Question
Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.
✓
Answer
Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2. P(X = 0) = P(no bad egg) $=\frac{\text{}^{10}\text{C}_3}{\text{}^{12}\text{C}_3}$ $=\frac{120}{220}$ $=\frac{6}{11}$ P(X = 1) = P(1 bad egg) $=\frac{\text{}^{2}\text{C}_1\times\text{}^{10}\text{C}_2}{\text{}^{12}\text{C}_3}$ $=\frac{90}{220}$ $=\frac{9}{22}$ P(X = 2) = P(2 bad eggs) $=\frac{\text{}^{2}\text{C}_2\times\text{}^{10}\text{C}_1}{\text{}^{12}\text{C}_3}$ $=\frac{10}{220}$ $=\frac{1}{22}$ Thus, the probability distribution of X is given by
Computation of mean
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{2}$
| $\text{X}$ | $\text{P}(\text{X})$ |
| $0$ | $\frac{6}{11}$ |
| $1$ | $\frac{9}{22}$ |
| $2$ | $\frac{1}{22}$ |
| $\text{x}_\text{i}$ | $\text{p}_\text{i}$ | $\text{x}_\text{i}\text{p}_\text{i}$ |
| $0$ | $\frac{6}{11}$ | $0$ |
| $1$ | $\frac{9}{22}$ | $\frac{9}{22}$ |
| $2$ | $\frac{1}{22}$ | $\frac{1}{11}$ |
| $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{2}$ |
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