1. Use the addition rule to calculate the probability of one event or some other event occurring.
2. Use contingency tables to describe relationships between two variables.
3. Use the complement rule to calculate probabilities.

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.

A contingency table is a way to summarize bivariate data, or data from two variables.

• 5136 is the count of people who lived AND were not inoculated. 
• 6224 is the total number of observations.
• 244 is the total number of people who were inoculated.
• 5374 is the total number of people who lived.

• These are basically two-variable frequency distributions.
• We can convert to proportions by dividing each count by the total number of observations.

• 0.8252 is the proportion of people who lived AND were not inoculated. 
• 1.000 is the proportion of total number of observations. Think of this as 100% of the observations.
• 0.0392 is the proportion of people who were inoculated.
• 0.8634 is the proportion of people who lived.

• The row and column totals are marginal probabilities.
• The probability of two events together (\(A\) and \(B\)) is a joint probability.

Find the probability an individual was inoculated OR lived.

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.

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

Use the complement rule to find the probability a person was NOT inoculated.

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. \(P(A)\)