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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 9 maths textbook solution

Answers (1)

Answer: Required probability distribution is 

X 0 1 2 3 4
P\left ( X \right ) \frac{48c_{4}}{52c_{4}} \frac{48c_{3}\times 4c_{1}}{52c_{4}} \frac{48c_{2}\times 4c_{2}}{52c_{4}} \frac{48c_{1}\times 4c_{3}}{52c_{4}} \frac{ 4c_{4}}{52c_{4}}

Hint: Let x denotes numbers of aces drawn out of 4 cards drawn, there are four aces in a pack of 52.

So,x: 0,1,2,3,4

Given:  Cards are drawn simultaneously from a well shuffled pack of 52 playing cards.

Solution:

\begin{aligned} &P(x=0)=\frac{48 c_{4}}{52 c_{4}} \\ &P(x=1)=\frac{48 c_{3} \times 4 c_{1}}{52 c_{4}} \\ &P(x=2)=\frac{48 c_{2} \times 4 c_{2}}{52 c_{4}} \\ &P(x=3)=\frac{48 c_{1} \times 4 c_{3}}{52 c_{4}} \\ &P(x=4)=\frac{4 c_{4}}{52 c_{4}} \end{aligned}

So, required probability distribution is 

X 0 1 2 3 4
P\left ( X \right ) \frac{48c_{4}}{52c_{4}} \frac{48c_{3}\times 4c_{1}}{52c_{4}} \frac{48c_{2}\times 4c_{2}}{52c_{4}} \frac{48c_{1}\times 4c_{3}}{52c_{4}} \frac{ 4c_{4}}{52c_{4}}
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