SOLUTION: Twenty-five percent of the employees of a large company are minorities. A random sample of 7 employees is selected. (round the answer to 4 decimal places if needed!) a) Wh

Question 1119163: Twenty-five percent of the employees of a large company are minorities. A random sample of 7 employees is selected.
(round the answer to 4 decimal places if needed!)
a) What is the probability that the sample contains exactly 4 minorities?
b) What is the probability that the sample contains fewer than 2 minorities?
c) What is the probability that the sample contains exactly 1 non-minority?
d) What is the expected number of minorities in the sample?
e) What is the variance of the minorities?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
a. 7C4(0.25)^4*(0.75)^3. Notice how the exponents add to 7 and they are split by how many of each are needed. The 7C4 are the number of ways those 4 can occur in each sample.
=0.0577
b. Fewer than 2 are 0 and 1
0: 0.75^7=0.1335
1: 7*0.25*0.75^6=0.3115
Their sum is 0.4450
c. Exactly one non-minority is the same as exactly 6 minorities
7C6*0.25^6*0.75=0.0013
d. Expected number is np=7*0.25=1.75
d. variance is np(1-p)=7*.25*.75=1.3125

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