Probability

Probability starts with counting

The probability of an event is the number of favorable outcomes divided by the total number of equally likely outcomes in the sample space. A bag with 5 red marbles and 3 blue marbles gives $P(\text{blue}) = \frac{3}{8}$. Nearly every probability question on the ACT, however wordy, is a translation job: turn the described collection of items into a count of favorable outcomes and a count of total outcomes, then divide.

Always check that the outcomes you are counting really are equally likely. The sum of two dice is not equally likely to be 2 or 7, but each of the 36 ordered pairs of dice is. Count pairs, not sums.

Independent and mutually exclusive events

Two rules do most of the work, and the first job is to pick the right one.

• Independent events do not affect each other. The probability that both occur is the product: $P(A \text{ and } B) = P(A) \times P(B)$.
• Mutually exclusive events cannot happen at the same time. The probability that either occurs is the sum: $P(A \text{ or } B) = P(A) + P(B)$.

Notice the pairing: and usually goes with multiplying, or usually goes with adding. A sum of two probabilities that exceeds 1 is always a warning sign that you added when you should have multiplied.

The general rules

The simple addition rule only works when the events cannot overlap. The general version works either way:

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

The overlap is subtracted once because it was already counted inside both $P(A)$ and $P(B)$. When $A$ and $B$ cannot happen together, $P(A \text{ and } B) = 0$ and the rule collapses to simple addition.

The multiplication rule broadens in the same way. For events in sequence, $P(A \text{ and } B) = P(A) \cdot P(B \mid A)$, where $P(B \mid A)$ is the probability of $B$ given that $A$ already happened. If the events are independent, $P(B \mid A) = P(B)$ and you are back to the simple product. So the first question to ask is not "multiply or add?" It is whether the two events overlap, and whether the second event's probability depends on the first.

Worked example. In a group of students, $P(\text{soccer}) = 0.5$, $P(\text{tennis}) = 0.3$, and $P(\text{both}) = 0.15$. Find $P(\text{soccer or tennis})$.

The events overlap, so use the general addition rule: $0.5 + 0.3 - 0.15 = 0.65$.

The complement

The complement of an event is everything that is not that event, and $P(\text{not } A) = 1 - P(A)$. Whenever a question says "at least one," consider the complement: it is often easier to find the probability of "none" and subtract from 1.

Conditional probability from a two-way table

A conditional probability restricted to one row or one column of a two-way table divides by that row's or column's total, not the grand total. Reading which restriction applies is the real skill; the division is easy once you pick the right numbers.

The probability that a random student buys lunch is $\frac{54}{100} = 0.54$. But the probability that a Grade 9 student buys lunch is $\frac{30}{50} = 0.6$, because "given Grade 9" restricts us to the Grade 9 row. And the probability that a student who packs a lunch is in Grade 10 is $\frac{26}{46}$, because "given packs lunch" restricts us to that column. The word after "given" or "of" names the total you divide by.

Combinations versus permutations

Counting selections depends on whether order matters. A combination counts groups where order does not matter, such as choosing a committee. A permutation counts arrangements where order does matter, such as filling distinct offices. Using the wrong one is the most common error on this topic.

${}_nP_r = n(n-1)(n-2)\cdots(n-r+1) \qquad {}_nC_r = \frac{{}_nP_r}{r!}$

Both start by multiplying $r$ descending numbers from $n$; a combination then divides by $r!$ to remove the duplicate orderings of the same group.

Worked example. How many ways can a 3-person subcommittee be selected from 6 people?

Order does not matter for a subcommittee, so this is a combination: ${}_6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

If instead the three people were to become president, treasurer, and secretary, order would matter and the answer would be ${}_6P_3 = 6 \times 5 \times 4 = 120$, which is exactly $3! = 6$ times larger.

Expected value

Expected value multiplies each possible outcome by its probability, then adds the results. It gives the long-run average payout of a game. Losses count as negative numbers, and if the problem says it costs money to play, include that cost as part of the outcomes.

Worked example. A game pays $20 with probability 0.25 and costs $4 (a loss) with probability 0.75. The expected value is $(20)(0.25) + (-4)(0.75) = 5 - 3 = 2$, or $2 per play. Check that the probabilities add to 1 before you trust the result.

Traps to avoid

• Adding probabilities of independent events instead of multiplying (look for answers above 1).
• Using a permutation when order does not matter, or the reverse. Ask: would swapping two chosen items create a different outcome?
• Dividing by the grand total of a table when the question restricts you to a row or column.
• Forgetting to subtract the overlap when two events can both happen.
• Dropping the sign of a loss in an expected value problem.