# What is the probability that a randomly formed committee includes both Amy and Betty? Amy as chair and Betty as an unranked member?

Five people are to be chosen to form a committee from a group of 100 students. The committee has a Chair, Vice Chair, and three other unranked members. Suppose Amy and Betty are among the group of 100 students. What is the probability that a randomly formed committee includes both Amy and Betty? What is the probability that a randomly formed committee includes Amy as Chair and Betty as an unranked member?

I've done $$^{100}P_2×\binom{98}3$$ to see how many ways the committee of $$5$$ can be selected. I'm having trouble visualizing the ways Amy and Betty can both be on the committee, well the number of ways at least.

## 1 Answer

What is the probability that a randomly formed committee includes Amy and Betty?

Method 1: If we work with the sample space you have chosen, we have to consider each of the following possibilities:

• Amy is the Chair and Betty is the Vice Chair.
• Amy is the Chair and Betty is an unranked member.
• Amy is the Vice Chair and Betty is the Chair.
• Amy is the Vice Chair and Betty is an unranked member.
• Amy is an unranked member and Betty is the Chair.
• Amy is an unranked member and Betty is the Vice chair.
• Both Amy and Betty are unranked members.

Amy is the Chair and Betty is the Vice Chair: There are $$\binom{98}{3}$$ ways to select the three unranked members.

Amy is the Chair and Betty is an unranked member: There are $$98$$ ways to select the Vice Chair and $$\binom{97}{2}$$ ways to select the other unranked members.

Amy is the Vice Chair and Betty is the Chair: There are $$\binom{98}{3}$$ ways to select the three unranked members.

Amy is the Vice Chair and Betty is an unranked member: There are $$98$$ ways to select the Chair and $$\binom{97}{2}$$ ways to select the other unranked members.

Amy is an unranked member and Betty is the Chair: There are $$98$$ ways to select the Vice Chair and $$\binom{97}{2}$$ ways to select the other unranked members.

Amy is an unranked member and Betty is the Vice chair: There are $$98$$ ways to select the Chair and $$\binom{97}{2}$$ ways to select the other unranked members.

Both Amy and Betty are unranked members: There are $$98$$ ways to select the Chair, $$97$$ ways to select the Vice Chair, and $$96$$ ways to pick the other unranked member.

Since these cases are mutually exclusive and exhaustive, the number of favorable cases is $$\binom{98}{3} + 98\binom{97}{2} + \binom{98}{3} + 98\binom{97}{2} + 98\binom{97}{2} + 98\binom{97}{2} + 98 \cdot 97 \cdot 96$$ which can be simplified to $$2\binom{98}{3} + 4 \cdot 98\binom{97}{2} + 98 \cdot 97 \cdot 96$$ You have corrected calculated that the number of elements in your sample space is $$100 \cdot 99\binom{98}{3}$$ Hence, the probability that Amy and Betty are both selected to serve on the committee is $$\frac{2\dbinom{98}{3} + 4 \cdot 98\dbinom{97}{2} + 98 \cdot 97 \cdot 96}{100 \cdot 99\dbinom{98}{3}}$$

Method 2: We take as our sample space the $$\binom{100}{5}$$ subsets of five of the $$100$$ students who could serve on the committee.

If Amy and Betty both serve on the committee, then three of the remaining $$98$$ students must also serve on the committee. Hence, there are $$\binom{98}{3}$$ favorable cases.

Thus, the probability that Amy and Betty both serve on the committee is $$\frac{\dbinom{98}{3}}{\dbinom{100}{5}}$$

As you can verify, this gives the same result as the first method.

Notice that since we only care about who serves on the committee, we can ignore the roles of the individual members for this problem.

What is the probability that a randomly formed committee includes Amy as Chair and Betty as an unranked member?

Since it matters who serves in what role on the committee, we use your sample space. We showed above that the number of favorable cases is $$98\binom{97}{2}$$ Hence, the desired probability is $$\frac{98\dbinom{97}{2}}{100 \cdot 99\dbinom{98}{3}}$$

• For the last part, why wouldn't the desired probability be over 100*99*choose(98,3)? – Pierre Oct 12 '20 at 16:29
• @Pierre It is. Thank you for pointing out the error. – N. F. Taussig Oct 12 '20 at 18:54