MATH SOLVE

2 months ago

Q:
# Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefore P(D)=0.009. A test for this type of cancer has a false positive rate of 6%. The test correctly confirms the presence of the cancer with an accuracy of 91%. Suppose that the test indicates that a person has cancer. What is the probability that the person actually does have cancer? Round your answer to four decimal places, if necessary

Accepted Solution

A:

Answer:There is a 12.13% probability that the person actually does have cancer.Step-by-step explanation:We have these following probabilities.A 0.9% probability of a person having cancerA 99.1% probability of a person not having cancer.If a person has cancer, she has a 91% probability of being diagnosticated.If a person does not have cancer, she has a 6% probability of being diagnosticated.The question can be formulated as the following problem:What is the probability of B happening, knowing that A has happened.It can be calculated by the following formula[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.In this problem we have the following questionWhat is the probability that the person has cancer, given that she was diagnosticated?SoP(B) is the probability of the person having cancer, so [tex]P(B) = 0.009[/tex]P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So [tex]P(A/B) = 0.91[/tex]P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So[tex]P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765[/tex]What is the probability that the person actually does have cancer?[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213[/tex]There is a 12.13% probability that the person actually does have cancer.