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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 8 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 8 maths textbook solution

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Answer: Required probability distribution is

X 0 1 2 3
P\left ( X \right ) \frac{1}{8} \frac{3}{8} \frac{3}{8} \frac{1}{8}

 

Hint: Probability of getting a head in one throws of a coin=\frac{1}{2}

\begin{aligned} &P(H)=\frac{1}{2} \\ &P(T)=1-\frac{1}{2} \\ &P(T)=\frac{1}{2} \end{aligned}

Given: Probability distribution of the number of heads, when three coins are tossed.

Solution: Let x denotes the numbers of heads obtained is 3 throws of a coin than x=0,1,2,3

\begin{aligned} &P(x=0)=P(T) P(T) P(T)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8} \\ &P(x=1)=P(H) P(T) P(T)+P(T) P(H) P(T)+P(T) P(T) P(H) \\ &P(x=1)=\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right) \\ &P(x=1)=\frac{3}{8} \\ &P(x=2)=P(H) P(H) P(T)+P(H) P(T) P(H)+P(T) P(H) P(H) \\ &P(x=2)=\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right) \end{aligned}

\begin{aligned} &P(x=2)=\frac{3}{8} \\ &P(x=3)=P(H) P(H) P(H) \\ &P(x=3)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \\ &P(x=3)=\frac{1}{8} \end{aligned}

So, the required probability distribution is

X 0 1 2 3
P\left ( X \right ) \frac{1}{8} \frac{3}{8} \frac{3}{8} \frac{1}{8}

 

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