1. Use contingency tables to describe relationships between two variables.
2. Find and interpret joint and marginal probabilities.
3. Find and interpret conditional probabilities.
4. Use the multiplication rule to find the probability of two events occurring together.

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.

What can we learn about the result of smallpox if we already know something about inoculation status?

• For example, given that a person is inoculated, what is the probability of death?
• To figure this out, we restrict our attention to the 244 inoculated cases. Of these, 6 died. So the probability is 6/244.

Conditional probability: the probability of some event \(A\) if we know that event \(B\) occurred (or is true): \[P(A|B) = \frac{P(A\text{ and }B)}{P(B)}\] where the symbol | is read as “given”.

• For death given inoculation, \[\begin{align} P(\text{death}|\text{inoculation}) &= \frac{P(\text{death and inoculation})}{P(\text{inoculation})} \\ &= \frac{0.0010}{0.0392} = 0.0255 \end{align}\]
• We could also write this as \[\begin{align} P(\text{death}|\text{inoculation}) &= \frac{P(\text{death and inoculation})}{P(\text{inoculation})} \\ &= \frac{6/6224}{244/6224} = \frac{6}{244} \end{align}\]

If knowing whether event \(B\) occurs tells us nothing about event \(A\), the events are independent. For example, if we know that the first flip of a (fair) coin came up heads, that doesn’t tell us anything about what will happen next time we flip that coin.

We can test for independence by checking if \(P(A|B)=P(A)\).

If \(A\) and \(B\) are independent events, then \[P(A \text{ and }B) = P(A)P(B).\]

• We can extend this to more than two events: \[P(A \text{ and }B \text{ and } C \text{ and } \dots) = P(A)P(B)P(C)\dots.\]
• Note that if \(P(A \text{ and }B) \ne P(A)P(B)\), then \(A\) and \(B\) are not independent.

Find the probability of rolling a \(6\) on your first roll of a die and a \(6\) on your second roll.

Let \(A=\) (rolling a \(6\) on first roll) and \(B=\) (rolling a \(6\) on second roll). For each roll, the probabiltiy of getting a \(6\) is \(1/6\), so \(P(A) = \frac{1}{6}\) and \(P(B) = \frac{1}{6}\).

Then, because each roll is independent of any other rolls, \[P(A \text{ and }B) = P(A)P(B) = \frac{1}{6}\times\frac{1}{6} = \frac{1}{36}\]

If \(A\) and \(B\) are any two events, then \[P(A \text{ and }B) = P(A|B)P(B).\]

• This is just the conditional probability formula, rewritten in terms of \(P(A \text{ and }B)\)!

Suppose we know that 38.4% of US households have dogs and that among those with dogs, 23.1% have cats. Find the probability that a US household has both dogs and cats.

Section 4.4 exercises 1-4