Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differ

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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…

Accepted Answer

Given: There are 3 defective oranges and 7 good oranges. 2 oranges are drawn at random without replacement. Let x be the number of defective oranges drawn. Find: The variance of x. Using the probability distribution of x: P(x=0)= 72 102= 2145= 715 P(x=1)= 31 71 102= 2145= 715 P(x=2)= 32 70 102= 345= 115 Now compute the expectation: E(x)=0 715+1 715+2 115= 915= 35 And, E(x^2)=0^2 715+1^2 715+2^2 115= 1115 Therefore, Var(x)=E(x^2)-[E(x)]^2= 1115- ( 35)^2= 1115- 925= 5575- 2775= 2875 Therefore, the correct option is C.

Given: The drawing is without replacement, so $x$ follows a hypergeometric distribution.

Find: The variance of the number of defective oranges drawn.

The possible values of $x$ are $0,1,2$ . Count each case by combinations:

For $x=0$ , both oranges are good.
For $x=1$ , one defective and one good are selected.
For $x=2$ , both oranges are defective.

Thus,

P(x=0)=\frac{\binom{7}{2}}{\binom{10}{2}}, \qquad P(x=1)=\frac{\binom{3}{1}\binom{7}{1}}{\binom{10}{2}}, \qquad P(x=2)=\frac{\binom{3}{2}}{\binom{10}{2}}

which gives

P(x=0)=\frac{7}{15}, \qquad P(x=1)=\frac{7}{15}, \qquad P(x=2)=\frac{1}{15}

Now apply the variance formula through expectations:

E(x)=\sum xP(x)=\frac{3}{5}, \qquad E(x^2)=\sum x^2P(x)=\frac{11}{15}

Hence,

\operatorname{Var}(x)=E(x^2)-[E(x)]^2=\frac{11}{15}-\frac{9}{25}=\frac{28}{75}

So the variance is $\frac{28}{75}$ .

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