Counting the outcomes in "A or B" worked in the last lesson because we could list them. When all we know are the probabilities of A, of B and of both, we need a rule.
Key idea
$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$ If A and B are mutually exclusive (they cannot happen together), $P(A \cap B) = 0$ and $P(A \cup B) = P(A) + P(B)$.
Why it works
Adding $P(A)$ and $P(B)$ adds up the two circles of the Venn diagram. The overlap belongs to both circles, so it is counted twice. Subtracting $P(A \cap B)$ once leaves every outcome counted exactly once. When the circles do not overlap, there is nothing to subtract.
Worked example: Counts from a survey
Of 120 students, 45 play a sport, 30 are in the band, and 12 do both. What is the probability that a randomly chosen student plays a sport or is in the band?
- Add, then remove the double count: $45 + 30 - 12 = 63$ students.
- $P = \dfrac{63}{120} = \dfrac{21}{40} = 0.525$.
Answer: $0.525$
Worked example: With probabilities
$P(A) = 0.45$, $P(B) = 0.30$, $P(A \cap B) = 0.12$. Find $P(A \cup B)$ and the probability of neither.
- $$P(A \cup B) = 0.45 + 0.30 - 0.12 = 0.63$$
- "Neither" is the complement of "A or B": $1 - 0.63 = 0.37$.
Answer: $0.63$ and $0.37$.
Worked example: Mutually exclusive events
A spinner lands on red with probability $0.25$ and on blue with probability $0.4$. It cannot land on both. Find $P(\text{red or blue})$.
- Red and blue cannot happen together, so the overlap is 0.
- $P = 0.25 + 0.4 = 0.65$.
Answer: $0.65$
The same idea extends to three events: add the three probabilities, subtract the three pairwise overlaps, then add back the triple overlap, which was removed once too often. This is called inclusion and exclusion.
Common mistake
Adding $P(A) + P(B)$ without subtracting the overlap. Check: a probability can never exceed 1. If $P(A) = 0.7$ and $P(B) = 0.6$, simple adding gives 1.3, which is impossible; the events must overlap by at least 0.3.