Six stand up comics, A,B,C,D,E and F are to perform on a single
evening at a comedy club. The order of performance is determined
by random selection.
With no restrictions 6 comedians can perform in 6! = 720 ways. This
will be the denominators of each of the probabilities.
Find the probability that
A.Comic C will perform first
C _ _ _ _ _
We can fill the 5 blanks with the 5 letters B,C,D,E,F in
5! = 120 ways
The probability of that is 120 ways out of 720 or 120/720 or 1/6.
[This could also be reasoned by saying each of the 6 has an equal
chance of performing first, which give us 1/6 immediately]
B. Comic B will perform last and Comic F will perform third
_ _ F _ _ B
We can fill the 4 blanks with the 4 letters A,C,D,E in
4! = 24 ways.
The probability of that is 24 ways out of 720 or 24/720 or 1/30.
C. The comedians will perform in the following order C,B,F,A,D,E
C B F A D E
There is just 1 way that can happen.
The probability of that is 1 way out of 720 or 1/720.
D.Comic D or E will perform fourth.
_ _ _ D _ _
There are 5! = 120 ways to fill those 5 blanks with the letters A,B,C,E,F.
In addition to those, there are another 5! = 120 ways to fill these 5 blanks
_ _ _ E _ _
with the letters A,B,C,D,F.
That's 2(5!) = 2(120) = 240 ways.
The probability of one of those is 240 ways out of 720 or 240/720 or 1/3.
Edwin