Answer :
Answer
a. 1/6
b. 1/30
c. 1/720
d. 1/3
Step-by-step explanation
a. First, we need to calculate the number of arrangements (like A,B,C,D,E,F; A,C,D,B,F,E, etc.) we can make with six stand-up comics. To calculate this we need to calculate the number of permutations of n = 6 people chosen r = 6 people at a time, as follows:
[tex]\begin{gathered} nPr=\frac{n!}{(n-r)!} \\ _6P_6=\frac{6!}{(6-6)!}=6! \end{gathered}[/tex]If comic D performs fourth, then we have 5 people (A, B, C, E, and F) and 5 places to order them (1st, 2nd, 3rd, 5th, and 6th). To find the number of ways these arrangements can be made, we need to calculate the number of permutations with n = 5 people chosen r = 5 people at a time, as follows:
[tex]_5P_5=\frac{5!}{(5-5)!}=5![/tex]Finally, the probability that Comic D will perform fourth is:
[tex]\begin{gathered} P(\text{ Comic D will perform fourth })=\frac{number\text{ of ways 5 comics can be ordered}}{number\text{ of ways 6 comics can be ordered}} \\ P(\text{ Comic D will perform fourth })=\frac{5!}{6!} \\ P(\text{ Comic D will perform fourth })=\frac{5!}{6\cdot5!} \\ P(\text{ Comic D will perform fourth })=\frac{1}{6} \end{gathered}[/tex]b. If Comic C performs last and Comic B performs fifth, then we need to arrange 4 people in 4 places, that is, we need to calculate the number of permutations with n = 4 people chosen r = 4 people at a time.
[tex]_4P_4=\frac{4!}{(4-4)!}=4![/tex]In consequence, the probability that Comic C will perform last and Comic B will perform fifth is:
[tex]\begin{gathered} P(\text{ Comic C will perform last and Comic B will perform fifth})=\frac{number\text{ of ways 4 com}\imaginaryI\text{cs can be ordered}}{number\text{ o fways 6 com}\imaginaryI\text{cs can be ordered}} \\ P(\text{ Comic C will perform last and Comic B will perform fifth})=\frac{4!}{6!} \\ P(\text{ Comic C will perform last and Comic B will perform fifth})=\frac{4!}{6\cdot5\operatorname{\cdot}4!} \\ P(\text{Com}\imaginaryI\text{c C w}\imaginaryI\text{ll perform last and Com}\imaginaryI\text{c B w}\imaginaryI\text{ll perform f}\imaginaryI\text{fth})=\frac{1}{30} \end{gathered}[/tex]c. The order: D, C, B, A, E, F is only one of the total possible arrangements. Then, the probability that the comedians will perform in this order is:
[tex]\begin{gathered} P(D,C,B,A,E,F\text{ order})=\frac{number\text{ o fways this order can be made}}{number\text{ of ways 6 com}\imaginaryI\text{cs can be ordered}} \\ P(D,C,B,A,E,F\text{ order})=\frac{1}{6!} \\ P(D,C,B,A,E,F\text{ order})=\frac{1}{720} \end{gathered}[/tex]d. The probability that Comic E or Comic F will perform first is computed as follows:
[tex]P(Comic\text{ E will perform first or Comic F will perform first })=P(Comic\text{ E will perform first\rparen+ P\lparen Comic F will perform first\rparen}[/tex]The probability that Comic E (or F) will perform first is numerically equivalent to the probability that Comic D will perform fourth (which was calculated in item a). Therefore,
[tex]\begin{gathered} P(Comic\text{ E will perform first or Comic F will perform first })=\frac{1}{6}+\frac{1}{6} \\ P(Com\imaginaryI c\text{ E w}\imaginaryI\text{ll perform f}\imaginaryI\text{rst or Com}\imaginaryI\text{c F w}\imaginaryI\text{ll perform f}\imaginaryI\text{rst})=\frac{2}{6} \\ P(Com\imaginaryI c\text{ E w}\imaginaryI\text{ll perform f}\imaginaryI\text{rst or Com}\imaginaryI\text{c F w}\imaginaryI\text{ll perform f}\imaginaryI\text{rst})=\frac{1}{3} \end{gathered}[/tex]