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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 17 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 17 mathsA textbook solution

Answers (1)

Answer:

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


Given: two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of acesHint:  use probability formula

Solution: while reading this question you might have observed that cards are being drawn successively and sometimes in other question you might have get cards are drawn simultaneously.

You have to be careful regarding these two words, while solving the question. Both have a different meaning

Let’s solve the problem now

Note: in our problem, it is given that after every draw we are not replacing the card so our sample space will change in the case

In a deck of 52 cards, there are 4 aces each of one suit respectively

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

∴ x can take value 0,1 or 2

P(x=0)=\frac{48}{51} \times \frac{47}{51}=\frac{188}{221}

{As we have to select 1 card at a time such that no ace is there so that first probability is \frac{48}{52} and as the drawn card is not  replaced, used time again probability is \frac{47}{51}because now our card is our sample space and that card is not from group of ace, so now are card are 47}

Similarly we proceed for other cases

P(x=1)=\frac{4}{52} \times \frac{48}{51}+\frac{48}{52} \times \frac{4}{51}=\frac{32}{221}

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

Similarly,

P(x=2)=\frac{4}{52} \times \frac{3}{51}=\frac{1}{221}      

Now we have X and P(X)

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

The following table gives probability

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

 

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