in a clinical study , volunteers are tested for a gene that has been found to increase the risk for a disease the probability that a person carries the gene is 0.2
1) what is the probablilty that 8 or more people will have to be tested before 3 with the gene are detected
2) how many people are expected to be tested before 3 ith gene are detected
+118608 in a clinical study , volunteers are tested for a gene that has been found to increase the risk for a disease the probability that a person carries the gene is 0.2
p=0.2 q=0.8
1) what is the probablilty that 8 or more people will have to be tested before 3 with the gene are detected
P(8 or more need to be tested before 3 are detected)
=P(2 or less will have it if 7 people are tested)
=P(0 out 7 have it) + P(1 out 7 have it) + P(2 out 7 have it)
\(= 7C0 * 0.8^7 + 7C1 * 0.2*0.8^6 +7C2 * 0.2^2 * 0.8^5\\ = 0.8^7 + 7* 0.2*0.8^6 +21 * 0.04 * 0.8^5\\\)
0.8^7 + 7* 0.2*0.8^6 +21 * 0.04 * 0.8^5 = 0.851968
approx 85.2%
2) ??
This is not the right answer. It’s not even close.
I have the correct answer but I don’t know how to do the workings.
+118608 You have not even said what you think the answer is.
Perhaps your are reading the answer section of your book upside down?
I stand by the answer that I have provided.
but if someone can see a flaw in my logic I will certainly listen to their reasoning.
+2441 I think I can answer the second one.
2) Well, if the probability of the of someone carrying this particular is 0.2, then the reciprocal would represent the number of times I would expect before I hit a match. Since 3 people need to be detected, multiply this by three.
\(3*\frac{1}{0.2}=15\text{ people}\)
