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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 6 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 6 maths textbook solution

Answers (1)

Answer: The probability distribution of x is

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

Hint: \sum P\left ( x=n\right )=1

Given:

\begin{aligned} &P(x=0)=P(x>0)=P(x<0) \\ &P(x=-3)=P(x=-2)=P(x=-1) \\ &P(x=1)=P(x=2)=P(x=3) \end{aligned}

Solution: Here \begin{aligned} &P(x=0)=P(x>0)=P(x<0) \\ \end{aligned}

Let  \begin{aligned} &P(x=0)=k \end{aligned}

\begin{aligned} \Rightarrow P\left ( x> 0 \right )=k=P\left ( x< 0 \right ) \end{aligned}

Since: \sum p\left ( x \right )=1

\begin{aligned} &\Rightarrow P(x<0)+P(x=0)+P(x<0)=1 \\ &\Rightarrow k+k+k=1 \\ &\Rightarrow 3 k=1 \\ &\Rightarrow k=\frac{1}{3} \end{aligned}

So, P(x<0)

\begin{aligned} &\Rightarrow P(x=-1)=P(x=-2)=P(x=-3)=\frac{1}{3} \\ &\Rightarrow 3 P(x=-1)=\frac{1}{3} \end{aligned}                 \begin{aligned} \quad[\text { as }, P(x=-1)=P(x=-2)=P(x=-3)] \end{aligned}

\begin{aligned} &\Rightarrow P(x=-1)=\frac{1}{9} \\ \end{aligned}

\begin{aligned} &\Rightarrow P(x=-1)=P(x=-2)=P(x=-3)=\frac{1}{9} \\ \end{aligned}                ....(i)

\begin{aligned} &\Rightarrow P(x=0)=\frac{1}{3} \end{aligned}                                                                              .....(ii)

\begin{aligned} And P(x>0)=k \end{aligned}

P(x=1)+P(x=2)+P(x=3)=\frac{1}{3}

\begin{aligned} &\Rightarrow 3 P(x=1)=\frac{1}{3} \\ \end{aligned}                                               [\text { as, } P(x=1)=P(x=2)=P(x=3)]

\begin{aligned} &\Rightarrow P(x=1)=\frac{1}{9} \\ &\Rightarrow P(x=1)=P(x=2)=P(x=3)=\frac{1}{9} \end{aligned}\\                                   .....(iii)

From equation \left ( i \right ),\left ( ii \right )\left ( iii \right )

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