Question

A new medical test has been designed to detect the presence of the mysterious Brainlesserian disease. Among those who have the disease, the probability that the disease will be detected by the new test is 0.82.

However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.06. It is estimated that 13 % of the population who take this test have the disease.

If the test administered to an individual is positive, what is the probability that the person actually has the disease?

Answer 1

Solution:

Given:Among those who have the disease, the probability that the disease will be detected by the new test is 0.82.

Let

TP = Test Positive

TN = Test Negative

and

D = Disease

ND = No Disease

Thus we have:

P( TP | D) = 0.82

The probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.06.

That means

P( Test Positive | Do not have Disease) = 0.06

That is:
P( TP | ND) = 0.06

We are also given that: 13 % of the population who take this test have the disease.

That is:

P( D) = 0.13

then

P(ND) = 1 - P( D)

P(ND) = 1 - 0.13

P(ND) = 0.87

We have to find: the probability that the person actually has the disease given that the test administered to an individual is positive.

That is find:

P( D | TP ) = ............?

Using Bayes rule:

P(TPD) PD P(TPD) X P(D) + P(TPND) X P(ND)

0.82 x 0.13 0.82 x 0.13 +0.06 x 0.87

0.1066 P(DTP) = 7 0.1066 + 0.0522

P(DTP) = 0.1066 0.1588

P(DTP) = 0.671285

P(DTP) = 0.6713

( Note: round final answer to specified number of decimal places)

Answer 2

Solution:

Given:Among those who have the disease, the probability that the disease will be detected by the new test is 0.82.

Let

TP = Test Positive

TN = Test Negative

and

D = Disease

ND = No Disease

Thus we have:

P( TP | D) = 0.82

The probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.06.

That means

P( Test Positive | Do not have Disease) = 0.06

That is:
P( TP | ND) = 0.06

We are also given that: 13 % of the population who take this test have the disease.

That is:

P( D) = 0.13

then

P(ND) = 1 - P( D)

P(ND) = 1 - 0.13

P(ND) = 0.87

We have to find: the probability that the person actually has the disease given that the test administered to an individual is positive.

That is find:

P( D | TP ) = ............?

Using Bayes rule:

P(TPD) PD P(TPD) X P(D) + P(TPND) X P(ND)

0.82 x 0.13 0.82 x 0.13 +0.06 x 0.87

0.1066 P(DTP) = 7 0.1066 + 0.0522

P(DTP) = 0.1066 0.1588

P(DTP) = 0.671285

P(DTP) = 0.6713

( Note: round final answer to specified number of decimal places)