Conditional Probability vs 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.

Every probability is a number in the interval from 0 to 1, and the probabilities of all the outcomes in the sample space add to exactly 1. $P(A)$, read "the probability of A", is the long-run relative frequency of the event $A$: the share of trials on which $A$ happens, once the number of trials is large. When every outcome is equally likely you can reach that number by counting, dividing the outcomes in $A$ by the total, but counting is a special case rather than the definition.

Draw a single digit at random from 0 through 9, each with probability 0.1. Let $A$ be "the digit is 0, 1, or 2". Then $P(A) = 3/10 = 0.30$.

"A probability of 0.30 means three out of every ten." Read ten digits from a random digit table and the chance that exactly three of them fall in $A$ is 0.2668, so that reading is wrong about 73 percent of the time, and roughly 2.8 percent of ten-digit runs contain none from $A$ at all. A probability is a rate the results approach over a long run, not a quota each block of ten has to fill. What the long run does and does not promise is the content of the law of large numbers.

Two boundaries are worth naming. Counting outcomes gives the right answer only when the outcomes really are equally likely, which rules out a loaded die or a thumbtack. And probability 0 does not always mean impossible: for a continuous random variable every single exact value has probability 0, yet the variable still lands somewhere.

Probability is introduced in topic 2.4, Introduction to Probability, after simulation has already given you an empirical version of the same quantity.