Re: probability word problem

Say the number of people infected is x. Then 98% of them will test positive, so 0.98x.

Say the number of people not infected is y. Then 3% of them will test positive (falsely), so 0.03y.

There is no guarantee that together you get the total number of people who are infected. It's just saying that of those infected, 98% will show up positive on the test, and from those not infected, 3% will receive a false positive.

Now your question is asking "If a person selected at random responds positively to the test, what is the probability that the person actually has the disease?"

Another way of saying this is "What is the probability of a person actually having the disease given that the person tested positive?"

This is a conditional probability, so you need to use the conditional probability formula.

Re: probability word problem

OK I get the bit from "now your question is asking"

but I just can't seem to grasp the first bit don't know why (doing my head in).

Re: probability word problem

Let's put some numbers to this problem. Suppose there are 100,000 people. 100 of them have the disease (1/1000 of 100,000 = 100). Of those 100 people, 98 of them test positive for the disease (98% of 100). Of the 99,900 people who do not have the disease, 2,997 of them test positive for the disease (3% of the 99,900 people who do not have the disease). So, if a person tests positive for the disease, they are one of 3,095 people (98+2,997 = 3,095). 98 out of the 3,095 people who tested positive actually have the disease. So, 98/3,095 = 0.03166 (which is the answer to the problem).