Suppose that 1% of all people have a particular disease. A test for the disease

Question

Suppose that 1% of all people have a particular disease. A test for the disease is 99% accurate. This means that a person who has the disease has a 99% chance of testing positive for the disease, while a person who doesn't have the disease has a 99% chance of testing negative for the disease. If a person tests positive for the disease, what is the chance (rounded to the nearest hundredth) that he or she actually has the disease? 0.99, 0.40, 0.50, 0.45, None of the above

I think the answer is 99% but am is it a trick question and too easy to be 99%. What do you think the answer is and why. thank you.

Explanation / Answer

Test is 99% accurate

Pr(Positive l when have disease) = 0.99

Pr(Negative l when don't have disease) = 0.99 => Pr(positive l when don''t have disease) = 0.01

Pr(A random person has the disease) = 0.01

Pr(A random person doesn't have the disease) = 0.99

A person is tested positive so we have to find that he is really affected.

Pr(Diseased l Tested Positive) = Pr(Tested Positive l diseased) * Pr(Diseased)/ [Pr(Tested Positive l diseased) * Pr(Diseased) + Pr(Tested Positive l not diseased) * Pr(not diseased) ]

= (0.99 * 0.01)/ (0.99 * 0.01 + 0.01 * 0.99)

= 0.50

so there are 50% chances that a random person who is tested p0sitive will have the disease.

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