Question

A committee consists of 9 members, and 3 of them are to be selected at random to represent the committee at an upcoming conference.two of the members of the committee are named Kristen and Luke. What is the probability that Kristen will be selected but Luke will not?

Enter your answer as a decimal to three decimal places. For example, if your answer is 1/3, enter 0.333

Answer 1

2/9=0.22

Answer 2

Is that right

Answer 3

The total number of committees that can be made is:
\[\left(\begin{matrix}9 \\ 3\end{matrix}\right)=\frac{9\cdot8\cdot7}{1\cdot2\cdot3}=84\]

The number of committees with Kristen on it and Luke not on it are:

(7)(6)=42

because when picking people for the 3 person committee, the first is already picked (Kristen), for the second we only have 7 people to pick from (because we cant pick Kristen or Luke), and for the 3rd person we have 6 people to choose from (because we cant pick Kristen, Luke, or the person we picked for the second place).

This give 1*7*6=42

So the probability would be:\[\frac{42}{84}=\frac{1}{2}\]

Answer 4

There's one way to pick Kristen, but for the other 2 spots there are 7 choose 2 ways to fill it.\[\frac{\left(\begin{matrix}1 \\ 1\end{matrix}\right)\left(\begin{matrix}7 \\ 2\end{matrix}\right)}{\left(\begin{matrix}9 \\ 3\end{matrix}\right)} = \frac{(1)(21)}{(84)}=\frac{1}{4}\]If you think about it intuitively, the odds of getting Kristen on the committee is 1/3. Some, but not all, of those committees will have Luke on them, and we'll have to subtract that number of committees. So the answer must be less than 1/3.