# Homework Help: Probability with 2 factors

1. Feb 6, 2013

### neubreed

1. The problem statement, all variables and given/known data
The test for an illness is 90% accurate (meaning it gives a positive result 90% of the time if the patient has gingivitis and a negative reading 90% of the time if the patient doesn't). 4% of all people have this illness.

a) calculate the probability the test will be positive for a random patient
b) once the patient tests positive, what is the probability that he actually has the illness?

2. Relevant equations

3. The attempt at a solution

a. P(positive):
b. P(has disease given test is positive):

I'm completely lost.

For a, I would think that the chances of the patient being actually ill are 0.04 and then him test positive is .9 of that, sooo the answer would be 0.036, but I have no idea how to make that look like an actual probability problem or if the answer is even correct...

For b, I don't even know where to start.

If someone could point me in the right direction, I would be much obliged. Thanks!

2. Feb 6, 2013

### HallsofIvy

Suppose you have a pool of 1000000 people. If "4% of all people have this illness." Okay, how many have the illness and how many do not?

"The test for an illness is 90% accurate (meaning it gives a positive result 90% of the time if the patient has gingivitis and a negative reading 90% of the time if the patient doesn't)."
So of the number of people who have the illness, how many will get a positive result? Of of the number of people who do not have the illness, how many will get a positive result?

P(positive) is the total number of positive results divided by the total number of people.

P(has disease given test is positive) is the number of people who actually have the disease and got a positive result divided by the total number of people who got a positive result.

3. Feb 8, 2013

### neubreed

so b actually turned out to be pretty simple and I'm certain of the answer, but a...

i can't figure out whether it would be: (0.04)(0.9 x 0.04 / 0.04) + 0.96(0.9 x 0.96 / 0.96)

or just (0.9 x 0.04 / 0.04) + (0.9 x 0.96 / 0.96)

Last edited: Feb 8, 2013
4. Feb 8, 2013

### Dick

Neither of those is right. P(testing positive)=P(sick)P(sick testing positive)+P(healthy)P(healthy testing positive).

5. Feb 8, 2013

### neubreed

On second thought, it turns out that I solved a as though the events were independent, which they weren't...

so the p(A given B)=p(A&B) / p(B)
but p(A&B)= p(A) x P(B given A)

I'm lost. I don't have either of the given probabilities, how can I solve the problem??

I had originally written 0.04 x 0.9/ 0.9 for a, but now I know that's not it...

6. Feb 8, 2013

### neubreed

Right, yeah, I just realized that I've been calculating the givens incorrectly, that's where my issue is...

7. Feb 8, 2013

### neubreed

Any suggestions? This is due in a couple of hours... :( I don't understand how to make the calculations without knowing either the p of A given B or vice versa.

8. Feb 8, 2013

### Staff: Mentor

Just begin with that. Absolute numbers are often easier to understand.
How many of them are ill? How many are healthy?

If you test all those ill people, how many will get a positive result?
If you test all those healthy people, how many will get a positive result?

9. Feb 8, 2013

### neubreed

4,000 people are ill, 96,000 are healthy.
out of the 4,000 3,600 will get a positive result.
out of the 96,000 9,600 will get a positive result.

I can do ratios, but I don't really understand how this helps me.

I just need to know how to plug the right numbers into the probability theorems, namely that whole circular A given B stuff.

10. Feb 8, 2013

### Staff: Mentor

How many will get a positive result in total, and which fraction of the original 100000 (was 1000000 before, but does not matter) is that?
Of those who get a positive result, how many are ill, and which fraction is that?

This is really a step-by-step explanation now...

11. Feb 8, 2013

### HallsofIvy

That depends upon what question you want to answer!
"a) calculate the probability the test will be positive for a random patient"
of the 100000 patients, 3600+ 9600= 13200 have a positive result.

"b) once the patient tests positive, what is the probability that he actually has the illness?"
Of the 13200 with a positive result, 3600 actually have the illness.