Question

Imagine you have a 3% chance of getting a disease. A medical test is performed to check it, the test has a 5% chance of throwing a wrong result. If the test is done twice and both times he was negative (that is, he is not sick), he calculates the probability that he will actually be ill. R:

Answer #1

The test is done twice and same result is obtained so the result can be obtained by multiplying the value of probability of one case.

P(getting a disease ) = 3/100

= 0.03

Said that test give 5% of wrong results .Wrong result mean that the test give positive result but he actually don't have disease .Also the test shows negative result but actually have the disease. The probabilities are given. ,

P( positive / disease ) = 95/100

= 0.95

P(positive / no disease) = 5/100

= 0.05

P( negative / disease ) = 5/100

= 0.05

P(negative / no disease ) = 95/100

= 0.95

We know Bayes theorem equation as ,

Probability that he is ill given that the test was negative ,

P( disease / negative ) =【 P(negative / disease) × P(disease) 】/ (P( negative )

= 0.05×0.03 / ( 0.03×0.05)+(0.97 ×0.95)

= 0.0015 / ( 0.0015+ 0.9215)

= 0.0015 / 0.923

= 0.0016

Then the total probability can be calculated as

Answer = P(disease/ negative )^{2}

= 0.0016^{2}

= 0.00000256

Answer is 0.00025% chance he will actually be ill .

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