A diagnostic test has a 95% probability of giving a positive result when given to a person who has a certain disease. It has a 10% probability of giving a (false) positive result when given to a person who doesn’t have the disease. It is estimated that 15% of the population suffers from this disease. (a) What is the probability that a test result is positive?
(b) A person recieves a positive test result. What is the probability that this person actually has the disease? (probability of a true positive)
(c) A person recieves a positive test result. What is the probability that this person doesn’t actually have the disease? (probability of a false negative)

a)0.2275   b)95/105=19/21    c)10/105= 2/21

Step-by-step explanation:

a) The case "The test result is positive" consists in 2 parts.

The 1st one is "The person has the desease (15%=0.15)  and the test's  result is positive (95%=0.95)

The probability of that is P(desease, positive) = 0.15*0.95=0.1425

The 2nd one is "The person has no the desease (100%-15%=85%=0.85). However the test result is positive (10%=0.1)

The probability of that is P(not desease, positive)=0.85*0.1=0.085

The total probability that test is positive is the sum of 1st and 2-nd parts of the case: P(pos) = 0.1425+0.085=0.2275

b) As it has been shown in a) The test result can be positive in case that the person is really has the desease (95%) and in case the person has no the desease (10%).  This actually means that 95 persons from 105  having positive test result are really has the desease.

So the probability that the test result is positive and person has the desease is P (desease/positive)= 95/105

c) It's clearly seen that the sum of  probabilities of b) and c)  equal 1.

Both events make full group of events.

If the test result is positive the person can have the desease or can have not the desease. So ( no desease/positive)= 1-95/105=10/105

the answer is a) y= 2 csc [0.5 (x+pi/2)]-1

step-by-step explanation: