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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 16 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 16 maths textbook solution

Answers (1)

Answer: 

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


Given:  two cards are drawn successively with replacement from a well shuffled pack of 52 cards, find the probability distribution of the number of kingsHint: use permutation formula

Solution: Lets us solve the problem now:

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

In a deck 52 cards, there are 4 kings each of our suit respectively

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

∴ X can take value 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 king is there so that first probability is \frac{48}{52} and as the drawn card is replaced, used 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 a king in the first card or second card, so both probabilities are added}

 Similarly

P(x=2)=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}

Now we have X and P(X)

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

The following table gives probability

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

 

 

 

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