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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions3 Marks
Question
Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings drawn.
✓

Answer

Let $X$ denote the number of kings in a draw of two cards. $X$ is a random variable which can assume the values 0,1 or 2 .
Then
$
P(X=0)=P(\text { no card is king })=\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{188}{221}
$
Then $\quad P(X=1)=P($ exactly one card is king $)=\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4 \times 48 \times 27}{52 \times 51}=\frac{32}{221}$
Then $\quad P(X=2)=P($ both cards are king $)=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}$
$
\begin{gathered}
\mu=\mathrm{E}(X)=\sum_{i=1}^n x_i p_i=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}=\frac{34}{221} \\
\begin{aligned}
& \operatorname{Var}(X)=\left(\sum_{i=1}^n x_i^2 p_i\right)-\left(\sum_{i=1}^n x_i p_i\right)^2=\left(0^2 \times \frac{188}{221}+1^2 \times \frac{32}{221}+2^2 \times \frac{1}{221}\right)-\left(\frac{34}{221}\right)^2 \\
&=\frac{36}{221}-\frac{1156}{48841}=\frac{6800}{48841}=0 \cdot 1392 \\
& \sigma=\sqrt{\operatorname{Var}(X)}=\sqrt{0 \cdot 1392}
\end{aligned}
\end{gathered}
$

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