Question ID: #450
Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If X denotes the number of defective oranges, then the variance of X is:
- (1) $28/75$
- (2) $14/25$
- (3) $26/75$
- (4) $18/25$
Solution:
Total oranges = 10 (3 Defective, 7 Good).
Two oranges are drawn. Let $X$ be the number of defective oranges.
$X$ can take values 0, 1, 2.
Total ways to select 2 oranges = ${}^{10}C_2 = 45$.
Probability Distribution:
1. $P(X=0)$ (Both Good):
$$ \frac{{}^7C_2}{{}^{10}C_2} = \frac{21}{45} = \frac{7}{15} $$
2. $P(X=1)$ (1 Good, 1 Defective):
$$ \frac{{}^7C_1 \times {}^3C_1}{{}^{10}C_2} = \frac{7 \times 3}{45} = \frac{21}{45} = \frac{7}{15} $$
3. $P(X=2)$ (Both Defective):
$$ \frac{{}^3C_2}{{}^{10}C_2} = \frac{3}{45} = \frac{1}{15} $$
Calculate Mean $\mu = E(X) = \sum x_i p_i$:
$$ \mu = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7+2}{15} = \frac{9}{15} = \frac{3}{5} $$
Calculate $E(X^2) = \sum x_i^2 p_i$:
$$ E(X^2) = 0^2 \cdot \frac{7}{15} + 1^2 \cdot \frac{7}{15} + 2^2 \cdot \frac{1}{15} = \frac{7+4}{15} = \frac{11}{15} $$
Calculate Variance $\sigma^2 = E(X^2) – (\mu)^2$:
$$ \sigma^2 = \frac{11}{15} – \left(\frac{3}{5}\right)^2 $$
$$ = \frac{11}{15} – \frac{9}{25} $$
LCM of 15 and 25 is 75.
$$ = \frac{55}{75} – \frac{27}{75} = \frac{28}{75} $$
Ans. (1)
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