Question

A committee will be formed with two managers and four engineers selected randomly without replacement from nine managers and 18 engineers.

(a)How many different committees are possible?

(b) What is the probability that the specific engineer Pat and the specific manager Chris are on the committee?

(c) What is the probability that specific engineer Pat or specific manager Chris are on the committee?

(d) Each of the engineers differ in years of experience. What is the probability that the most experienced and least experienced engineers are on the committee?

Answer 1

a)number of different committee =N(selecting 2 manager from 9 and 4 engineer from 18)

=$\binom{9}{2}\binom{18}{4}$ =36*3060=110160

b)

as Pat and Chris are in the committee ; therefore we need to select 1 manager from 8 and 3 engineer from 17)

hence probability=$\binom{8}{1}\binom{17}{3}$/110160 =8*680/110160=5440/110160=4/81

c)

P(Pat or Chris are in the comittee )=1-P(Pat and Chris are not in the committee)

=1-$\binom{8}{2}\binom{17}{4}$/110160 =1-28*2380/110160=1-833/1377=544/1377

d)P( most experienced and least experienced engineers are on the committee )

=P(sleecting 2 manager from 9 and 2 engineer from remaing 16)

=$\binom{9}{2}\binom{16}{2}$/110160=36*120/110160=2/51