Question

There was a team of ten people working on the building, including three electricians and two plumbers. The architect (not included) called a meeting with five of the team, and randomly selected people to attend. Calculate the probability that exactly 2 electricians and one plumber were called to the meeting.

Answer 1

Pick electrician, electrician, plumber 3/10 * 2/9 * 2/8 +
electrician, plumber, electrician 3/10 * 2/9 * 2/8 +
plumber, electrician, electrician 2/10 * 3/9 * 2/8

> 3/10 * 2/9 * 2/8 + 3/10 * 2/9 * 2/8 + 2/10 * 3/9 * 2/8
[1] 0.05

Answer 2

Does the other group of people . Like the other 2 in the team matter?

Answer 3

My answer is wrong, I forgot that is a meeting of 5 people, the other two doesn't have to be electrician or plumber (lets call them "other")

He have to take the permutations of:
other other electrician electrician plumber

each pick diminished total people by one
-> x1/10 * x2/9 * x3/8 * x4/7 * x5/6
pick two different from electrician or plumber (other) makes that any two from xi = 1,.., 5 take 5 and 4, for two electricians: 3 and 2, and 2 for the plumber

There are 30 different permutations from take 2 others, 2 electricians and 1 plumber
5!/(2!2!1!) (see http://mdm4u1.wetpaint.com/page/4.3+Permutations+with+Some+Identical+Elements)

> 30 * ((5*4*3*2*2)/(10*9*8*7*6))
[1] 0.2380952
>

Answer 4

My PREVIOUS answer WAS wrong, I mean to say