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AMC 10 & 12

At Least / At Most

A meta-lesson: recognizing 'at least/at most/exactly' phrasing and picking between complement counting, direct casework, and inclusion-exclusion.

This isn't a new formula — it's a decision, and making it well is often the difference between a two-line solution and a ten-line one. Whenever a problem says “at least,” “at most,” or “exactly,” pause and consider these three routes:

  • Complement counting — often fastest for “at least 1,” since the complement (“none”) is usually a single clean case.
  • Direct casework — sum the counts for each specific case (exactly 1, exactly 2, etc.). Sometimes easier than the complement, especially when “none” isn't simple either.
  • Inclusion-exclusion — needed when “at least one of several possible events” involves overlapping conditions.
P(at least k)=1−P(fewer than k)P(\text{at least } k) = 1 - P(\text{fewer than } k)

A quick gut check: if “none” or “zero” is a single easy-to-count case, complement first. If the direct cases are each about equally easy to count (and there aren't too many of them), direct casework can be just as fast and sometimes clearer.

Worked example

Problem: A 5-card hand is dealt from a standard 52-card deck. Find the probability it contains at least one ace.

Solution (complement): “No aces” means all 5 cards come from the 48 non-ace cards:

P(no aces)=(485)(525)P(\text{no aces}) = \dfrac{\binom{48}{5}}{\binom{52}{5}}

So P(at least one ace)=1−(485)(525)P(\text{at least one ace}) = 1 - \dfrac{\binom{48}{5}}{\binom{52}{5}}, which computes to about 0.3410.341.

Solution (direct casework, as a check): Sum over exactly 1, 2, 3, or 4 aces:

∑k=14(4k)(485−k)(525)\sum_{k=1}^{4} \dfrac{\binom{4}{k}\binom{48}{5-k}}{\binom{52}{5}}

This gives the same value — but required four separate terms instead of one, confirming the complement was the better route here.

Practice problems

1.A fair coin is flipped 4 times. Find the probability of getting at least one head.

2.Three dice are rolled. Find the probability that at least one shows a 6.

3.A committee of 4 is chosen from 6 men and 5 women. Find the probability it has at least 1 woman.

4.Explain why complement counting would be a poor choice for finding “the probability of getting exactly 2 heads in 10 coin flips.”

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