## Section Project: Unordered Samples and Probability, Inclusion and Exclusion

*Unordered Samples and Probability*

###### Problem 3.49.

Suppose that \(A\text{,}\) \(B\text{,}\) and \(C\) are sets such that \(B \subseteq A\text{,}\) \(C \subseteq A\text{,}\) \(B \cap C = \emptyset\text{,}\) \(|A| = 60\text{,}\) \(|B| = 25\text{,}\) and \(|C| = 20\text{.}\)

How many four-element subsets does \(A\) have?

How many five-element subsets of \(A\) contain no elements of \(B\text{?}\) How many contain no elements of \(C\text{?}\) How many contain no elements of \(B\) and no elements of \(C\text{?}\)

How many six-element subsets of \(A\) contain an element of \(B\text{?}\) How many contain an element of \(B\) or an element of \(C\text{?}\)

How many seven-element subsets of \(A\) contain at least three elements of \(B\text{?}\)

###### Definition 3.50.

Suppose that \(A\) and \(B\) are sets and \(B \subseteq A\text{.}\) If we randomly choose an element of \(A\text{,}\) then the probabilty that the element we pick is in \(B\) is \(|B|/|A|\text{.}\)

We may now rephrase our last problem in terms of probabilities.

###### Problem 3.51.

Suppose that \(A\text{,}\) \(B\text{,}\) and \(C\) are sets such that \(B \subseteq A\text{,}\) \(C \subseteq A\text{,}\) \(B \cap C = \emptyset\text{,}\) \(|A| = 60\text{,}\) \(|B| = 25\text{,}\) and \(|C| = 20\text{.}\)

Suppose I choose a five-element set from \(A\text{.}\) What is the probability that the set contains no elements of \(B\text{?}\) What is the probability that the set contains no elements of \(C\text{?}\) What is the probability that the set contains no elements of \(B\) and no elements of \(C\text{?}\)

Suppose I choose a six-element subset of \(A\text{.}\) What is the probability that the set contains an element of \(B\text{?}\) What is the probability that the set contains an element of \(B\) or an element of \(C\text{?}\)

Suppose I choose a seven-element subset of \(A\text{.}\) What is the probability that it contains at least three elements of \(B\text{?}\)

Consider the simplest game of poker where we start with a fifty-two card deck and draw hands of five cards. The goal is to have the “best” hand. For example, three sevens beats two kings. Similarly three fours and two jacks (a full house) beats three aces (three of a kind). The order, from weakest hand to strongest hand, goes like this: High Card \(\lt\) One Pair \(\lt\) Two Pair \(\lt\) Three of a Kind \(\lt\) Straight \(\lt\) Flush \(\lt\) Full House \(\lt\) Four of a Kind \(\lt\) Straight Flush. We won't define and discuss all these possibilities. Rather, we will just pick two and ask the relevant question related to probability.

Why should three of a kind beat two pair? Well it must be that you are more likely to find in a five card hand a pair of threes and a pair of kings than three queens. The next two problems will demonstrate this, first by counting and then by probability.

###### Problem 3.52.

Suppose we have a fifty-two card deck and a five card hand from that deck.

How many five card hands are there?

How many five card hands have two pair in them?

How many five card hands have three of a kind in them?

We may now rephrase our last problem in terms of probabilities.

###### Problem 3.53.

Suppose we have a fifty-two card deck and draw a five card hand from that deck.

What is the probability that there are two pair in the hand?

What is the probability that there are three of a kind in the hand?

*The Principle of Inclusion and Exclusion*

###### Theorem 3.54.

If A and B are finite sets, then \(|A \cup B| = |A| + |B| - |A \cap B|\text{.}\)

###### Problem 3.55.

Generalize Theorem 3.54 to three sets. That is, find a nice formula for \(|A \cup B \cup C| = \dots\text{.}\)

###### Problem 3.56.

Generalize Theorem 3.54 to four sets.