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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards.
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Answer

Number of red cards = 26Let X be a random variable which can take values
0, 1, 2 where X is the no. of red cards selected
$\therefore$   X = 0 means 0 red card
P(X = 0) = $\frac{26_{\text{c}_{2}}}{\text{52}_{\text{c}_{2}}}=\frac{26\times25}{52\times51}=\frac{25}{102}$
P(X = 1) = $\frac{26_{\text{c}_{1}}\times26_{\text{c}_{1}}}{\text{52}_{\text{c}_{2}}}=\frac{26\times26\times2}{52\times51}=\frac{52}{102}$
P(X = 2) = $\frac{26_{\text{c}_{2}}}{\text{52}_{\text{c}_{2}}}=\frac{26\times25}{52\times51}=\frac{52}{102}$
Probability distribution of random variable X is
X 0 1 2
P(X) 25/102 52/102 25/102
Mean = $\Sigma$ x P(X) = $\frac{52+50}{102}=1$
variance = $\Sigma\text{x}^{2}\text{P(X)-(}\Sigma\text{xP(X))}^{2}=\frac{152}{102}-1=\frac{50}{102}\text{ OR }\frac{25}{51}$.

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