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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 0.1. What is the probability that 4 will have to be tested before 2 with the gene are detected?

yuhki  Nov 21, 2014

Best Answer 

 #1
avatar+92624 
+5

The probability that a person carries the gene is 0.1. What is the probability that 4 will have to be tested before 2 with the gene are detected?

 

I think this really means that there is one sucess in the first 3 and then the forth one is a success.

P(1 succes in first 3)=3C1*(0.1)^1(0.9)^2

$${\mathtt{3}}{\mathtt{\,\times\,}}{\left({\mathtt{0.1}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{0.9}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{243}}}{{\mathtt{1\,000}}}} = {\mathtt{0.243}}$$

 

P(1 succes in first 3 FOLOWED by a success) = 0.243*0.1 = 0.0243

 

Once again - I think that is correct  

Melody  Nov 22, 2014
 #1
avatar+92624 
+5
Best Answer

The probability that a person carries the gene is 0.1. What is the probability that 4 will have to be tested before 2 with the gene are detected?

 

I think this really means that there is one sucess in the first 3 and then the forth one is a success.

P(1 succes in first 3)=3C1*(0.1)^1(0.9)^2

$${\mathtt{3}}{\mathtt{\,\times\,}}{\left({\mathtt{0.1}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{0.9}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{243}}}{{\mathtt{1\,000}}}} = {\mathtt{0.243}}$$

 

P(1 succes in first 3 FOLOWED by a success) = 0.243*0.1 = 0.0243

 

Once again - I think that is correct  

Melody  Nov 22, 2014

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