In the past few years, outsourcing overseas has become more frequently used than ever before by U.S. companies. However, outsourcing is not without problems. A recent survey by Purchasing indicates that 20% of the companies that outsource overseas use a consultant. Suppose 15 companies that outsource overseas are randomly selected.
(a) What is the probability that exactly five companies that outsource overseas use a consultant?
(b) What is the probability that more than nine companies that outsource overseas use a consultant?
(c) What is the probability that none of the companies that outsource overseas use a consultant?
(d) What is the probability that between four and seven (inclusive) companies that outsource overseas use a consultant?

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=15, p=0.2)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Part a

[tex]P(X=5)=(15C5)(0.2)^5 (1-0.2)^{15-5}=0.103[/tex]

Part b

[tex]P(X> 9)=P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)[/tex]

[tex]P(X=10)=(15C10)(0.2)^{10} (1-0.2)^{15-10}=0.0001[/tex]

[tex]P(X=11)=(15C11)(0.2)^{11} (1-0.2)^{15-11}=0.0000114[/tex]

[tex]P(X=12)=(15C12)(0.2)^{12} (1-0.2)^{15-12}=9.54x10^{-7}[/tex]

[tex]P(X=13)=(15C13)(0.2)^{13} (1-0.2)^{15-13}=5.505x10^{-8}[/tex]

[tex]P(X=14)=(15C14)(0.2)^{14} (1-0.2)^{15-14}=1.96x10^{-9}[/tex]

[tex]P(X=15)=(15C15)(0.2)^{15} (1-0.2)^{15-15}=3.27x10^{-11}[/tex]

And adding we got: 0.000113

Part c

[tex]P(X=0)=(15C0)(0.2)^{0} (1-0.2)^{15-0}=0.0352[/tex]

Part d

[tex]P(4 \leq X\leq 7)=P(X=4)+P(X=5)+P(X=6)+P(X=7)[/tex]

[tex]P(X=4)=(15C4)(0.2)^{4} (1-0.2)^{15-4}=0.1876[/tex]

[tex]P(X=5)=(15C5)(0.2)^{5} (1-0.2)^{15-5}=0.1032[/tex]

[tex]P(X=6)=(15C6)(0.2)^{6} (1-0.2)^{15-6}=0.043[/tex]

[tex]P(X=7)=(15C7)(0.2)^{7} (1-0.2)^{15-7}=0.0138[/tex]

And adding we got: 0.3476