Respuesta :
Answer:
0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Test indicates cancer.
Event B: Person does not have cancer.
Probability of a test indicating cancer.
98% of 2%(those who have).
1% of 100 - 2 = 98%(those who do not have). So
[tex]P(A) = 0.98*0.02 + 0.01*0.98 = 0.0294[/tex]
Probability of a test indicating cancer and person not having.
1% of 98%. So
[tex]P(A \cap B) = 0.01*0.98 = 0.0098[/tex]
What is the probability of a randomly chosen person not having cancer given that the test indicates cancer?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0098}{0.0294} = 0.3333[/tex]
0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer
