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

Hint:

For binomial distribution, \( E(X) = n \cdot p \); verify p.m.f. sums to 1 if needed.

Answer 1

The Correct Option is A

This is a binomial distribution with \( n = 4 \), \( p = \frac{5}{9} \), \( q = \frac{4}{9} \). Expected value: \[ E(X) = n \cdot p = 4 \cdot \frac{5}{9} = \frac{20}{9}. \]