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One in two hundred people in a population have a particular disease. A test is developed for the disease. The test gives a false positive $3\%$ of the time (i.e. reports that a person has the disease even when he/she does not), and a false negative $2\%$ of the time (i.e. reports that a person does not have the disease even when he/she does). Rahul takes the test and the report comes positive. What is the probability that Rahul has the disease?
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3 Answers

+2 votes

Consider the following events

P: The report comes positive that is the report tells that a person has a disease.

D: The person has the disease.

According to the given conditions in the problem we have P ( P | D' ) = 3 / 100 [ D' is the complement of D that is the person don't have any disease]

and P ( P' | D ) = 2/100  [P' is the complement of P that is the result is negative]

Here we have to find P (D | P) = P(D ∩ P) / P (P) = P (P ∩ D) / P (P) = ( P (P|D) * P(D) )  / P(P) ..................(1)

Now, it is given that P (P' | D) = 2/100

=> 1 - P (P|D) = 2/100

=> P (P|D) = 1- 2/100 = 98/100 ..................(2)

Again the probability that a person has a disease is 1/200 that is P(D) = 1/200 ..............(3)

Hence P(D') = 199/200
Now, P (P) = P (D) * P(P|D) + P (D') * P(P|D')

Hence P (P) = 1/200 * 98/100 + 199/200 * 3/100 = 695/ ( 2* 104 ) .....................(4)

Now, putting the values from 2,3 and 4 in 1 we get

So, P(D|P) = (98/100 * 1/200 ) / ( 695 / ( 2* 104 )  = 98/695.

Thus the probability Rahul has the disease is 98/695.

answered by Loyal (9k points)
edited
0
But P(D) =1/200
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Is The answer 1/200 ?
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I don't know the correct answer
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I am not getting what do you mean by P(D) = 1/200
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It is given that "One in two hundred people in a population have a particular disease" . So prob of a person having disease is P(D)=1/200...
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Yes you are right P(D) should be 1/200. Edited my answer check it now.
0 votes
Report is true in $\frac{95}{100}$ cases

Test result should be +ve in $\frac{1}{200}$ cases

Test result will be false for $\frac{5}{100}$ cases

Now it is given reports that a person has the disease even when he/she does not in 3% cases

and it is also given reports that a person does not have the disease even when he/she does in 2% cases

So, Rahul takes the test and the report comes positive

Now it could be a true report and test result is positive =$\frac{95}{100}\times \frac{1}{200}=\frac{95}{20000}$

otherwise it could be false report and result is positive=$\frac{3}{100}$

So, probability that rahul has the desease=$\frac{95}{20000}$
answered by Veteran (110k points)
0 votes
Such questions must be dealt with straight forward $Bayes Theorem$.

Let $D$ be the event that person has the disease. Let $P+$ be the event that the test comes positive.

We are required to calculate $P(D|P+)$. Using the theorem we arrive at:

$P(D|P+)$ = $\frac{P(P+|D)*P(D)}{P(P+|D)*P(D) + P(P+|D')*P(D')}$

Now the values for various probabilities are:

$P(D) = 1/200$ , $P(D') = 199/200$

$P(P+|D) = 100\% - 2\% = 98\%$

$P(P+|D') = 3\%$

So put the values , the answer is $98/695$
answered by Junior (879 points)

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