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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 39

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 39

Answers (1)

Answer:

 \frac{200}{231}

Hint:

 Use Baye’s theorem.

Given:

There are three categories of student in class 60 student.

A very hardworking, B = regular but not hardworking, C = careless and irregular, 10 student are category A, 30 in B, rest in C. if found that probability of student of category A unable to get good marks in final year, probability of student in exam is 0.002 of category B is 0.02 and category C is 0.20.

Solution:

Let

E1= Student select category A

E2= Student select category B

E3= Student select category C

S=Student not get good marks

\begin{aligned} &P(E_1)=\frac{1}{6}\\ &P(E_2)=\frac{3}{6}\\ &P(E_3)=\frac{2}{6}\\ &P\left ( \frac{S}{E_1} \right )=0.002\\ &P\left ( \frac{S}{E_2} \right )=0.002\\ &P\left ( \frac{S}{E_3} \right )=0.2\\ \end{aligned}

Using Baye’s theorem we get

Required probability

\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{S}{E_3} \right )}{P(E_1)\times P\left ( \frac{S}{E_1} \right )+P(E_2)\times P\left ( \frac{S}{E_2} \right )+P(E_3)\times P\left ( \frac{S}{E_3} \right )}\\ &=\frac{{\frac{2}{6}\times 0.2 }}{\frac{1}{6}\times 0.002+\frac{3}{6}\times 0.02+\frac{2}{6}\times 0.2}\\ &=\frac{200}{231} \end{aligned}

Posted by

Gurleen Kaur

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