Question

Let the probability mass function (p.m.f.) of a random variable \( X \) be \(P(X = x) = \binom{4}{x} \left( \frac{5}{9} \right)^x \left( \frac{4}{9} \right)^{4-x}\), for \(x = 0, 1, 2, 3, 4\), then \(E(X)\) is equal to 
 

• A. \( \frac{20}{9} \)
• B. \( \frac{9}{20} \)
• C. \( \frac{12}{9} \)
• D. \( \frac{9}{25} \)

Hint:

For a binomial distribution, the expected value is simply \( n \cdot p \), where \( n \) is the number of trials and \( p \) is the probability of success.

Answer 1

The Correct Option is A

Step 1: Understanding the problem. 
We are given the probability mass function (p.m.f.) for the random variable \( X \) in the form of a binomial distribution. For a binomial random variable \( X \), the expected value is given by: \[ E(X) = n \cdot p \] where \( n \) is the number of trials and \( p \) is the probability of success in each trial. In this case, \( n = 4 \) and \( p = \frac{5}{9} \).

Step 2: Calculate the expected value. 
Substitute the values into the formula for \( E(X) \): \[ E(X) = 4 \cdot \frac{5}{9} = \frac{20}{9} \]

Step 3: Conclusion. 
Thus, the expected value \( E(X) = \frac{20}{9} \), so the correct answer is (A).