Conditional probability

Conditional probability shrinks the sample space. $P(A \mid B)$, read the probability of A given B, is defined as $\frac{P(A \cap B)}{P(B)}$ whenever $P(B) > 0$: the outcomes where both happen, divided by the outcomes where the condition holds. The event after the bar is the denominator. Nothing in that definition requires $B$ to happen first in time, only that you are counting inside it.

Take 1,000 drivers followed for a year. 200 texted at the wheel and 800 did not; 40 of the texting drivers had an accident and 10 of the others did.

Both conditional probabilities are built from the same 40. $P(\text{texting} \mid \text{accident}) = 40/50 = 0.80$, while $P(\text{accident} \mid \text{texting}) = 40/200 = 0.20$. Same numerator, different denominator, a factor of four apart.

Here is the sentence readers actually get wrong: 80 percent of accidents involved a texting driver, so texting gives you an 80 percent chance of crashing. The first clause is $P(\text{texting} \mid \text{accident})$ and the second is $P(\text{accident} \mid \text{texting})$, which is 0.20. Flipping the bar is not a rephrasing, it is a different quantity, and news copy makes that swap constantly.

As long as both events can occur together, the two directions agree only when the events are equally likely, since dividing the definitions gives $\frac{P(A \mid B)}{P(B \mid A)} = \frac{P(A)}{P(B)}$. Here that ratio is $0.05/0.20 = 0.25$, matching $0.20/0.80$. So the gap is widest exactly where it does the most damage: a rare outcome conditioned on a common behavior, or the reverse. Topic 2.6 is where the course takes this up.