Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 32 maths

Answers (1)

Answer:

 \frac{110}{221}

Hint:

 Use Baye’s theorem.

Given:

 Probability that T.B is detected when a person is actually suffering is 0.99. The probability that doctor diagnose incorrectly that person has T.B on basis of x-ray is 0.001. In certain city 1 in 1000 person suffer from T.B.

Solution:

Let

E1= Event that person has T.B

E2=Event that person does not have T.B

E=Event that person is diagnose to have T.B

\begin{aligned} &P(E_1)=\frac{1}{1000}=0.001\\ &P(E_2)=\frac{999}{1000}=0.999\\ &P\left ( \frac{E}{E_1} \right )=0.99\\ &P\left ( \frac{E}{E_2} \right )=0.001\\ \end{aligned}

And

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_1}{E} \right )=\frac{P(E_1).P\left ( \frac{E}{E_1} \right )}{P(E_1)\times P\left ( \frac{E}{E_1} \right )+P(E_2)\times P\left ( \frac{E}{E_2} \right )}\\ &=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.001} \\ &=\frac{990}{990+999}\\ &=\frac{110}{110+111}\\ &=\frac{110}{221} \end{aligned}

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