Math Core

Most AMC probability problems are counting problems wearing a costume. If every outcome is equally likely, a probability is just a ratio of two counts, and everything from the last three modules applies. The skill is choosing a sample space where the outcomes really are equally likely and both counts are easy.

Probability as a ratio

Equally likely outcomes

If an experiment has $N$ equally likely outcomes and $G$ of them are "good,"

$P(\text{good}) = \frac{G}{N}.$

Complement rule: $P(\text{not } E) = 1 - P(E)$ . Use it whenever "at least one" appears.

The phrase "equally likely" matters. When you roll two dice, the sums $2$ through $12$ are not equally likely, but the $36$ ordered pairs $(\text{first}, \text{second})$ are. Always build your sample space from outcomes that are clearly symmetric: ordered pairs of dice, sequences of coin flips, all $\dbinom{n}{k}$ subsets, all $n!$ orderings.

Choosing the sample space

You can often choose between counting ordered outcomes and unordered ones. Either works as long as you use the same choice for the good outcomes and the total. For drawing $3$ cards, you can count $\dbinom{52}{3}$ unordered hands or $52 \cdot 51 \cdot 50$ ordered draws; the ratio comes out the same.

A powerful shortcut is to count only what matters. For "the probability that Ann and Bo sit next to each other when $10$ people sit in a row at random," you don't need all $10!$ seatings. Ann and Bo occupy a random pair of seats, all $\dbinom{10}{2} = 45$ pairs equally likely, and $9$ of those pairs are adjacent. So the probability is $\dfrac{9}{45} = \dfrac15$ .

Conditional probability

"Given that $B$ happened" means you shrink the sample space to the outcomes in $B$ :

$P(A \mid B) = \frac{\#(A \text{ and } B)}{\#(B)}.$

For equally likely outcomes, just list or count the outcomes in $B$ , then count how many of those are also in $A$ .

Worked example: Dice and primes

Two fair dice are rolled. What is the probability that the sum is prime?

The $36$ ordered pairs are equally likely. The prime sums are $2, 3, 5, 7, 11$ , which occur in $1, 2, 4, 6, 2$ ways. So $G = 15$ and the probability is $\dfrac{15}{36} = \dfrac{5}{12}$ .

Worked example: Complement

Three different numbers are chosen at random from $\{1, 2, \dots, 10\}$ . What is the probability that their product is even?

The product is odd only if all three numbers are odd. There are $\dbinom{5}{3} = 10$ all-odd choices out of $\dbinom{10}{3} = 120$ . So

$P(\text{even}) = 1 - \frac{10}{120} = \frac{11}{12}.$

Worked example: Conditional

Three fair dice are rolled, and the sum is $10$ . What is the probability that at least one die shows a $6$ ?

Count ordered triples with sum $10$ . With $y_i = d_i - 1$ , we need $y_1 + y_2 + y_3 = 7$ with each $y_i \le 5$ : $\dbinom{9}{2} - 3\dbinom{3}{2} = 36 - 9 = 27$ .

Now count those with a $6$ . Two dice can't both be $6$ (that's already $12$ ). If one specific die is $6$ , the other two sum to $4$ : $(1,3), (2,2), (3,1)$ , so $3$ ways. Three choices of which die: $9$ triples.

The probability is $\dfrac{9}{27} = \dfrac13$ .

Common mistake

Don't treat unequal outcomes as equal. "Two coins: the outcomes are $0$ , $1$ or $2$ heads, so $P(1 \text{ head}) = \tfrac13$ " is wrong. The equally likely outcomes are HH, HT, TH, TT, so $P(1 \text{ head}) = \tfrac24 = \tfrac12$ . Label your dice, coins and people so they're distinguishable, even when the problem doesn't.

Worked example: Geometry of a polygon

Three vertices of a regular octagon are chosen at random. What is the probability that they form a right triangle?

A triangle inscribed in a circle is a right triangle exactly when one side is a diameter. The octagon has $4$ diameters (pairs of opposite vertices), and each can be paired with any of the other $6$ vertices, giving $24$ right triangles. No triangle contains two diameters, so there's no double counting. The total is $\dbinom{8}{3} = 56$ , so the probability is $\dfrac{24}{56} = \dfrac37$ .

Practice

Practice 1

Six fair coins are flipped. What is the probability of exactly three heads?

Practice 2

Two fair dice are rolled. What is the probability that the product of the two numbers is even? Give your answer as a fraction.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Two different numbers are chosen at random from $\{1, 2, \dots, 9\}$ . What is the probability that their sum is divisible by $3$ ?

Practice 4

Four people are each equally likely to have been born on any of the seven days of the week, independently. What is the probability that all four were born on different days of the week?

Practice 5

Three different numbers are chosen at random from $\{1, 2, \dots, 10\}$ . What is the probability that no two of them are consecutive? Give your answer as a fraction.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Two fair dice are rolled. Given that at least one die shows a $6$ , what is the probability that the sum is at least $10$ ?

Practice 7

Twelve people, including Cal and Dee, are seated at random around a round table with $12$ seats. What is the probability that Cal and Dee sit next to each other?

Practice 8

Integers $a$ and $b$ are chosen independently and at random from $\{1, 2, \dots, 20\}$ ( $a = b$ is allowed). What is the probability that $a^2 + b^2$ is divisible by $5$ ?

Practice 9

Five different numbers are chosen at random from $\{1, 2, \dots, 10\}$ . What is the probability that the median of the five numbers is $6$ ?

Real contest practice

2022 AMC 10B, Problem 12: an "at least once" dice question, made easy by the complement.
2021 AMC 10B, Problem 22: a harder probability that combines counting with inclusion-exclusion.