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  • Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31 point 1 question 3 sub question 2 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 3 sub question 2 maths textbook solution

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Answer:  P(x<3)=\left(\frac{1}{9}\right), P(x \geq 3)=\frac{8}{9}, P(0<x<5)=\frac{8}{27}

Hint: \sum P\left ( x=n \right )=1

Given:

X 0 1 2 3 4 5 6 7 8
P\left ( X \right ) A 3a 5a 7a 9a 11a 13a 15a 17a

 

Solution:

 P(x<3)=P(0)+P(1)+P(2)

P(x<3)=a+3 a+5 a

P(x<3)=9 a

P(x<3)=9\left(\frac{1}{81}\right)                      [ using the previous solution ]

\therefore P(x<3)=\left(\frac{1}{9}\right)

\therefore P(x \geq 3)=1-P(x<3)=1-\frac{1}{9}=\frac{8}{9}

P(0<x<5)=P(1)+P(2)+P(3)+P(4)

P(0<x<5)=3 a+5 a+7 a+9 a

P(0<x<5)=24 a

P(0<x<5)=24\left(\frac{1}{81}\right)

\therefore P(0<x<5)=\frac{8}{27}

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