Question

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 F will perform fourth
(B) Comic D will perform fourth and Comic B will perform second.
(C) The comedians will perform in the following​ order: F, D, B, A, C, E.
(D) Comic C or Comic E will perform last.

Answer 1

Answer:

a) 1/6

b) 1/36

c) 1/720

d) 1/3

Step-by-step explanation:

a) Any of the six comics can perform in the fourth place, so there is one chance in six that Comic F is the one that performs fourth.

$P(F=4th)=1/6$

b) In this case, we have two conditions. Both are independent of each other, so the probability of both happening is the product of the probabilities of each happening individually:

$P(D=4th \&B=2nd)=P(D=4th)\cdot P(B=2nd)=(1/6)\cdot(1/6)=1/36$

c) This combination is one in all possible orders of perform. The amount of combinations of orders is n!=6!=720 possible combinations. So the probability of this specific order is:

$P(F, D, B, A, C, E)=1/720$

d) In this case, of the six possible comics performing in the last place, we calculate the probability of 2 of them being in that place. So the probability is:

$P(C=6th\,or\, E=6th)=2/6=1/3$