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. find the probability that a) comic e will be performed first b) comic c will perform fifth and comic b will perform last c) the comedians will perform in the following order d, e, c, a, b, f d) comic a or comic b will perform first

a) comic E will be performed first 

With no restrictions 6 comedians can perform in 6! = 720 ways

E _ _ _ _ _
we can fill the 5 blanks with the 5 letters A,B,C,D,F in 5! = 120 ways The probability of that is 120 ways out of 720 or 120/720 or 1/6.

b) comic C will perform fifth and comic B will perform last

_ _ _ _C B
 We can fill the 4 blanks with the 4 letters A,D,E,F 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 D, E,C, A, B, F

D E C A B F
 There is just 1 way that can happen. The probability of that is 1 way out of 720 or 1/720.

d) comic A or comic B will perform first

_ _ _ A _ _
   There are 5! = 120 ways to fill those 5 blanks with the letters B,C,D,E,F.
In addition to those, there are another 5! = 120 ways to fill these 5 blanks
   _ _ _ B _  with the letters A,C,D,E,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.

Yes she is correct. when you do 1/5 the tbe is their to represent the fraction. so yes, kathy got it