please help ... explanation required?

Question

A new test is developed to test for a certain disease, giving "positive" or "negative" results to indicate that the person does or does not have the disease. For a person who actually has the disease, the test will give a positive result with probability 0.95. For a person who does not have the disease, the test will give a positive result with probability 0.05. Furthermore, only one person in 1000 actually has the disease. If a randomly selected person is given the test and tests positive, what is the probability that the person actually has the disease?

Is there anyone who can explain the answer in detail

Answer 1

First thing with these probability questions: define your events

P = event that the test is positive
D = event that a person actually has the disease.

Given:
P(D) = 1/1000
P(P \ D) = 0.95
P(P \ D') = 0.05
That's the information we've been given

RTF: P(D \ P)

The formula for conditional probability is:
P(D \ P) = P(D n P) / P(P)
We don't know either term on teh RHS but we can find them.

P(P) = P(P n D) + P(P n D')
Draw a Venn diagram and you'll see why this is true.
P(PnD) = P(P\D)*P(D) = 0.95/1000
P(PnD') = P(P\D')*P(D') = 0.05*999/1000

So now we have the information we need:
P(D \ P) = 0.95/1000 / [0.05*999/1000 + 0.95/1000]
= 19/1018 = 0.01866

Answer 2

1 in a 1000

Answer 3

chances of testing positive.

there are 999 people without. x0.05 probability means that 49.95 people will test positive but not have it.

1 person with x 0.95 probability means that 0.95 people will test positive and have it.

so for every 1,000 there are 50.90 people who will test positive.

if this person is a genuine positive then he is 1 : 50.90 or 0.02
if this person is a fake positive then he is 49.90:50.90 or 0.98

0.98 + 0.02 = 1 so our probabilities add up (either he is genuine positive or he isnt)

so the answer is 0.02

Answer 4

P(test positive and is accurate) = 0.95
P(test positive and is inaccurate) = 1 - 0.95 = 0.05
P(disease in population) = 1/1000 = 0.001
P(random person selected, tested positive and has disease) = P(test positive and is accurate) x P(disease in population)
= 0.95 x 0.001 = 0.00095

Ironduke, You need to consider the test being inaccurate as well, hence 0.00095

Answer 5

19/1018 or 0.018664
it's been a while, so i might be wrong
here's what i did:

let-
P(yes)=probability of having the disease
P(no)=probability of not having the disease
P(pos)=probability of a positive test retult
P(neg)=probability of a negative test result

so
P(yes/pos)
= P(yes AND pos) / [ P(yes AND pos)+P(no AND pos) ]