Probability Help Please

NTMK

hi
Im stuck on a question of a past exam college paper

A new test was developed to diagnose a particular Disease.If a person has the disease, the test has a 95% chance of identifying them a having the disease. if the person does not have the disease the test has a 1% chance of identifying them as having the disease. 5% of the pop. have the disease. Suppose we select a person at random from the population

i) What is the probability That the test will identify them as having the disease?

ii) What is the prob. that the person has the disease given that the test identifies them as having the disease

I ve got part one out but not sure how to do part two
The answer is .833

any help on how to do the question would be great

redarmyblues

From a sample of 2000 people 20 healthy will test positive and 100 will have the disease (95 of those will test positive 5 will test negative) I think your answer may be wrong because there will be 115 positive tests and 20 of these will be false 20/115= .0869 being the probabilty of infection, but then I got a pass C in my leaving in maths.

If there were no false negatives .833 would be correct.

Fremen

It's just a matter of setting up the problem right, then applying Bayes' theorem.

Let A be the event that a person has the disease.
Let B be the event that the test is positive.

Now, we need to figure out what information we're given:

If a person has the disease, the test has a 95% chance of identifying them a having the disease.

In other words, P(B | A) = .95

if the person does not have the disease the test has a 1% chance of identifying them as having the disease.

so P(B | A compliment) = .01

5% of the pop. have the disease

so P(A) = .05

In the first part, you're looking for P(B), in the second part you're looking for P(A | .

Now go forth and apply Bayes!

redarmyblues

Fremen wrote: »

It's just a matter of setting up the problem right, then applying Bayes' theorem.

Let A be the event that a person has the disease.
Let B be the event that the test is positive.

Now, we need to figure out what information we're given:



In other words, P(B | A) = .95



so P(B | A compliment) = .01


so P(A) = .05

In the first part, you're looking for P(B), in the second part you're looking for P(A | .

Now go forth and apply Bayes!

I don't understand maths nomenclature, It was annoying me in bed last night and the answer was tantalisingly close but I couldn't figure it out, can you explain the error in my logic, it seemed sound but there was a nagging doubt.

L'prof

NTMK wrote: »

I ve got part one

Is the answer 0.057?

hivizman

redarmyblues wrote: »

From a sample of 2000 people 20 healthy will test positive and 100 will have the disease (95 of those will test positive 5 will test negative) I think your answer may be wrong because there will be 115 positive tests and 20 of these will be false 20/115= .0869 being the probabilty of infection, but then I got a pass C in my leaving in maths.

If there were no false negatives .833 would be correct.

You are actually double-counting the 100 people with the disease when you say "From a sample of 2000 people 20 healthy will test positive". From a sample of 2000 people, we would expect 100 to have the disease, so only 1900 are healthy. Of these, 19 will test positive. So you have 95+19=114 positive tests, of which 19 are false. So the probability of being ill given a positive test is 95/114, which simplifies to 5/6 or 0.833. As Fremen points out, this is just Bayes' Theorem, without the algebra.

By the way, 20/115 = 0.173. You actually calculated 10/115.

NTMK

Thanks for all the replys i figured it out in the end and it didnt even come up:rolleyes: