Consider a six-sided die. \[P(\text{roll a 1 or 2}) = \frac{\text{2 ways}}{\text{6 outcomes}} = \frac{1}{3}\] We get the same result by taking \[P(\text{roll a 1})+P(\text{roll a 2}) = \frac{1}{6}+\frac{1}{6} = \frac{1}{3}\] It turns out this is widely applicable!

If \(A_1\) and \(A_2\) are disjoint outcomes, then the probability that one of them occurs is \[P(A_1 \text{ or } A_2) = P(A_1)+P(A_2)\] This can also be extended to more than two disjoint outcomes: \[P(A_1 \text{ or } A_2 \text{ or } \dots \text{ or } A_k) = P(A_1)+P(A_2)+\dots + P(A_k)\] for \(k\) disjoint outcomes.

• Let \(A\) be the event that a card drawn is a diamond.
• Let \(B\) be the event it is a face card.

\(A\): \(\quad 2\diamondsuit\) \(3\diamondsuit\) \(4\diamondsuit\) \(5\diamondsuit\) \(6\diamondsuit\) \(7\diamondsuit\) \(8\diamondsuit\) \(9\diamondsuit\) \(10\diamondsuit\) \(J\diamondsuit\) \(Q\diamondsuit\) \(K\diamondsuit\) \(A\diamondsuit\)

\(B\): \(\quad J\heartsuit\) \(Q\heartsuit\) \(K\heartsuit\) \(J\clubsuit\) \(Q\clubsuit\) \(K\clubsuit\) \(J\diamondsuit\) \(Q\diamondsuit\) \(K\diamondsuit\) \(J\spadesuit\) \(Q\spadesuit\) \(K\spadesuit\)

The collection of cards that are diamonds or face cards (or both) is

Looking at these cards, I can see that there are 22 of them, so \[P(A \text{ or } B) = \frac{22}{52}\]

• If I try to apply the addition rule for disjoint outcomes, \[P(A)=\frac{13}{52}\] and \[P(B)=\frac{12}{52}\]
• And we get \(\frac{13+15}{52} = \frac{25}{52}\).
• But we know \(P(A \text{ or } B) = \frac{22}{52}\).

What happened?

• When I added these, I double counted \(J\diamondsuit\), \(Q\diamondsuit\), and \(K\diamondsuit\)
  • (the cards that are in both \(A\) and \(B\)).
• I need to subtract off the double count \[\frac{13}{52}+\frac{12}{52}-\frac{3}{52}\]

For any two events \(A\) and \(B\), the probability that at least one will occur is \[P(A \text{ or } B) = P(A)+P(B)-P(A \text{ and }B).\]

• The general addition rule applies to any two events, even disjoint events.
  • For disjoint events, \(P(A \text{ and } B) = 0\).

• When we say “or”, we include the situations where:
  • A is true
  • B is true
  • both A and B are true.
• This is an inclusive or.

The complement of an event is all of the outcomes in the sample space that are not in the event. For an event \(A\), we denote its complement by \(A^c\).

• For a single roll of a six-sided die, the sample space is all possible rolls: 1, 2, 3, 4, 5, or 6.
• Let event \(A\) be rolling a 1 or a 2.
• Then \(A^c\) is rolling a 3, 4, 5, or 6.

In probability notation: \[S = \{1, 2, 3, 4, 5, 6\}\] \[A=\{1,2\}\] \[A^c=\{3, 4, 5, 6\}\]

\[P(A \text{ or } A^c)=1\]

• Using the addition rule, \[P(A \text{ or } A^c) = P(A)+P(A^c) = 1\]
• Make sure you can convince yourself that \(A\) and \(A^c\) are always disjoint.

\[P(A) = 1-P(A^c)\]

Consider rolling 2 six-sided dice and taking their sum.

The event of interest \(A\) is a sum less than 12. Find

1. \(A^c\)
2. \(P(A^c)\)
3. \(P(A)\)

Hint #1: how many ways can you get a sum of at least 12?

Hint #2: there are 36 possible outcomes.