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

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}

Posted by

Gurleen Kaur

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