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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 22 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 22 maths textbook solution

Answers (1)

Answer:

X 0 1 2 3
P\left ( X \right ) \frac{64}{343} \frac{144}{343} \frac{108}{343} \frac{27}{343}


Hint: use probability formula

Given: An contains 4 red and 3 blue balls, find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement

Solution: Let X denote the number of blue balls drawn in a random draw of 3 balls.

∴X can take value 0,1,2,3

Since bag contains 4 red and 3 blue balls i.e. at a total of 7 balls

∴ Balls are drawn with replacement for selecting 0 blue balls, we will select all 3 red balls.

=\frac{4}{7} \times \frac{4}{7} \times \frac{4}{7}=\frac{64}{343}

For selecting 1 blue ball, we will select all 2 red balls and 1 blue

\therefore P(x=1)=\frac{3}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{4}{7} \times \frac{3}{7}=\frac{144}{343} 

For selecting 2 blue balls, we will select all 1 red balls and 2 blue.

\therefore P(x=2)=\frac{3}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{3}{7} \times \frac{4}{7} \times \frac{3}{7}=\frac{108}{343} 

For selecting 3 blue balls, we will select all 3 blue and 0 red balls

\therefore P(x=3)=P(\text { selecting all blue balls })=\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}=\frac{27}{343}

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{64}{343} \frac{144}{343} \frac{108}{343} \frac{27}{343}
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