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Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 7 maths textbook solution

Answers (1)

Answer: The probability distribution of the number of aces is

X 0 1 2
P\left ( X \right ) \frac{188}{221} \frac{32}{221} \frac{1}{221}


Hint: Let x denotes numbers of aces in a sample of 2 cards drawn, there are four aces in a pack of 52 cards so, x can have values 0,1,2.

Given: Two cards are drawn from a well shuffled pack of 52 cards.

Solution:

Now,

\begin{aligned} &P(x=0)=\frac{48_{C_{2}}}{52_{C_{2}}}=\frac{48 \times 47}{2}=\frac{2}{52 \times 51}=\frac{188}{221} \\ &P(x=1)=\frac{48 c_{1} \times 4 c_{1}}{52 c_{2}}=\frac{48 \times 4 \times 2}{52 \times 51}=\frac{32}{221} \\ &P(x=2)=\frac{4 c_{2}}{52 c_{2}}=\frac{4 \times 3}{2}=\frac{2}{52 \times 51}=\frac{1}{221} \end{aligned}

So,

X 0 1 2
P\left ( X \right ) \frac{188}{221} \frac{32}{221} \frac{1}{221}
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