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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 7

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 7

Answers (1)

Answer:

 0.016

Hint:

 To solve this we use total probability formula.

Given:

A=production 60%, 2% defective
B=production 40%, 1% defective

Solution:

\begin{aligned} &A=60=\frac{6}{10}\times \frac{2}{100} \text { (that item is defective and coming from machine A )}\\ &B=40=\frac{4}{10}\times \frac{1}{100} \text { (that item is defective and coming from machine B )}\\ \end{aligned}

Using total probabilty theorem

\begin{aligned} & \text { Required probability }=\frac{6}{10}\times \frac{2}{100}+\frac{4}{10}\times \frac{1}{100}\\ &=\frac{12+4}{1000}=\frac{16}{1000}=0.016 \end{aligned}

Posted by

Gurleen Kaur

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