Basic Probability

What you need to know

• Equally likely outcomes. Write $P(E)$ for the probability of an event $E$. Then $P(E) = \dfrac{\text{outcomes in } E}{\text{all outcomes}}$, but only when every outcome is truly equally likely. For two dice, count the $36$ ordered pairs, not the $11$ possible sums. Example: a sum of $7$ comes from $6$ of the $36$ pairs, so its probability is $\frac16$, not $\frac1{11}$.
• The opposite event (complementary counting). $P(E) = 1-P(\text{not } E)$. This is the key to "at least one" questions. Example: at least one $6$ in $4$ rolls: $1-\left(\frac56\right)^4 = 1-\frac{625}{1296} = \frac{671}{1296}$.
• Independent events multiply. For events that do not affect each other (separate coins or dice, or draws with replacement), $P(A\text{ and }B) = P(A)\cdot P(B)$. Without replacement, multiply step by step, updating the bag after each draw. Example: $2$ reds in a row from $5$ red and $3$ blue marbles, without replacement: $\frac58\cdot\frac47 = \frac{5}{14}$.
• Common sample spaces (the list of all outcomes): two dice give $36$, $n$ coins give $2^n$, and choosing $k$ of $n$ objects gives $\binom{n}{k}$.
• Symmetry. In draws without replacement, each later draw (second, third, …) has the same chances as the first, as long as you don't know what the earlier draws were. Example: the third card dealt from a shuffled deck is an ace with probability $\frac{4}{52}$, the same as the first card.

How AMC 10 tests it

• Problems 4–12: dice, coins, spinners, marbles from a bag, with outcomes few enough to count directly.
• Problems 10–18: the probability that a random arrangement or subset has a property, so the real work is the counting underneath.
• Problems 12–20: "at least one" or "two the same" questions designed for complementary counting.
• Problems 18–25: random whole numbers from $1$ to $n$ with a number theory condition (the product is even, or the numbers are relatively prime, meaning they share no factor bigger than $1$), which need casework on remainders.
• The choices are fractions in lowest terms, built to catch counts that are off by one.

Standard approaches

1. Write down the sample space and check that its outcomes are equally likely. Then count the outcomes you want.
2. If the event says "at least" or "not all the same," find the probability of the opposite and subtract from $1$.
3. Multiply along a sequence of draws, shrinking the counts as objects leave the bag. Or use $\binom{n}{k}$ if order does not matter.
4. Use symmetry: "the first ball is red" and "the last ball is red" have the same probability.
5. Do casework on the first object or the largest value when the condition depends on it.

Worked example

A bag contains $4$ red, $3$ blue, and $2$ green marbles. Three marbles are drawn at random without replacement. What is the probability that the three marbles are not all of different colors?

(A) $\frac{2}{7}$ (B) $\frac{3}{7}$ (C) $\frac{4}{7}$ (D) $\frac{5}{7}$ (E) $\frac{6}{7}$

Solution. "Not all of different colors" is the opposite of "one of each color," so find the probability of the opposite and subtract it from $1$.

Step 1: all draws. Order does not matter, so there are $\binom93 = \frac{9\cdot8\cdot7}{3\cdot2\cdot1} = 84$ equally likely draws.

Step 2: one of each color. There are three colors, so we need one red, one blue, and one green: $4\cdot3\cdot2 = 24$ draws.

Step 3: subtract from $1$. All different colors has probability $\frac{24}{84} = \frac27$. So "not all different" has probability $1-\frac27 = \frac57$.

The answer is $\boxed{\textbf{(D)}\ \frac{5}{7}}$.

Pitfalls

• Treating outcomes that are not equally likely as if they were, such as the dice sums $2$ through $12$.
• Forgetting to shrink the bag after each draw without replacement: two reds from $5$ red and $3$ blue is $\frac58\cdot\frac47$, not $\frac58\cdot\frac58$.
• Counting the outcomes you want in order but the total without order, or the other way around. Pick one way and keep it. For two reds, $\frac{5\cdot4}{\binom82} = \frac{20}{28}$ is wrong; $\frac{5\cdot4}{8\cdot7} = \frac{\binom52}{\binom82} = \frac{5}{14}$ is right.
• Answering the opposite of what was asked. Reread the last sentence before you choose.

Traps that recur

• Forgetting that rerolling all three dice (15/216) beats keeping a 4 or 5, or failing to count ordered outcomes when listing the favorable rolls. (2020 AMC 10A #25)
• Forgetting to multiply the 3 completions by the 3 · 2 choices for the first two moves (giving 3/2187 = 1/729, choice (B)), or using 3^8 as the total. (2006 AMC 10A #25)
• Adding up only the straight-up path 0 to 1 to 2 to 3 to 4, giving (1/2)(1/4)^3 = 1/128 and a guess at the smallest choice (A); the backtracking paths are exactly what turn 128 into 97. (2025 AMC 10B #24)
• Only forbidding immediate backtracking (probability 125/216) without requiring the path to stay in a unit cube, i.e. that a repeated axis be traversed back the other way. (2024 AMC 10A #24)