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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 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 $X$ denotes 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 $P(X=0)=P($ no card is king $)$
$ =\frac{{ }^{48} C _2}{{ }^{52} C _2}$
$=\frac{48 \times 47}{52 \times 51}$
$=\frac{188}{221} $
$P(X=1)=P($ exactly one card is king $)$
$ =\frac{{ }^4 C _1 \times{ }^{48} C _1}{{ }^{52} C _2}$
$=\frac{4 \times 48 \times 2}{52 \times 51}$
$=\frac{32}{221} $
$P(X=2)=P($ both cards are king $)$
$=\frac{{ }^4 C _2}{{ }^{52} C _2}$
$=\frac{4 \times 3}{52 \times 51}$
$=\frac{1}{221}$
$E ( X )=\sum_{ i =1}^{ n } x_{ i } p _{ i }$
$=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}$
$=\frac{34}{221}$
$\operatorname{Var}( X )=\left(\sum_{ i =1}^{ n } x_{ i }^2 p _{ i }\right)-\left(\sum_{ i =1}^{ n } x_{ i } p _{ i }\right)^2$
$=\left(0^2 \times \frac{188}{221}+1^2 \times \frac{32}{221}+2^2 \times \frac{1}{221}\right)-\left(\frac{34}{221}\right)^2$
$=\frac{36}{221}-\frac{1156}{48841}$ $=\frac{6800}{48841}$
$=0.1392$
$\sigma=\sqrt{\operatorname{Var}( X )}$
$=\sqrt{0.1392}$
$=0.3730$

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