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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 29 maths textbook solution

Answers (1)

Answer: (i) K=\frac{8}{15}     (i i) P(x \leq 2)=\frac{14}{15}, P(x>2)=\frac{1}{15}=1   (iii) P(x \leq 2)+P(x>2)=1        

Hint: Use probability formula

Given: the probability distribution of a random variable x is given below:

    

X 0 1 2 3
P\left ( X \right ) K \frac{k}{2} \frac{k}{4} \frac{k}{8}


i }Determine the value of k

ii} determine P\left ( x\leq 2 \right ) and P\left ( x> 2 \right )

ii} Find P\left ( x\leq 2 \right )+P\left ( x> 2 \right )

Solution: the key point to solve the problem is, if a probability distribution is given the as per definition sum of probability associated with each value of a random variable of given distribution is equal to 1

i.e.  \varepsilon \left ( p_{1} \right )=1

Given distribution is:

X 0 1 2 3
P\left ( X \right ) K \frac{k}{2} \frac{k}{4} \frac{k}{8}

\begin{aligned} &\therefore k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8}=1 \\ &\Rightarrow 8 k+4 k+2 k+k=8 \\ &\Rightarrow k=\frac{8}{15}(\text { ans }) \end{aligned}

ii \}P(x=2)=\frac{k}{4}$ and $P(x=3)=\frac{k}{8}

\begin{aligned} &\therefore P(x>2)=P(x=3)=\frac{k}{8}=\frac{\frac{8}{15}}{8}=\frac{1}{15} \\ &P(x \leq 2)=P(x=0)+P(x=1)+P(x=2) \\ &\Rightarrow k+\frac{k}{2}+\frac{k}{4} \\ &\Rightarrow \frac{7 k}{4}=\frac{7}{4} \times \frac{8}{15}=\frac{14}{15} \end{aligned}

iii \} \quad \therefore P(x \leq 2)+P(x>2)=\frac{14}{15}+\frac{1}{15}=1

 

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