Please help in this probability problem...?

Question

The probability that a person has a deadly virus is 5 in one thousand. A test will correctly diagnose this disease 95% of the time and incorrectly on 20% of occasions. find the probability of this test giving a correct diagnosis. (b) given that the test diagnose the patient as having the disease, what is the probability that the patient does not have the disease? (c) given that the test diagnose the patient as not having the disease, what is the probability that the patient does have the disease?

Answer 1

a. 80%

b.20%

c. 20%

I am not confident of the answer, if the answer of this question involves the complex mathematical formula then i am incorrect else with my simple logic of set i have tried my best

Answer 2

Let D be the event of having the disease

Let ~D be the event of not having the disease

Let + be the event of testing positive

Let - be the event of testing negative

we know from the question that

P(D) = 5/1000

P( + | D ) = 0.95

P( + | ~D ) = 0.20

a) find P( + )

use the law of total probability

P( + ) = P( + | D) * P(D) + P(+ | ~D) * P(~D)

P( + ) = 0.95 * 5/1000 + 0.20 * (1 - 5/1000)

P( + ) = 0.20375

(b)

find:

P( D | + ) = P( + | D ) * P(D) / P( + )

P( D | + ) = (0.95 * 5/1000) / 0.20375

P( D | + ) = 0.02331288

This is an exercise in conditional probability.  You can use Bayes' Rule to solve this.  However, I find that it is much easier to just use the definition of conditional probability.  In short I derived Bayes' Rule every time I do this but it prevents me from making errors in more complex problems.

This is an exercise in conditional probability.  For any two events A and B, where P(B) ≠ 0, you have the conditional probability:

P( A | B ) = P( A ∩ B ) / P( B ) = P( B | A) * P(A) / P(B)

the above is read as: the probability of A given B is equal to the probability of A and B divided by the probability of B.

Use The Law of Total Probability to find P(B)

For a set of events A1, A2, A3, ... , An where the Ai's are mutually exclusive and exhaustive events and for any other event B

P(B)

= P(B and A1) + P(B and A2) + ... + P(B and An)

= P(B | A1) * P(A1) + P(B | A2) * P(A2) + ... + P(B | An) * P(An)