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.