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Sagot :
Answer:
8.33% probability that she does, in fact, have the disease
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: Positive test
Event B: Has the disease
Probability of a positive test:
10% of 100-1 = 99%
90% of 1%
So
[tex]P(A) = 0.1*0.99 + 0.9*0.01 = 0.108[/tex]
Positive test and having the disease:
90% of 1%
[tex]P(A \cap B) = 0.9*0.01 = 0.009[/tex]
What is the conditional probability that she does, in fact, have the disease
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.009}{0.108} = 0.0833[/tex]
8.33% probability that she does, in fact, have the disease
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