Math Core

Probability is counting with a denominator. Almost every probability problem on MATHCOUNTS and the AMC 8 comes down to two counts: how many outcomes are possible, and how many of them are the ones you want. Everything from the earlier modules (casework, the multiplication principle, combinations, complements) goes to work here.

Equally likely outcomes

Probability

If all outcomes are equally likely, the probability of an event is

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

A probability is always between $0$ (impossible) and $1$ (certain).

Rolling one fair die: $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ . Drawing one card from $1$ to $20$ : $P(\text{prime}) = \dfrac{8}{20} = \dfrac{2}{5}$ (the primes are $2, 3, 5, 7, 11, 13, 17, 19$ ).

Two dice: use the $6 \times 6$ grid

When two dice are rolled, think of them as different colors so that $(2, 5)$ and $(5, 2)$ are different outcomes. There are $6 \cdot 6 = 36$ equally likely outcomes. The table shows the sum for each one.

+	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

A sum of $7$ appears $6$ times (the diagonal), so $P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$ .

Common mistake

The $11$ possible sums ( $2$ through $12$ ) are not equally likely, so $P(\text{sum} = 7)$ is not $\dfrac{1}{11}$ . Always count equally likely outcomes, like the $36$ ordered pairs.

Complements and "and"

Two more tools:

Complement: $P(\text{not } A) = 1 - P(A)$ . For "at least one," find the chance of "none" and subtract from $1$ .
Independent events: if one event doesn't affect the other, $P(A \text{ and } B) = P(A) \cdot P(B)$ .

Worked example: At least one head

Three fair coins are flipped. What is the probability of at least one head?

The complement is "no heads," which is $TTT$ : probability $\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{8}$ . So $P(\text{at least one head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$ .

Worked example: Drawing without replacement

A bag holds $3$ red and $5$ blue marbles. Two marbles are drawn at random without replacement. What is the probability both are red?

Counting: there are $\dbinom{8}{2} = 28$ equally likely pairs, and $\dbinom{3}{2} = 3$ of them are both red. The probability is $\dfrac{3}{28}$ .

Step by step: the first is red with probability $\dfrac{3}{8}$ . Then $2$ reds remain among $7$ marbles, so the second is red with probability $\dfrac{2}{7}$ . Multiply: $\dfrac{3}{8} \cdot \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}$ .

Geometric probability

When a point is chosen at random from a segment (or region), probability is a ratio of lengths (or areas).

Worked example: Closer to 3 than to 8

A number $x$ is chosen at random between $0$ and $10$ . What is the probability that $x$ is closer to $3$ than to $8$ ?

The point halfway between $3$ and $8$ is $5.5$ . Numbers less than $5.5$ are closer to $3$ .

The favorable part has length $5.5$ out of a total length $10$ , so the probability is $\dfrac{5.5}{10} = \dfrac{11}{20}$ .

Practice

Practice 1

A card is drawn at random from cards numbered $1$ to $20$ . What is the probability that its number is a multiple of $3$ or a multiple of $5$ ? Express your answer as a common fraction.