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  • Explain Solution R.D. Sharma Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise Very Short Answer type Question 7 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 31 Mean and Variance of a Random Variable  Exercise Very Short Answer type  Question 7 maths Textbook Solution.

Answers (1)

Answer: - \frac{7}{10}

Hint: - You must know the rules for solving probability distribution problems.

Given: -

                                 \begin{array}{|c|c|c|c|c|} \hline X=x_{i} & 1 & 2 & 3 & 4 \\ \hline P\left(X=x_{i}\right) & k & 2 k & 3 k & 4 k \\ \hline \end{array}

Solution: - We know that the sum of probabilities in probability distribution function is always 1.

\sum_{i=1}^{n} p_{i}=1

\begin{aligned} &\therefore P(x=1)+P(x=2)+P(x=3)+P(x=4)=1 \\ \end{aligned}

\begin{aligned} &k+2 k+3 k+4 k=1 \\ \end{aligned}

\begin{aligned} &10 k=1 \\ \end{aligned}

\begin{aligned} &k=\frac{1}{10} \\ \end{aligned}

\begin{aligned} &P(x \geq 3)=P(x=3)+P(x=4) \\ \end{aligned}

\begin{aligned} &=3 k+4 k \\ \end{aligned}

\begin{aligned} &=3 \times \frac{1}{10}+4 \times \frac{1}{10} \\ \end{aligned}

\begin{aligned} &=\frac{3}{10}+\frac{4}{10} \\ &=\frac{7}{10} \end{aligned}

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