Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 33 maths textbook solution

Answers (1)

Answer:

 \frac{90}{589}

Hint:

 Use Baye’s theorem.

Given:

 The test will correctly detect the disease 1% of time probability large population of which an estimate 0.2% has disease a person.

Solution:

Let A,E1 and E2  denote the event that the person suffer from the disease, the test detecte the disease correctly and the test does not detect the disease correctly, respectively.

\begin{aligned} &P(E_1)=0.002\\ &P(E_2)=0.998\\ \end{aligned}

Now

\begin{aligned} &P\left ( \frac{A}{E_1} \right )=0.90\\ &P\left ( \frac{A}{E_2} \right )=0.01\\ \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{90}{100}\times \frac{2}{1000} }}{\frac{90}{100}\times \frac{2}{1000}+\frac{1}{100}\times \frac{998}{1000}}\\ &=\frac{180}{180+998}\\ &=\frac{180}{1178}\\ &=\frac{90}{589} \end{aligned}

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