Addition Rule vs Multiplication Rule

The general addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, where $\cup$ is the union sign, read "A or B", and $\cap$ is the intersection sign, read "A and B". It holds for every pair of events, with nothing to check first. The subtraction is there because any outcome living in both events was counted once inside $P(A)$ and a second time inside $P(B)$, so it has to come back out once. "Or" is inclusive: $A \cup B$ means at least one of the two, including the outcomes where both happen.

Draw one card from a standard 52-card deck. There are 13 hearts and 4 kings, and exactly one card, the king of hearts, is both. So $P(\text{heart or king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} \approx 0.3077$. Listing them agrees: 13 hearts plus the 3 kings that are not hearts is 16.

"The two events are different, so add them: $13/52 + 4/52 = 17/52$." That counts the king of hearts twice, and the list has only 16 cards in it. Plain addition is legal only when $P(A \cap B) = 0$, which is what mutually exclusive events give you: they share no outcome at all, so the subtracted term was zero anyway. The plain sum is a consequence of the general rule, never the rule itself.

For three events the pattern extends: add the three single probabilities, subtract the three pairwise intersections, then add the triple intersection back. Roll one fair die with $A = \{2, 4, 6\}$, $B = \{4, 5, 6\}$ and $C = \{2, 3, 5\}$, so the pairs hold 2, 1 and 1 outcomes and no outcome is in all three. The rule gives $\frac{3 + 3 + 3 - 2 - 1 - 1 + 0}{6} = \frac{5}{6} \approx 0.8333$, and the union is $\{2, 3, 4, 5, 6\}$, five outcomes. Reverse those signs, adding the pairs and subtracting the triple, and you get $13/6 \approx 2.17$. A probability above 1 is the tell.

Unions of events are topic 2.7.

For independent events the rule is $P(A \cap B) = P(A)P(B)$, where $\cap$ is the intersection symbol, read A and B. Independence means learning that one event happened leaves the probability of the other unchanged, so the second factor never has to be adjusted. Flipping a fair coin and rolling a fair die qualify: listing all 12 equally likely pairs shows exactly one is heads with a 6, and $0.5 \times 1/6 = 1/12 \approx 0.0833$ agrees.

The product form is a shortcut, not the rule. The general multiplication rule $P(A \cap B) = P(A) \cdot P(B \mid A)$ holds for any two events, and the shortcut is what it collapses to when $P(B \mid A) = P(B)$. Write the general form first and let it simplify rather than choosing between two formulas.

The error worth naming: a shelf holds 8 batteries of which 2 are dead, so the chance that two batteries taken at random are both dead is $0.25 \times 0.25 = 0.0625$. Taken without replacement, the first battery removes itself from the shelf, so the second pick faces 1 dead among 7 and the answer is $\frac{2}{8} \cdot \frac{1}{7} = \frac{2}{56} \approx 0.0357$. The shortcut overstates it by 75 percent. Enumerating all 56 ordered pairs of distinct batteries confirms that 2 of them are dead pairs.

Mutually exclusive is not independent, and the product rule is at its worst there. If $P(A) = 0.3$, $P(B) = 0.5$, and the two events cannot both occur, then $P(A \cap B) = 0$ while the product returns 0.15.

The gap closes as the population grows. Take 2 batteries from 1,000 of which 250 are dead and the exact value is $\frac{250}{1000} \cdot \frac{249}{999} \approx 0.06231$ against 0.0625 from the shortcut, a difference of 0.3 percent. That is what the 10 percent condition formalizes: sampling without replacement may be treated as approximately independent when the sample is at most 10 percent of the population.