Question

Maria, Sergio, Tyrone, Katrina, Kim, 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 Maria arrive first and Sarah last?c. Find the probability that Maria will arrive first and Sarah last.a.b.(Type an integer.)(Type an integer.)c-Type a fraction. Simplify your answer.).Time Remaining:02 20:22NextNextModule 8 Homework

Answer 1

a. 720

b. 24

c. 1/30

Explanation

a. To find in how many ways can they arrive we need to calculate the number of permutations.

The number of permutations of n things chosen r at a time is found using:

`nPr=(n!)/((n-r)!)`

In this case, we have n = 6 people, and we have to choose r = 6 of them. The number permutations is:

`_6P_6=(6!)/((6-6)!)=(6!)/(0!)=(6!)/(1)=720`

b. If Maria arrives first and Sarah last, then there are 4 places to select randomly (from 2nd to 5th) and 4 people to be ordered. Therefore, we have n = 4 people, and we have to choose r = 4 of them. The number permutations is:

`_4P_4=(4!)/((4-4)!)=4!=24`

c. The probability that Maria will arrive first and Sarah last is calculated as follows:

`\begin{gathered} p=\frac{number\text{ of ways Maria arrives first and Sarah last}}{\text{ Total }number\text{ of ways people can arrive}} \\ p=(24)/(720) \\ p=(1)/(30) \end{gathered}`