Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 9 maths textbook solution

Answers (1)

Answer:

 \frac{3}{7}

Hint:

Use Baye’s theorem. 

Given:

 In class 5% of boy and 10% of girl have 10 of more than 150. In this class 60% of students boys.

If a student is selected at random and found to have 10 of more than 150. Find probability of student is boy.

Solution:

Let E1 is the student chose boy and E2 is the student chose is girl.

Thus

\begin{aligned} &P(E_1)=\frac{60}{100}\\ &P(E_2)=\frac{40}{100}\\ \end{aligned}

Event E1 and E2 are mutually exclusive event A. a student have 10 more than 150.

\begin{aligned} &P\left ( \frac{A}{E_1} \right )=P(\text { a boy student has 10 more than 150 })=\frac{5}{100}\\ &P\left ( \frac{A}{E_2} \right )=P(\text { a girl student has 10 more than 150 })=\frac{10}{100}\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{\frac{60}{100}\times \frac{5}{100}}{\frac{60}{100}\times \frac{5}{100}+\frac{4}{100}\times \frac{10}{100}}\\ \end{aligned}

Multiplying every term by 10000

\begin{aligned} &=\frac{60\times 5}{60\times 5+40\times 10}\\ &=\frac{300}{700}\\ &=\frac{3}{7} \end{aligned}

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