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Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 20 maths

Answers (1)

Answer:

 \frac{2}{23}

Hint:

 Use Baye’s theorem.

Given:

 A factory has three machine A, B, C which produce 100, 200, 300 item of particular type. The machine produce 2%, 3% and 5% defective item respectively.

Solution:

Let E1,E2,E3  machine A, B and C.

D = defective item

Total production = 100 + 200 + 300 = 600

\begin{aligned} &P(E_1)=\frac{100}{600}=\frac{1}{6}\\ &P(E_2)=\frac{200}{600}=\frac{1}{3}\\ &P(E_3)=\frac{300}{600}\\ &P\left ( \frac{D}{E_1} \right )=0.02\\ &P\left ( \frac{D}{E_2} \right )=0.03\\ &P\left ( \frac{D}{E_3} \right )=0.05\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_1}{D} \right )=\frac{P(E_1).P\left ( \frac{D}{E_1} \right )}{P(E_1)\times P\left ( \frac{D}{E_1} \right )+P(E_2)\times P\left ( \frac{D}{E_2} \right )+P(E_3)\times P\left ( \frac{D}{E_3} \right )}\\ &=\frac{{\frac{1}{6}\times 0.02 }}{\frac{1}{6}\times 0.02+\frac{1}{3}\times 0.03+\frac{1}{2}\times 0.05}\\ &=\frac{2}{23} \end{aligned}

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Gurleen Kaur

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