Three defective oranges are accidentally mixe | vedclass

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(A) Total number of oranges = $10$. Number of defective oranges = $3$. Number of good oranges = $7$. Two oranges are drawn at random. Let $x$ be the n

Vedclass · Accepted Answer

$28 / 75$

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(A) Total number of oranges = $10$. Number of defective oranges = $3$. Number of good oranges = $7$. Two oranges are drawn at random. Let $x$ be the number of defective oranges. The possible values of $x$ are $0, 1, 2$.The probability distribution is as follows:$x_i$$P(x_i)$$0$$\frac{^7C_2}{^{10}C_2} = \frac{21}{45} = \frac{42}{90}$$1$$\frac{^7C_1 	imes ^3C_1}{^{10}C_2} = \frac{21}{45} = \frac{42}{90}$$2$$\frac{^3C_2}{^{10}C_2} = \frac{3}{45} = \frac{6}{90}$Mean $\mu = E(x) = \sum x_i P(x_i) = 0 	imes \frac{42}{90} + 1 	imes \frac{42}{90} + 2 	imes \frac{6}{90} = \frac{42 + 12}{90} = \frac{54}{90} = \frac{3}{5} = 0.6$.Now,$E(x^2) = \sum x_i^2 P(x_i) = 0^2 	imes \frac{42}{90} + 1^2 	imes \frac{42}{90} + 2^2 	imes \frac{6}{90} = \frac{42 + 24}{90} = \frac{66}{90} = \frac{11}{15}$.Variance $\sigma^2 = E(x^2) - [E(x)]^2 = \frac{11}{15} - (\frac{3}{5})^2 = \frac{11}{15} - \frac{9}{25} = \frac{55 - 27}{75} = \frac{28}{75}$.

Three defective oranges are accidentally 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:

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