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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise 31.2 Question 15 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise 31.2  Question 15  Maths Textbook Solution.

Answers (1)

Answer:

\begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline p(x) & \frac{28}{143} & \frac{70}{143} & \frac{40}{143} & \frac{5}{143} \\ \hline \end{array}

Hint:

            Let x is denote the number of defective bulb in a sample of 3 bulb drawn from a bag drawn out to be defective

Given:

            A box contains 13 bulb, out of which 5 are defective, 3 bulb are randomly drawn, one by one without replacement from the box

Solution:

Now as x can take the values: 0, 1, 2 and 3

P(x=0) = P (No defective bulb)

\begin{aligned} &=\frac{5 C_{0} \times 8 C_{3}}{13 C_{3}} \\ &=\frac{8 \times 7 \times 6}{13 \times 12 \times 11} \\ &=\frac{28}{143} \end{aligned}

P(x=1) = P (One defective bulb)

\begin{aligned} &=\frac{5 C_{1} \times 8 C_{2}}{13 C_{3}} \\ &=\frac{5 \times \frac{8 \times 7}{2 \times 1}}{\frac{13 \times 12 \times 11}{3 \times 2 \times 1}}=\frac{70}{143} \end{aligned}

P(x=2) = P (Two defective bulb)

\begin{aligned} &=\frac{5 C_{2} \times 8 C_{1}}{13 C_{3}} \\ &=\frac{\frac{5 \times 4}{2 \times 1} \times 8}{\frac{13 \times 12 \times 11}{3 \times 2 \times 1}}=\frac{40}{143} \end{aligned}

P(x=3) = P (Three defective bulb)

\begin{aligned} &\begin{aligned} &=\frac{5 C_{3} \times 8 C_{0}}{13 C_{3}} \\ &=\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{143} \end{aligned}\\ \end{aligned}

\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline p(x) & \frac{28}{143} & \frac{70}{143} & \frac{40}{143} & \frac{5}{143} \\ \hline \end{array} \end{aligned}

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