Question

By examining the chest x ray, the probability that tb is detected when a person is actually suffering is 0.99. the probability of an healthy person diagnosed to have tb is 0.001. in a certain city, 1 in 1000 people suffers from tb. a person is selected at random and is diagnosed to have tb. what is the probability that he actually has tb

Answer 1

this is Ur required result

Answer 2

Answer:

Probability that he actually has tb is 0.49

Step-by-step explanation:

Given : By examining the chest x ray, the probability that tb is detected when a person is actually suffering is 0.99. the probability of an healthy person diagnosed to have tb is 0.001. in a certain city, 1 in 1000 people suffers from tb.

To find : What is the probability that he actually has tb?

Solution :

Let $E_1$ be the event that the person suffer from tb.

$P(E_1)=\frac{1}{1000}=0.001$

Let $E_2$ be the event that the person not suffer from tb.

$P(E_2)=\frac{999}{1000}=0.999$

Let A be the event doctor diagnose correctly.

Then, $P(A/E_1)=0.99$

$P(A/E_2)=0.001$

Now, Required probability is given by,

$P(E_1/A)=\frac{P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}$

$P(E_1/A)=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.001}$

$P(E_1/A)=\frac{0.00099}{0.001989}$

$P(E_1/A)=0.4977$

Therefore, Probability that he actually has tb is 0.49