retagged by
889 views
0 votes
0 votes
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?
retagged by

3 Answers

2 votes
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.

1 votes
1 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 $98/695$
0 votes
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=$\frac{95}{20000}$

Related questions

2 votes
2 votes
1 answer
3
admin asked Oct 21, 2023
2,498 views
A fair coin is flipped twice and it is known that at least one tail is observed. The probability of getting two tails is $\frac{1}{2}$ $\frac{1}{3}$ $\frac{2}{3}$ $\frac{...
0 votes
0 votes
0 answers
4
Debargha Mitra Roy asked Sep 26, 2023
175 views
Determine the geometric distribution for which the mean is 3 and variance is 4.