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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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