Question

A fruit box contains 6 apples and 4 oranges. A person picks out a fruit three times with replacement. Find:
(i) The probability distribution of the number of oranges he draws.
(ii) The expectation of the number of oranges.

Hint:

In a binomial distribution \( B(n, p) \), the expectation is simply \( E(X) = np \). Use the binomial formula for exact probabilities.

Answer 1

Probability of orange in one draw: \( p = \frac{4}{10} = 0.4 \)
Let $X$ be the number of oranges in 3 draws. Since replacement is done, $X$ follows a binomial distribution: \[ X \sim B(n=3, p=0.4) \] (i) Probability distribution: \[ \begin{aligned} P(X = 0) &= \binom{3}{0} (0.4)^0 (0.6)^3 = 1 \cdot 1 \cdot 0.216 = 0.216\\ P(X = 1) &= \binom{3}{1} (0.4)^1 (0.6)^2 = 3 \cdot 0.4 \cdot 0.36 = 0.432\\ P(X = 2) &= \binom{3}{2} (0.4)^2 (0.6)^1 = 3 \cdot 0.16 \cdot 0.6 = 0.288\\ P(X = 3) &= \binom{3}{3} (0.4)^3 (0.6)^0 = 1 \cdot 0.064 \cdot 1 = 0.064\\ \end{aligned} \] (ii) Expectation of a binomial variable: \[ E(X) = np = 3 \cdot 0.4 = 1.2 \]