Question

A committee of 5 people is to be formed from 11 civil employees and 8 small business people. Calculate the following :

a) The probability that none of the people selected are small business people
b) The probability at least 2 people are small business people
c) The expected number of small business people on the committee

Answer 1

Let $X=$ the number of small business people on the committee.

There are ₁₉C₅ ways to select 5 people (committee members) out of a total of 19 people (11 civil employees and 8 small business people).

$$\binom{19}{5}={19! \over 5!(19-5)!}={19(18)(17)(16)(15) \over 1(2)(3)(4)(5)}=11628$$

a) The probability that none of the people selected are small business people  

$$P(X=0)={ \dbinom{11}{5-0}\dbinom{8}{0}\over \dbinom{19}{8}}=\dfrac{\dfrac{11!}{5!(11-5)!}\cdot\dfrac{8!}{0!(8-0)}}{11628}=$$ $$={462\cdot1 \over 11628}={77 \over 1938}\approx0.0397$$

b) The probability at least 2 people are small business people  

$$P(X\geq2)=1-P(X=0)-P(X=1)=$$ $$=1-{ \dbinom{11}{5-0}\dbinom{8}{0}\over \dbinom{19}{8}}-{ \dbinom{11}{5-1}\dbinom{8}{1}\over \dbinom{19}{8}}=$$ $$=1-\dfrac{\dfrac{11!}{5!(11-5)!}\cdot\dfrac{8!}{0!(8-0)}}{11628}-\dfrac{\dfrac{11!}{4!(11-4)!}\cdot\dfrac{8!}{1!(8-1)}}{11628}=$$

$$=1-{462\cdot1 \over 11628}-{330\cdot8 \over 11628}={1421 \over 1938}\approx0.7332$$

c) The expected number of small business people on the committee.

$$P(X=0)={77 \over 1938}$$

$$P(X=1)={440 \over 1938}$$

$$P(X=2)={ \dbinom{11}{5-2}\dbinom{8}{2}\over \dbinom{19}{8}}=\dfrac{\dfrac{11!}{3!(11-3)!}\cdot\dfrac{8!}{2!(8-2)}}{11628}=$$ $$={165\cdot28 \over 11628}={770 \over 1938}$$

$$P(X=3)={ \dbinom{11}{5-3}\dbinom{8}{3}\over \dbinom{19}{8}}=\dfrac{\dfrac{11!}{2!(11-2)!}\cdot\dfrac{8!}{3!(8-3)}}{11628}=$$ $$={55\cdot56 \over 11628}={770 \over 2907}$$

$$P(X=4)={ \dbinom{11}{5-4}\dbinom{8}{4}\over \dbinom{19}{8}}=\dfrac{\dfrac{11!}{1!(11-1)!}\cdot\dfrac{8!}{4!(8-4)}}{11628}=$$ $$={11\cdot70 \over 11628}={385 \over 5814}$$

$$P(X=5)={ \dbinom{11}{5-5}\dbinom{8}{5}\over \dbinom{19}{8}}=\dfrac{\dfrac{11!}{0!(11-0)!}\cdot\dfrac{8!}{5!(8-5)}}{11628}=$$ $$={1\cdot56 \over 11628}={14\over 2907}$$

$$E(X)=\displaystyle\sum_{i=0}^5x_iP(X=x_i)=0\cdot{77 \over 1938}+1\cdot{440 \over 1938}+$$

$$+2\cdot{770 \over 1938}+3\cdot{770 \over 2907}+4\cdot{385 \over 5814}+5\cdot{14 \over 2907}={40 \over 19}$$

$$E(X)={40 \over 19}\approx2.1053$$