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

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}

Posted by

Gurleen Kaur

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