Question

A box containing 8 light bulbs of which 3 are defective. A bulb is selected from the
box and tested. If it is defective, another bulb is selected and tested until a nondefective
bulb is chosen. Find the expected number and variance of bulbs chosen.

Answer 1

Answer on Question #45599 – Math – Statistics and Probability

Question. A box containing 8 light bulbs of which 3 are defective. A bulb is selected from the box and tested. If it is defective, another bulb is selected and tested until a nondefective bulb is chosen. Find the expected number and variance of bulbs chosen.

Solution. Let $\xi$ be the number of bulbs chosen. Then

$$P(\xi = 1) = \frac{C_5^1}{C_8^1} = \frac{5!}{4!} \cdot \frac{7!}{8!} = \frac{5}{8},$$ $$\begin{array}{l} P(\xi = 2) = P(\text{the first is defective, the second is nondefective}) = \frac{C_3^1}{C_8^1} \cdot \frac{C_5^1}{C_7^1} = \\ = \frac{3!}{2!} \cdot \frac{7!}{8!} \cdot \frac{5!}{4!} \cdot \frac{6!}{7!} = \frac{3}{8} \cdot \frac{5}{7} = \frac{15}{56}, \end{array}$$ $$\begin{array}{l} P(\xi = 3) = P(\text{the first and the second are defective, the third is nondefective}) = \\ = \frac{C_3^1}{C_8^1} \cdot \frac{C_2^1}{C_7^1} \cdot \frac{C_5^1}{C_6^1} = \frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6} = \frac{5}{56}, \end{array}$$ $$\begin{array}{l} P(\xi = 4) = P(\text{the first, the second and the third are defective}) = \frac{C_3^1}{C_8^1} \cdot \frac{C_2^1}{C_7^1} \cdot \frac{C_1^1}{C_6^1} = \\ = \frac{3}{8} \cdot \frac{2}{7} \cdot \frac{1}{6} = \frac{1}{56}. \end{array}$$

Therefore $\xi$ has the next distribution:

The expected number of bulbs chosen is $E\xi = 1 \cdot \frac{5}{8} + 2 \cdot \frac{15}{56} + 3 \cdot \frac{5}{56} + 4 \cdot \frac{1}{56} = \frac{3}{2}$.

Calculate

$$E\xi^2 = 1 \cdot \frac{5}{8} + 4 \cdot \frac{15}{56} + 9 \cdot \frac{5}{56} + 16 \cdot \frac{1}{56} = \frac{39}{14},$$

The variance of bulbs chosen is

$$Var\xi = E\xi^2 - (E\xi)^2 = \frac{39}{14} - \frac{9}{4} = \frac{15}{28}.$$

Answer. $E\xi = \frac{3}{2}, Var\xi = \frac{15}{28}$.

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