(a) First choose the team of 2 people. This can be done in \[ \binom {12}{2} \] ways. After those two people have been chosen, there are 10 people left, and we need to divide them into two teams of 5. If we choose 5 of them for one team, the remaining 5 automatically form the other team. This gives \(\binom {10}{5}\) choices, but the two teams of size 5 have no labels, so this counts each split twice. Therefore the number of possible splits is \[ \binom {12}{2}\frac {\binom {10}{5}}{2} = 66\cdot \frac {252}{2} = 8316. \]

(b) Now all three teams have the same size, so none of the teams is distinguished by its size. If the teams were labeled, we could choose 4 people for the first team, then 4 of the remaining 8 for the second team, and then the last 4 people would form the third team. This gives \[ \binom {12}{4}\binom {8}{4}\binom {4}{4}. \] Since the three teams are not labeled, each actual split has been counted \(3!\) times. Thus the desired number is \[ \frac {1}{3!}\binom {12}{4}\binom {8}{4}\binom {4}{4} = \frac {12!}{(4!)^3 3!} = 5775. \]

Imagine \(n+1\) people arranged from youngest to oldest. We want to choose a subgroup of \(k+1\) people. If we ignore the order of the people, the number of possible subgroups is simply \[ \binom {n+1}{k+1}. \]

Now count the same subgroups according to who the oldest chosen person is. If the oldest chosen person is the \((j+1)\)-st person in the age order, then the remaining \(k\) members of the subgroup must all be chosen from the \(j\) people younger than that person. This can be done in \[ \binom {j}{k} \] ways. The oldest chosen person must have at least \(k\) people younger than them, and can range up to the oldest person in the whole group. Hence \(j\) ranges from \(k\) to \(n\). Summing over all possible choices of the oldest selected person gives \[ \binom {k}{k} + \binom {k+1}{k} + \binom {k+2}{k} + \cdots + \binom {n}{k}. \]

Both expressions count the same collection of subgroups of size \(k+1\). Therefore

The addition rule gives \[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \] Since \(P(A\cup B)\le 1\), this identity implies \[ P(A)+P(B)-P(A\cap B)\le 1. \] Rearranging gives \[ P(A)+P(B)-1\le P(A\cap B). \] Equality holds here exactly when \(P(A\cup B)=1\), meaning that \(A\cup B\) occurs with probability one.

Next, since \[ A\cap B\subseteq A\cup B, \] monotonicity of probability gives \[ P(A\cap B)\le P(A\cup B). \] Equality holds exactly when the part of \(A\cup B\) outside \(A\cap B\) has probability zero. This part is the symmetric difference \[ (A-B)\cup (B-A). \] Thus equality holds exactly when \[ P\big ((A-B)\cup (B-A)\big )=0. \]

Finally, from the addition rule and the fact that \(P(A\cap B)\ge 0\), we get \[ P(A\cup B) = P(A)+P(B)-P(A\cap B) \le P(A)+P(B). \] Equality holds exactly when \[ P(A\cap B)=0. \]

Therefore all three inequalities hold, and the equality cases are as described above.

Since \(A\subseteq B\), the event \(B\) can be decomposed into two disjoint pieces, \[ B=A\cup (B-A). \] These two events are disjoint because \(B-A\) consists exactly of the outcomes in \(B\) that are not in \(A\). Therefore, by finite additivity, \[ P(B) = P(A\cup (B-A)) = P(A)+P(B-A). \] Rearranging this identity gives \[ P(B-A)=P(B)-P(A). \]

This proves the desired formula directly from the additivity axiom of probability.

There are \[ \binom {52}{13} \] equally likely 13-card hands. For each suit \(s\), let \(A_s\) be the event that the hand has no cards of suit \(s\). We want \[ P(A_{\spadesuit }\cup A_{\heartsuit }\cup A_{\diamondsuit }\cup A_{\clubsuit }). \]

We count this union by inclusion-exclusion. If one specified suit is missing, then all 13 cards must come from the remaining 39 cards, giving \[ \binom {39}{13} \] hands. There are 4 choices for the missing suit.

If two specified suits are missing, then all 13 cards must come from the remaining 26 cards, giving \[ \binom {26}{13} \] hands. There are \(\binom {4}{2}\) choices for the two missing suits.

If three specified suits are missing, then all 13 cards must come from a single suit. This is possible in \[ \binom {13}{13}=1 \] way for each choice of the three missing suits. There are \(\binom {4}{3}\) such choices. It is impossible for all four suits to be missing, since the hand has 13 cards.

Therefore the probability that the hand is void in at least one suit is \[ \frac { 4\binom {39}{13} - \binom {4}{2}\binom {26}{13} + \binom {4}{3}\binom {13}{13} }{ \binom {52}{13} }. \] Equivalently, \[ \frac { 4\binom {39}{13} - 6\binom {26}{13} + 4 }{ \binom {52}{13} }. \]

There are \(10+12+15=37\) students in the club, so the total number of possible committees of size 5 is \[ \binom {37}{5}. \]

(a) To have exactly 3 sophomores, we choose 3 of the 15 sophomores and then choose the remaining 2 committee members from the 22 students who are not sophomores. This gives \[ \binom {15}{3}\binom {22}{2} \] committees. Hence the desired probability is \[ \frac {\binom {15}{3}\binom {22}{2}}{\binom {37}{5}}. \]

(b) Now the committee must contain at least one senior, at least one junior, and at least one sophomore. We count by the triple \[ (s,j,p), \] where \(s\), \(j\), and \(p\) are the numbers of seniors, juniors, and sophomores on the committee. We need \[ s+j+p=5, \qquad s,j,p\ge 1. \] Thus the favorable committees are counted by \[ \sum _{\substack {s+j+p=5\\ s,j,p\ge 1}} \binom {10}{s}\binom {12}{j}\binom {15}{p}. \] Writing this out explicitly gives \[ \binom {10}{1}\binom {12}{1}\binom {15}{3} + \binom {10}{1}\binom {12}{2}\binom {15}{2} + \binom {10}{1}\binom {12}{3}\binom {15}{1} \] \[ + \binom {10}{2}\binom {12}{1}\binom {15}{2} + \binom {10}{2}\binom {12}{2}\binom {15}{1} + \binom {10}{3}\binom {12}{1}\binom {15}{1}. \] Dividing by the total number of committees gives the probability \[ \frac { \binom {10}{1}\binom {12}{1}\binom {15}{3} + \binom {10}{1}\binom {12}{2}\binom {15}{2} + \binom {10}{1}\binom {12}{3}\binom {15}{1} + \binom {10}{2}\binom {12}{1}\binom {15}{2} + \binom {10}{2}\binom {12}{2}\binom {15}{1} + \binom {10}{3}\binom {12}{1}\binom {15}{1} }{ \binom {37}{5} }. \]