Math  /  Data & Statistics

Questionmutant gene is 8%8 \% in the population. If the probability that a randomly chosen woman will develop breast cancer is 0.1 , then what is the probability that a randomly chosen breast cancer patient in this population will carry the mutant BRCA1 gene? \square
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DETAILS MY NOTES SCALCLS1 12.3.060. ASK YOUR TEACHER PRACTICE ANOTHER
DNA evidence is often used in criminal trials. Suppose a murder has been committed and the perpetrator's blood is found at the crime scene. DNA from the blood is analyzed and it is found that the probability of an innocent person having DNA that matches is 10910^{-9}. Let MM be the event 'DNA match' and II be the event 'innocent'. Therefore, we have P(MI)=109P(M \mid I)=10^{-9}. What is really of interest is the quantity P(IM)P(I \mid M), the probability that a person whose DNA matches is innocent. Calculate this quantity if the community has 1000 people, and therefore the probability that a randomly chosen person is innocent is P(I)=999/1000P(I)=999 / 1000. You can assume that the probability of a match for the guilty person is 1 (that is, P(MI)=1P\left(M \mid I^{\prime}\right)=1 ). (Round your answer to three decimal places.) \square ×107\times 10^{-7}
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Studdy Solution
Calculate P(GB) P(G \mid B) using Bayes' Theorem:
P(GB)=0.80.080.11 P(G \mid B) = \frac{0.8 \cdot 0.08}{0.11}
P(GB)=0.0640.11 P(G \mid B) = \frac{0.064}{0.11}
P(GB)0.582 P(G \mid B) \approx 0.582
The probability that a randomly chosen breast cancer patient carries the mutant BRCA1 gene is approximately:
0.582 \boxed{0.582}

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