Probability and Counting

Probability Measures How Likely Something Is

Probability is a number between $0$ and $1$ that measures how likely an event is to happen. An impossible event has probability $0$; a certain event has probability $1$. On the ACT Math section, probability shows up as pure counting problems, as word problems about drawing cards or picking marbles, and occasionally alongside data tables. The good news: nearly every ACT probability question rests on one idea.

The Basic Formula

For an experiment in which every outcome is equally likely,

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

If a bag holds $3$ red and $5$ blue marbles, the probability of drawing red is $\tfrac{3}{8}$, because $3$ of the $8$ equally likely marbles are red. Always count the total carefully — it is the denominator, and forgetting one outcome is the most common error.

The Complement Rule

The complement of an event is everything that is not that event. Because some outcome must occur, the two probabilities add to $1$:

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

The complement is a shortcut for "at least one" questions. Computing $P(\text{at least one})$ directly means adding many cases; computing $1 - P(\text{none})$ is usually a single line of work.

Compound Probability — "And" vs. "Or"

A compound event combines two or more simpler events, and the connecting word tells you the operation.

Two events are independent when one does not affect the other — a coin flip and a die roll, for instance. To find the chance that both occur, multiply: $P(\text{heads and a 3}) = \tfrac{1}{2} \cdot \tfrac{1}{6} = \tfrac{1}{12}$.

Events are mutually exclusive when they cannot both happen at once, such as rolling a $2$ or a $5$ on a single die. To find the chance that either occurs, add: $\tfrac{1}{6} + \tfrac{1}{6} = \tfrac{1}{3}$.

Without replacement problems are dependent: once you remove the first item, the total shrinks. Drawing two blue marbles from a bag of $3$ red and $5$ blue gives $\tfrac{5}{8} \cdot \tfrac{4}{7}$, because after one blue leaves, only $4$ blue remain out of $7$ total.

Counting — Multiply the Choices

Before you can find a probability you often must count outcomes, and the master tool is the Fundamental Counting Principle: if one stage has $m$ choices and a second independent stage has $n$ choices, the two stages together have $m \times n$ outcomes. A menu with $4$ entrées and $3$ desserts offers $4 \times 3 = 12$ meals.

Factorials

A factorial, written $n!$, is the product of every whole number from $n$ down to $1$:

$n! = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1$

So $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$. Factorials count the number of ways to arrange a full set in order: five different books can sit on a shelf in $5! = 120$ orders. By definition $0! = 1$.

Permutations vs. Combinations

The single most tested distinction here is order.

• A permutation counts arrangements where order matters (first, second, third place): $_nP_r = \frac{n!}{(n-r)!}$
• A combination counts selections where order does not matter (a committee, a hand of cards): $_nC_r = \frac{n!}{r!\,(n-r)!}$

To award gold, silver, and bronze to $3$ of $8$ runners, order matters, so use $_8P_3 = 8 \cdot 7 \cdot 6 = 336$. To simply pick a $3$-person committee from those same $8$ people, order does not matter, so use $_8C_3 = \tfrac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$. Notice the combination is smaller: it divides out the $r!$ ways each group could be ordered.

A Reliable Game Plan

Read the question and ask three things: Is this a counting problem or a probability problem? Does order matter? Are the events independent or does the pool change? Answer those, pick the matching formula, and the arithmetic is short. When in doubt on an "at least one" question, reach for the complement.