Question

The probability that a person has a certain disease is 0.03. Medical diagnostic tests are available to determine whether the person actually has the disease. If the disease is actually present, the probability that the medical diagnostic test will give a positive result (indicating that the disease is present) is 0.9. If the disease is not actually present, the probability of a positive test result (indicating that the disease is present) is 0.02.a) Suppose that the medical diagnostic test has given a positive result (indicating that the disease is present), what is the probability that the disease is actually present?b) What is the probability of a positive test result?

Answer 1

Answer:

a) 0.58

b) 0.0464

Step-by-step explanation:

A= The person is sick

B= The give a positive result

P(A)=0.03

P(A')=1-P(A)=1-0.03=0.97

P(B|A)=0.9

P(B|A')=0.02

P(B)=P(B|A)P(A)+P(B|A')P(A')=0.9*0.03+0.02*0.97=0.0464

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{0.9*0.03}{0.0464}=0.58$

Answer 2

You can sue the Bayes' theorem to find the needed probability.

The answers are (for A and B defined below)

a) P(A|B) = 0.5819 approximately

b) P(B) = 0.0464

What is Bayes' theorem?

Suppose that there are two events A and B.
Then suppose the conditional probability are:

P(A|B) = probability of occurrence of A given B has already occurred.

P(B|A) = probability of occurrence of B given A has already occurred.

Then, according to Bayes' theorem, we have:

$\rm P(A|B) = \dfrac{P(B|A)P(A)}{P(B)}$

(assuming the P(B) is not 0)

Using that theorem to find the needed probabilities

Let the events be

A = Person has disease

B = Person is diagnosed positive for disease test.

Thus, from the given data, we have

P(A) = 0.03

P(A') = 1- P(A) = 0.97

P(B|A) = 0.9

P(B|A') = 0.02

Then we have:

a) Probability that disease is actually present if its given that person is diagnosed diseased  = P(A|B)

Let A' be the event that person is not diseased. Then probability for B can be expressed as

P(B) = P(A')P(B|A') + P(A)P(B|A)

Thus,

$\rm P(A|B) = \dfrac{P(B|A)P(A)}{P(B)} = \dfrac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A')P(A')}$$\rm P(A|B) = \dfrac{0.9 \times 0.03}{0.9 \times 0.03 + 0.02 \times 0.97} \approx 0.5819$

b) Probability of positive test result =

P(B) = $P(A')P(B|A') + P(A)P(B|A) = 0.9 \times 0.03 + 0.02 \times 0.97 = 0.0464$

P(B) = 0.0464

Thus,

a) P(A|B) = 0.5819 approximately

b) P(B) = 0.0464

Learn more about Bayes' theorem here:

https://brainly.com/question/16038936