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 d will perform fifth. b. comic f will perform second and comic e will perform fifth. c. the comedians will perform in the following order: d, f, e, b, c, a. d. comic c or comic a will perform first. a. (type a fraction. simplify your answer.) b. (type a fraction. simplify your answer.) c. (type a fraction. simplify your answer.) d. (type a fraction. simplify your answer.)

Explanation:

Step1: Calculate total number of arrangements

The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 6\), so the total number of ways the 6 comics can perform is \(n!=6!=6\times5\times4\times3\times2\times 1=720\).

Step2: Solve part a

Fix comic D in the fifth - position. Then the remaining 5 comics can be arranged in \(5! = 5\times4\times3\times2\times1 = 120\) ways. The probability \(P(D\text{ fifth})=\frac{5!}{6!}=\frac{120}{720}=\frac{1}{6}\).

Step3: Solve part b

Fix comic F in the second - position and comic E in the fifth - position. Then the remaining 4 comics can be arranged in \(4! = 4\times3\times2\times1=24\) ways. The probability \(P(F\text{ second and }E\text{ fifth})=\frac{4!}{6!}=\frac{24}{720}=\frac{1}{30}\).

Step4: Solve part c

There is only 1 way to arrange the comics in the order D, F, E, B, C, A. So the probability \(P(D,F,E,B,C,A)=\frac{1}{6!}=\frac{1}{720}\).

Step5: Solve part d

The number of ways comic C can perform first is \(5!\) (since the remaining 5 comics can be arranged in the remaining 5 positions). Similarly, the number of ways comic A can perform first is \(5!\). The number of ways C or A can perform first is \(2\times5!\). The probability \(P(C\text{ or }A\text{ first})=\frac{2\times5!}{6!}=\frac{2\times120}{720}=\frac{1}{3}\).

Answer:

a. \(\frac{1}{6}\)
b. \(\frac{1}{30}\)
c. \(\frac{1}{720}\)
d. \(\frac{1}{3}\)