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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 5  maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 5 maths textbook solution

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Answer: The required probability distribution of x is.

X x_{1} x_{2} x_{3} x_{4}
P\left ( X \right ) \frac{15}{61} \frac{10}{61} \frac{30}{61} \frac{6}{61}


Hint: \sum P\left ( x=n \right )=1

Given:  2 P\left(X=x_{1}\right)=3 P\left(X=x_{2}\right)=P\left(X=x_{3}\right)=5 P\left(X=x_{4}\right)

Solution: 2 P\left(x_{1}\right)=3 P\left(x_{2}\right)=P\left(x_{3}\right)=5 P\left(x_{4}\right)

Let P\left(x_{3}\right)=a

\begin{aligned} &2 P\left(x_{1}\right)=P\left(x_{3}\right) \Rightarrow P\left(x_{1}\right)=\frac{a}{2} \\ &3 P\left(x_{2}\right)=P\left(x_{3}\right) \Rightarrow P\left(x_{2}\right)=\frac{a}{3} \\ &5 P\left(x_{4}\right)=P\left(x_{3}\right) \Rightarrow P\left(x_{4}\right)=\frac{a}{5} \end{aligned}

Since, P\left(x_{1}\right)+P\left(x_{2}\right)+P\left(x_{3}\right)+P\left(x_{4}\right)=1

\begin{aligned} &\Rightarrow \frac{a}{2}+\frac{a}{3}+\frac{a}{1}+\frac{a}{5}=1 \\ &\Rightarrow \frac{15 a+10 a+30 a+6 a}{30}=1 \\ &\Rightarrow 61 a=30 \\ &\Rightarrow a=\frac{30}{61} \end{aligned}

So,

X x_{1} x_{2} x_{3} x_{4}
P\left ( X \right ) \frac{15}{61} \frac{10}{61} \frac{30}{61} \frac{6}{61}

 

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