8.3 Conditional probability

The phrase “given that” changes a probability question. Once you know a result passed a filter, results outside that filter no longer belong in the denominator. This idea is needed for Bayes' theorem, classifier questions, and probability trees.

This is intermediate probability. Start with OpenStax's conditional-probability examples. A table of counts is often easier to read than a formula at first.

A basic idea: which group are we asking about?

Suppose a batch contains $100$ items, of which $20$ are faulty. If we choose an item at random without knowing anything else, the chance that it is faulty is $20/100=0.20$. That is the ordinary, or unconditional, probability: the denominator is the whole batch.

Now suppose we are told that the chosen item was flagged by a test. The table shows that only $10$ items were flagged: $8$ faulty and $2$ sound. The question has changed. We are no longer asking about faults among all $100$ items; we are asking about faults among the $10$ flagged items. That chance is $8/10=0.80$. The test did not change the item. The new information changed which group we consider possible.

Let $A$ mean “faulty” and $B$ mean “flagged.” We write this new question as $P(A\mid B)$, read “the probability of $A$ given $B$.” The event after the bar is the known condition. It supplies the new denominator; we then count how much of that smaller group also belongs to the event before the bar.

A quick tour of rules

If $P(B)>0$, then

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$$

Rearrange to get $P(A\cap B)=P(B)P(A\mid B)$. You may instead condition on $A$, but $P(A\mid B)$ and $P(B\mid A)$ generally differ. Always label which event is the known condition.

Application 1: a filtered die roll

A fair die shows an even number. What is the probability that it is greater than $3$?

1. The condition leaves $\{2,4,6\}$.
2. Of these three equally likely outcomes, $\{4,6\}$ are greater than $3$.
3. The answer is $2/3$, not $3/6$ from the original die.

Application 2: a flagged batch

Out of 100 items, 20 are faulty. A test flags 8 faulty items and 2 sound items. If an item chosen at random is flagged, what is the chance it is faulty?

1. The known condition is “flagged,” which occurs for $8+2=10$ items.
2. The overlap “faulty and flagged” contains $8$ items.
3. $P(\text{faulty}\mid\text{flagged})=8/10=0.8$.

By contrast, $P(\text{flagged}\mid\text{faulty})=8/20=0.4$. Reversing the condition changes the denominator.