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Sagot :
Answer:
0.9959 = 99.59% probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.
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: Tested negative
Event B: Does not carry an abdornal CF gene.
Probability of a negative test:
10% of 1/25 = 0.04 = 4%
100% of 100 - 4 = 96%. So
[tex]P(A) = 0.1*0.04 + 1*0.96 = 0.964[/tex]
Negative test and not carrying an abdornal gene:
100% of 96%. So
[tex]P(A \cap B) = 1*0.96 = 0.96[/tex]
Compute the probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.96}{0.964} = 0.9959[/tex]
0.9959 = 99.59% probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.
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