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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 19 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 19 maths textbook solution

Answers (1)

Answer:

X 0 1 2
P\left ( X \right ) \frac{64}{81} \frac{16}{81} \frac{1}{81}


Given: find the probability of y in two throws of two dice, where y represents the number of times a total of 9 appears when two fair dice are drawn.

Solution: total 36 possible outcome

In our question, we are throwing two dice 2 times

∴y Denotes the number of times a sum of two numbers appearing on dice is equal to 9

∴y Can take values 0,1 or 2

Means in both the throw sum of 9 is not obtained and both the throw of two die a sum of 9 is obtained

∴(3,6) (6,3) (5,4) and (4,5) are the combinations resulting in sum = 9

Let A denotes the event of getting a sum of 9 a throw of 2 dice

\therefore P\left ( A \right )=probability of getting the sum of 9 in a throw of 2 dice is

\frac{4}{36}=\frac{1}{9}

And probability of not getting the sum of 9 in the throw of 2 dice =\frac{32}{36}=\frac{8}{9}

Note: P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right ) where A and B are independent events

P\left ( Y=0 \right )=P\left ( {A}' \right )P\left ( {A}' \right )

\begin{aligned} &=\frac{8}{9} \cdot \frac{8}{9}=\frac{64}{81} \\ &P(Y=1)=P(A) P\left(A^{\prime}\right)+P\left(A^{\prime}\right) P(A) \\ &=\frac{8}{9} \times \frac{1}{9}+\frac{1}{9} \times \frac{8}{9}=\frac{16}{81} \\ &P(Y=2)=P(A) P(A) \\ &=\frac{1}{9} \times \frac{1}{9}=\frac{1}{81} \end{aligned}

Now we have X and P(X)

∴ the required  probability distribution of x

The following table gives probability

Distributions:

X 0 1 2
P\left ( X \right ) \frac{64}{81} \frac{16}{81} \frac{1}{81}

 

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