Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 38

Answers (1)

Answer:

 \frac{22}{133}

Hint:

 Use Baye’s theorem.

Given:

 A lab blood test 99% effective in detecting certain when its infection is present. However, the test yield a false positive result for 0.5% of healthy person 0.01% of population actually has disease.

Solution:

Let E1 , E2  be respective event that a person has disease and person not have disease.

Since, E1 , E2  are events complimentary to each other

\begin{aligned} &P(E_1)+P(E_2)=1\\ &P(E_2)=1-0.001=0.999 \end{aligned}

Let A be event that blood test positive.

\begin{aligned} &P(E_1)=1%=\frac{0.1}{100}=0.001\\ &P\left ( \frac{A}{E_1} \right )=99%=0.99\\ &P\left ( \frac{A}{E_2} \right )=\text { result positive no disease }=0.5%=0.005\\ \end{aligned}

Probability that a person have disease given that his result positive

\begin{aligned} &P\left ( \frac{E_1}{A} \right )\ \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{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}\\ &=\frac{0.00099}{0.00099+0.004995}\\ &=\frac{0.00099}{0.005985}\\ &=\frac{990}{5989}\\ &=\frac{110}{665}\\ &=\frac{22}{133} \end{aligned}

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