+0  
 
0
11
3188
1
avatar+82 

Suppose a researcher goes to a small college with 200 faculty members, 12 of which have blood type O-negative. She obtains a simple random sample of n = 20 of the faculty. Let the random variable X represent the number of faculty in the sample of size n = 20 that have blood type 0-negative.

a. What is the probability that 3 of the faculty have type O-negative?

b. What is the probability that at least one of the faculty has blood type O-negative?

 Nov 21, 2014

Best Answer 

 #1
avatar+23246 
+5

a)  The probability of randomly choosing someone with blood type O-negative is 12/200.

     The probability of randomly choosing someone without blood type O-negative is 188/200.

     The probability of randomly choosing three type-O and seventeen non-type-O is:

          (12/200)^3 x (188/200)^17 x the number of ways that this can occur.

     Since the order in which these are chosen is unimportant, it is a combination, so we been the combination of           choosing three out of twenty:  20nCr3

     Our probabiltiy has become:  20nCr3 x (12/200)^3 x (188/200)^17  =  0.086   (approx.)

b)  "At least one" means "one or more", so the easiest way is to find the probability that none have type O-              negative blood and subtract this from 1.000.

     The probability that none has type O-negative blood is:  20nCr0 x (12/200)^0 x (188/200)^20  =  0.290

     Subtracting this from 1.000 gives a probability of 0.710.

 Nov 22, 2014
 #1
avatar+23246 
+5
Best Answer

a)  The probability of randomly choosing someone with blood type O-negative is 12/200.

     The probability of randomly choosing someone without blood type O-negative is 188/200.

     The probability of randomly choosing three type-O and seventeen non-type-O is:

          (12/200)^3 x (188/200)^17 x the number of ways that this can occur.

     Since the order in which these are chosen is unimportant, it is a combination, so we been the combination of           choosing three out of twenty:  20nCr3

     Our probabiltiy has become:  20nCr3 x (12/200)^3 x (188/200)^17  =  0.086   (approx.)

b)  "At least one" means "one or more", so the easiest way is to find the probability that none have type O-              negative blood and subtract this from 1.000.

     The probability that none has type O-negative blood is:  20nCr0 x (12/200)^0 x (188/200)^20  =  0.290

     Subtracting this from 1.000 gives a probability of 0.710.

geno3141 Nov 22, 2014

3 Online Users

avatar
avatar