Question

tyrone, maria, simone, jim, and sarah have all been invited to a dinner party. they arrive randomly and each person arrives at a different time
a. in how many ways can they arrive?
b. in how many ways can tyrone arrive first and sarah last?
c. find the probability that tyrone will arrive first and sarah last.
a. \square (type an integer)

Explanation:

Step1: Calculate the number of arrival ways for part a

The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 5\) (Tyrone, Maria, Simone, Jim, and Sarah).

$$5! = 5\times4\times3\times2\times1=120$$

Step2: Calculate the number of arrival ways for part b

If Tyrone arrives first and Sarah last, then we only need to arrange the remaining \(5 - 2=3\) people (Maria, Simone, Jim). The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 3\)

$$3! = 3\times2\times1 = 6$$

Step3: Calculate the probability for part c

The probability \(P\) of an event is given by \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). The number of favorable outcomes (Tyrone first and Sarah last) is \(3!\) and the total number of outcomes is \(5!\)

$$P=\frac{3!}{5!}=\frac{6}{120}=\frac{1}{20}$$

Answer:

a. \(120\)
b. \(6\)
c. \(\frac{1}{20}\)