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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 21 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 21 maths textbook solution

Answers (1)

Answer:

X 0 1 2 3
P\left ( X \right ) \frac{27}{64} \frac{27}{64} \frac{9}{64} \frac{1}{64}


Hint: use probability formula

Given: Three cards are drawn successively with replacement from a well shuffled deck of 52 cards.

 A random variable x denotes the number of hearts in the three cards drawn. Determine the probability distribution of x

Solution: let us solve the problem now:

Note: in our problem, it is given that after every draw, we are replacing the card, so our sample spaces will no change.

In a desk of 52 cards, there are 13 hearts let x be the random variable denoting the number of hearts drawn for an event when 3 cards are drawn successively with replacement

∴x Can take value 0,1,2,3

P(x=0)=\frac{39}{52} \times \frac{39}{52} \times \frac{39}{52}=\frac{27}{64}

{As we have to select 1 card at a time such that no heart is there so the first probability is \frac{39}{52} and as the drawn card is replaced time again probability is \frac{39}{52} and again the same thing}

Similarly we proceed for other cases.

P(x=1)=\frac{13}{52} \times \frac{39}{52} \times \frac{39}{52}+\frac{39}{52} \times \frac{13}{52} \times \frac{39}{52}+\frac{39}{52} \times \frac{39}{52} \times \frac{13}{52}=\frac{27}{64}

 {We figure might get a heart is the first card or second card or third so probabilities in all 3 cases are added as they are mutually exclusive events}

Similarly

P(x=2)=\frac{13}{52} \times \frac{13}{52} \times \frac{39}{52}+\frac{39}{52} \times \frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{39}{52} \times \frac{13}{52}=\frac{9}{64}

{first two cards are heart and 3rd is non-heart  last 2 are hearts and so on cases as these  cases are mutually exclusive hence they are added }

Similarly

P(x=3)=\frac{13}{52} \times \frac{13}{52} \times \frac{13}{52}=\frac{1}{64}

Now we have  X and P(X)             

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

The following table gives probability distribution

X 0 1 2 3
P\left ( X \right ) \frac{27}{64} \frac{27}{64} \frac{9}{64} \frac{1}{64}

 

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