Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Cystic fibrosis (CF) is a hereditary lung disorder that often results in death. It can be inherited only if both parents are carriers of an abnormal gene. In 1989, the CF gene that is abnormal in carriers of cystic fibrosis was identified. The probability that a randomly chosen person of European ancestry carries an abnormal CF gene is 1/25. (The probability is less in other ethnic groups.) The CF20m test detects most, but not all, harmful mutations of the CF gene. The test is positive for 90% of people who are carriers. It is never positive for people who are not carriers.

Required:
Compute the probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

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.

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.