# Thread: Bayes Rule probability problem.

1. ## Bayes Rule probability problem.

Hi all,

I am not sure if this belongs to University Math help. It looks like it's basic enough but we are doing it in college. I hope I am not intruding in the wrong territory.

The problem is as follows:

A certain disease can be detected by a blood test in 95% of those who have it. Unfortunately, the test also has a 0.02 probability of showing that a person has the disease when in fact he or she does not. It has been estimated that 1% of those people who are routinely tested actually have the disease. If the test shows that a certain person has the disease, find the probability that the person actually has it.

This is how I did it:

For straight forwardness we will look at the population as being 10000 people.

95% of those who have it will test "yes" i.e. 95% of 1% = 0.95 x 0.01 = 0.0095 from all population, so from 10000 people it would be 95 people

2% of those who do not have it will also test "yes" i.e. 2% of 99% = 0.02 x 0.99 = 0.0198 from all population, so from 10000 people it would be 198 people

So, the total number of all "yes" tests (positive "yes" and negative "yes") would be 95+198=293 "yes" tests

Since we know that 1% of population actually have the disease i.e. 100 people then we can conclude that chance of having disease if the test was positive is 100/293=0.3413

As you can see, my answer came out as 0.3413. Is that correct? The problem is - my lecturer's answer is 0.324

2. Originally Posted by johammbass
Hi all,

I am not sure if this belongs to University Math help. It looks like it's basic enough but we are doing it in college. I hope I am not intruding in the wrong territory.

The problem is as follows:

A certain disease can be detected by a blood test in 95% of those who have it. Unfortunately, the test also has a 0.02 probability of showing that a person has the disease when in fact he or she does not. It has been estimated that 1% of those people who are routinely tested actually have the disease. If the test shows that a certain person has the disease, find the probability that the person actually has it.

This is how I did it:

For straight forwardness we will look at the population as being 10000 people.

95% of those who have it will test "yes" i.e. 95% of 1% = 0.95 x 0.01 = 0.0095 from all population, so from 10000 people it would be 95 people

2% of those who do not have it will also test "yes" i.e. 2% of 99% = 0.02 x 0.99 = 0.0198 from all population, so from 10000 people it would be 198 people

So, the total number of all "yes" tests (positive "yes" and negative "yes") would be 95+198=293 "yes" tests

Since we know that 1% of population actually have the disease i.e. 100 people then we can conclude that chance of having disease if the test was positive is 100/293=0.3413

As you can see, my answer came out as 0.3413. Is that correct? The problem is - my lecturer's answer is 0.324
Well I get $P(D|+)=\frac{P(+|D)P(D)}{P(+|D)P(D)+P(+|D^c)P(D^c) }=0.324232082$.

3. Dear Plato,

If you are not very busy, would you be able to the solution in explicitly because I am not clearly understanding all the terms. Would you be able to tell me where I went wrong with my reasoning?

Thank you

4. Originally Posted by johammbass
would you be able to the solution in explicitly
You should understand that this is not a tutorial service.
I will however give you the numbers.
$+$ means a positive test.
$D$ means that an individual actually has the disease.
So $P(+|D)=0.95,~P(+|D^c)=0.02,~\&~P(D)=0.01~.$

5. Hmm, I don't know, I thought the answer would have to be the probability of being actually diseased over the probability of a positive test but I guess I was wrong

6. Originally Posted by johammbass
Hmm, I don't know, I thought the answer would have to be the probability of being actually diseased over the probability of a positive test but I guess I was wrong
The question asks, "Given that the test is positive, what is the probability that the disease is actually present"?
That is $P(D|+)=\frac{P(D~\&~+)}{P(+)}$.
One needs to understand conditional probabilies.
If you have doubts about that, then that may be you problem.