P(X=0) \).
Given \( n=4 \) and \( P(X=0) = \frac{16}{81} \).
\( P(X=0) = {}^4 C_0 \cdot p^0 \cdot q^{4-0} = 1 \cdot 1 \cdot q^4 = q^4 \).
So, \( q^4 = \frac{16}{81} \).

Step 3: Solve for \(q\).
\( q^4 = \left(\frac{2}{3}\right)^4 \).
Since \(q\) is a probability, \( 0 \le q \le 1 \).
Thus, \( q = \frac{2}{3} \).

Step 4: Calculate \(p\).
\( p = 1-q = 1 - \frac{2}{3} = \frac{1}{3} \).

Step 5: Calculate \( P(X=4) \).
We need to find \( P(X=4) \) with \( n=4, p=1/3, q=2/3 \).
\( P(X=4) = {}^4 C_4 \cdot p^4 \cdot q^{4-4} = 1 \cdot p^4 \cdot q^0 = p^4 \).
\[ P(X=4) = \left(\frac{1}{3}\right)^4 = \frac{1^4}{3^4} = \frac{1}{81} \] This matches option (4).