Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 10

Answers (1)

Answer:

 0.1

Hint:

 Use Baye’s theorem.

Given:

 A factor has three machines X, Y, Z producing 1000, 2000, 3000, bolts per day respectively. The machine X produce 1% defective bolts, Y produce 1.5% defective and Z produce 2% defective bolt.

Solution:

Let E1,E2,E3 denote the event that machine X produce bolts, machine Y produce bolts and 2 produce bolt.

Let A be event that bolt is defective.

Total bolt = 1000 + 2000 + 3000 = 6000

\begin{aligned} &P(E_1)=\frac{1000}{6000}=\frac{1}{6}\\ &P(E_2)=\frac{2000}{6000}=\frac{1}{3}\\ &P(E_3)=\frac{3000}{6000}=\frac{1}{2}\\ &P\left ( \frac{A}{E_1} \right )=1%=\frac{1}{100}\\ &P\left ( \frac{A}{E_2} \right )=1.5%=\frac{15}{1000}\\ &P\left ( \frac{A}{E_3} \right )=2%=\frac{2}{100}\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{6}\times \frac{1}{100} }}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{15}{1000}+\frac{1}{2}\times \frac{2}{100}}\\ &=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{2}+1}\\ &=\frac{1}{6}\times \frac{1+3+6}{6}\\ &=\frac{1}{10} \end{aligned}

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