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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

May 29, 2018

#1
+100024
+3

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) ??

May 29, 2018
#2
0

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.

Jun 3, 2018
#3
+100024
+1

You have not even said what you think the answer is.

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.

Jun 3, 2018
#4
+2340
0

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.

$$3*\frac{1}{0.2}=15\text{ people}$$

TheXSquaredFactor  Jun 3, 2018
#5
0

It’s not what I think the answer is. It’s not a book, it’s an online homework practice site. If you enter a wrong answer in the box it gives the correct answer, but it doesn’t show how to get it.

Jun 3, 2018
#6
+2340
0