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  • Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 16 maths

Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 16 maths

Answers (1)

Answer:

 0, 2

Hint:

 Use Baye’s theorem.

Given:

 In factory machine produce 30% total item B produce 25% and C produce remaining output. If defective item produce by machine A, B, C is 1%. 12% and 2% respectively. Three machines working together produce 10,000 item in day. An item is drawn at random from day output and found defective. Find probability that was produce by machine A.

Solution:

Let E1, E2,E3  event as define

E1 =item is produce by machine A

E2= item is produce by machine B

E3 = item is produce by machine C

A=total event that item is defective

\begin{aligned} &P(E_1)=\frac{30}{100}\\ &P(E_2)=\frac{25}{100}\\ &P(E_3)=\frac{45}{100}\\ &P\left ( \frac{A}{E_1} \right )=\frac{1}{100}\\ &P\left ( \frac{A}{E_2} \right )=\frac{1.2}{100}\\ &P\left ( \frac{A}{E_3} \right )=\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{25}{100}\times \frac{1.2}{100} }}{\frac{30}{100}\times \frac{1}{100}+\frac{25}{100}\times \frac{1.2}{100}+\frac{45}{100}\times \frac{2}{100}}\\ &=\frac{30}{30+30+90}\\ &=\frac{1}{5}\\ &=0.2 \end{aligned}

Posted by

Gurleen Kaur

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