Conditional Probability vs Independent Events

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.

Events $A$ and $B$ are independent when $P(A \cap B) = P(A)P(B)$. Provided $P(B) > 0$, that is the same as $P(A \mid B) = P(A)$, read "the probability of A given B equals the probability of A", and the relation runs both ways: if $B$ tells you nothing about $A$, then $A$ tells you nothing about $B$. Independence is a numerical condition you check, not a verdict you reach from how the story sounds.

It does not require two separate physical actions. Roll one fair die, and let $A$ be "even", the set $\{2, 4, 6\}$, and $B$ be "at least 5", the set $\{5, 6\}$. Then $P(A) = 0.5$, $P(B) = 2/6 \approx 0.3333$, and $A \cap B$ is the single outcome 6, so $P(A \cap B) = 1/6 \approx 0.1667$. The product $0.5 \times 1/3 = 1/6$ matches it, so two events on the very same roll are independent, and indeed $P(A \mid B) = (1/6)/(2/6) = 0.5 = P(A)$. Widen $B$ to "at least 4" and it breaks: $P(A \cap B) = 2/6 \approx 0.3333$ against a product of $0.5 \times 0.5 = 0.25$.

"The two events cannot both happen, so one cannot affect the other, so they are independent." That gets it exactly backwards. Mutually exclusive events with nonzero probabilities are never independent, because $P(A \cap B) = 0$ while $P(A)P(B) > 0$, and those two cannot be equal. If $A$ and $B$ are disjoint, $B$ occurring forces $A$ not to occur, which is the strongest dependence available.

Selecting people without replacement destroys independence. In a class of 20 with 8 in band, picking two students at random gives $P(\text{second in band} \mid \text{first in band}) = 7/19 \approx 0.3684$, not the $8/20 = 0.40$ you started with, because the first pick shrinks the class. The ten percent condition is what lets such draws be treated as close enough to independent. Independence itself is topic 2.7.