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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 15 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 15 maths textbook solution

Answers (1)

Answer:

X 0 1 2
P\left ( X \right ) \frac{144}{169} \frac{24}{169} \frac{1}{169}


Hint: use permutation formula

Given:  2 cards are drawn successively with replacement from well shuffled pack of 52 cards find the probability distribution of aces

Solution: Let solve the problem now:

Note 1 in our problem, it is given that after every draw we are replacing the card.

So our sample space will be not change

In a deck of 52 cards, there are y aces each of our suit respectively.

Let x be the random variable denoting the number of aces for an event when 2 cards are drawn successfully.

\therefore x can take values 0,1 or 2 

P(x=0)=\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}

{ as we have to select 1 card at a time such that no ace is there so the first probability is \frac{48}{52}and as the drawn card is replaced , next time again probability is \frac{48}{52} }

Similarly, we proceed for other cases

P(x=1)=\frac{4}{52} \times \frac{48}{52}=\frac{24}{169}

{we might get are in the first card or second card, so both probabilities are added}

Similarly,

P(x=1)=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}

Now we have X and P(X)

∴ now we are ready to write the probability distribution of x

The following table gives probability Distributions:

X 0 1 2
P\left ( X \right ) \frac{144}{169} \frac{24}{169} \frac{1}{169}
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