A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

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Q.
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

1 $\frac{3}{5}$

2 $\frac{2}{15}$

3 $\frac{28}{75}$

4 $\frac{11}{15}$

Ans.
(3)

X X=0 X=1 x=2

P(x) $\frac{{}^{7}C_{2} \times {}^{3}C_{0}}{{}^{10}C_{2}} = \frac{7}{15}$ $\frac{{}^{7}C_{1} \times {}^{3}C_{1}}{{}^{10}C_{2}} = \frac{7}{15}$ $\frac{{}^{7}C_{0} \times {}^{3}C_{2}}{{}^{10}C_{2}} = \frac{1}{15}$

Mean, $E (X) = \sum_{}^{} X_{i} P (X_{i}) = 0 + \frac{7}{15} + \frac{2}{15} = \frac{3}{5}$

Now, variance $(X) = E (X^{2}) - (E {(X))}^{2} = \frac{28}{75}$

$= (0 + \frac{7}{15} + \frac{4}{15}) - {(\frac{3}{5})}^{2} = \frac{11}{15} - \frac{9}{25} = \frac{28}{75}$

Solution

Q.
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

1 $\frac{3}{5}$

2 $\frac{2}{15}$

3 $\frac{28}{75}$

4 $\frac{11}{15}$

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X	X=0	X=1	x=2
P(x)	$\frac{{}^{7}C_{2} \times {}^{3}C_{0}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{1} \times {}^{3}C_{1}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{0} \times {}^{3}C_{2}}{{}^{10}C_{2}} = \frac{1}{15}$

Solution

Q. A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

1 35

2 215

3 2875

4 1115

Ans. (3) X X=0 X=1 x=2 P(x) C27×C03C210=715 C17×C13C210=715 C07×C23C210=115 Mean, E(X)=∑XiP(Xi)=0+715+215=35 Now, variance (X)=E(X2)–(E(X))2=2875 =(0+715+415)–(35)2=1115–925=2875

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Q.
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

1 $\frac{3}{5}$

2 $\frac{2}{15}$

3 $\frac{28}{75}$

4 $\frac{11}{15}$