Probability: counting, complements and combined events

Probability is counting

When all outcomes are equally likely,

$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$

so many probability questions are counting questions with a division at the end; the tools in the counting lesson apply directly. A probability is always between 0 and 1, and the probabilities of all the separate possible outcomes add to 1. Both facts are quick checks on any answer.

The complement

$$P(\text{not } A) = 1 - P(A)$$

The complement is the fastest route whenever the question says "at least one". Instead of adding the chances of exactly one, exactly two, exactly three, find the chance of none and subtract it from 1. If each of three independent attempts succeeds with probability $0.4$, the chance that all three fail is $0.6^3 = 0.216$, so the chance of at least one success is $1 - 0.216 = 0.784$.

"And": multiply

The probability that $A$ and $B$ both happen is $P(A) \times P(B \text{ given } A)$. If the events are independent (one tells you nothing about the other, such as separate coin tosses or separate people's results), that is simply $P(A) \times P(B)$.

Draws without replacement are not independent: each draw changes what is left. From a bag of 5 red and 3 blue counters, the chance that two draws are both red is $\frac{5}{8} \times \frac{4}{7}$, because after one red is gone 4 of the remaining 7 are red. The counting route agrees: $\frac{\binom{5}{2}}{\binom{8}{2}}$.

"Or": add, then remove the overlap

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

If $A$ and $B$ cannot happen together (mutually exclusive), the last term is zero and the probabilities simply add. Otherwise the outcomes in both are counted twice and must be subtracted once, exactly as with overlapping sets. For an integer chosen from 1 to 60, 15 are divisible by 4 and 10 by 6; the 5 multiples of 12 are in both, so $\frac{15 + 10 - 5}{60} = \frac{1}{3}$ are divisible by 4 or 6.

"Exactly one": count every order

The chance of exactly one red in two draws is not $\frac{5}{8} \times \frac{3}{7}$. That is red then blue; blue then red is a separate outcome with the same probability. Add both, or count with combinations ($5 \times 3$ favourable pairs out of $\binom{8}{2}$), which handles order automatically.

Choosing a method

The question says	Reach for
at least one, not all, not	the complement
both, all, and, in a row	multiplication
either, or	addition, minus the overlap
exactly $k$ of $n$	multiplication, times the number of orders
a selection from a group	combinations over combinations

Sanity checks

Before you choose, ask whether the answer is plausibly large or small. "At least one" over several tries should be bigger than a single try's chance; "all of them" should be smaller. And the exactly-zero, exactly-one and exactly-two probabilities of a two-draw question must add to 1.