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six stand - up comics, a, b, c, d, e, and f, are to perform on a single…

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 first.
b. comic c will perform first and comic e will perform fifth.
c. the comedians will perform in the following order: d, c, e, b, f, a
d. comic f or comic a will perform fifth.
a. ( \frac{1}{6} )
(type a fraction. simplify your answer.)
b. ( square )
(type a fraction. simplify your answer.)

Explanation:

Step1: Calculate total number of permutations

The total number of ways to arrange \(n = 6\) comics is \(n!=\ 6! = 6\times5\times4\times3\times2\times1=720\) (using the formula for permutations of \(n\) distinct objects \(P(n,n)=n!\)).

Step2: Calculate number of favorable permutations for part b

If comic \(C\) performs first and comic \(E\) performs fifth:

  • The first - position is fixed for \(C\) (1 way)
  • The fifth - position is fixed for \(E\) (1 way)
  • The remaining \(6 - 2=4\) positions (positions 2,3,4,6) can be filled with the remaining \(6- 2 = 4\) comics. The number of ways to arrange these 4 comics is \(4!=\ 4\times3\times2\times1 = 24\) (using the formula \(P(k,k)=k!\) where \(k = 4\))

Step3: Calculate the probability for part b

The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P=\frac{4!}{6!}\)
Since \(6! = 6\times5\times4!\), then \(P=\frac{4!}{6\times5\times4!}=\frac{1}{30}\)

Answer:

\(\frac{1}{30}\)