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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
Question
An urn contains 5 red and 2 blcak balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X.
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Answer

X can assume values 0 to 2. Yes X is a random variable. P(X = 0) = (Probability of getting no black ball) $=\frac{\text{}^{2}\text{C}_0\times\text{}^{5}\text{C}_2}{\text{}^{7}\text{C}_2}=\frac{1\times\frac{5\times4}{2\times1}}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 1) = (Probability of getting one black ball) $=\frac{\text{}^{2}\text{C}_1\times\text{}^{5}\text{C}_1}{\text{}^{7}\text{C}_2}=\frac{1\times5}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 2) = (Probability of getting two black balls) $=\frac{\text{}^{2}\text{C}_2\times\text{}^{5}\text{C}_0}{\text{}^{7}\text{C}_2}=\frac{1\times1}{\frac{7\times6}{2\times1}}=\frac{2}{42}$ Thus, Probability distribution of random variable X is
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X}) $
$\frac{20}{42}$
$\frac{20}{42}$
$\frac{2}{42}$
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{X}_\text{i}^2$
$0$ $\frac{20}{42}$ $0$ $0$
$1$ $\frac{20}{42}$ $\frac{20}{42}$ $\frac{20}{42}$
$2$ $\frac{2}{42}$ $\frac{4}{42}$ $\frac{8}{42}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{2}{3}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{2}{3}-\Big(\frac{4}{7}\Big)^2=\frac{50}{147}$

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