Answer to Question #135101 in Statistics and Probability for Soong

Question

A medical school has discovered a new test for hepatitis. Experimentation has shown that the probability of a positive test is 0.85 given that a person has hepatitis. The probability is 0.11 that the test is positive given that the person does not have hepatitis. Assume that in the general population the probability that a person has hepatitis is 0.05. What is the probability that a person chosen at random will:
not have hepatitis and have a negative test?
have a positive test?
have hepatitis given that the test is positive?

Answer 1

1). The probability that person does not have hepatitis is 0.95. Assume that we have two events:

A: a test is negative; B: a person does not have a hepatitis. Then we have:

$0.89=P(A|B)=\frac{P(A\cap B)}{P(B)}$ .

From the latter we obtain: $P(A\cap B)=0.89\cdot P(B)=0.89\cdot0.95=0.8455$ .

2). We consider events: A -- the test is positive; $B_1$ - a person has a hepatitis; $B_2$ -

a person does not have a hepatitis. Due to the law of total probability we have:

$P(A)=P(A|B_1)P(B_1)+P(A|B_2)P(B_2)=0.85\cdot0.05+0.11\cdot0.95=0.0425+0.1045=0.147$

3). Consider events: A: a person has a hepatitis; B: a test is positive.

We have the following formula:

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B| A)P(A)}{P(B)}=0.85\cdot0.05/0.147=0.2891$

Answer:1). 0.8455; 2). 0.147; 3). 0.2891