Question

The mean and variance of a Binomial distribution are \(4\) and \(\frac43\) respectively, then the value of \(P(X\geq1)\) is:

• A. \(\frac{80}{81}\)
• B. \(\frac{63}{64}\)
• C. \(\frac{728}{729}\)
• D. \(\frac{665}{729}\)

Answer 1

The Correct Option is C

The mean μ of a binomial distribution is given by \( \mu = np \), where n is the number of trials and p is the probability of success. The variance σ² is given by \( \sigma^2 = np(1-p) \).

Given:

• Mean \( \mu = 4 \)
• Variance \( \sigma^2 = \frac{4}{3} \)

From the mean equation, we have:

\( np = 4 \) … (1)

From the variance equation, we have:

\( np(1-p) = \frac{4}{3} \) … (2)

Substitute equation (1) into equation (2):

\( 4(1-p) = \frac{4}{3} \)

Solving for p gives:

\( 4 - 4p = \frac{4}{3} \)

\( 4p = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3} \)

\( p = \frac{2}{3} \)

Substitute p back into equation (1):

\( n \cdot \frac{2}{3} = 4 \)

\( n = 4 \times \frac{3}{2} = 6 \)

From this calculation, we know the distribution is \( B(6, \frac{2}{3}) \).

We are asked to find \( P(X \geq 1) \), which can be calculated as:

\( P(X \geq 1) = 1 - P(X = 0) \)

Using the binomial probability formula:

\( P(X = 0) = \binom{6}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^6 = 1 \times 1 \times \left(\frac{1}{729}\right) = \frac{1}{729} \)

Thus,

\( P(X \geq 1) = 1 - \frac{1}{729} = \frac{728}{729} \)

Therefore, the correct answer is \( \frac{728}{729} \).