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2018 AMC 10B Problem 6

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2018 AMC 10B Problem 6AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 2 of 5 · about 2 min · position P1-10

A box contains 55 chips, numbered 1,2,3,4,1, 2, 3, 4, and 55 . Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 44 . What is the probability that 33 draws are required?

  1. A)

    115\frac{1}{15}

  2. B)

    110\frac{1}{10}

  3. C)

    16\frac{1}{6}

  4. D)

    15\frac{1}{5}

  5. E)

    14\frac{1}{4}

Answer

D

Key insight

Exactly three draws happen when the first two chips sum to at most 4, which means the unordered pair is {1,2} or {1,3}: 2 of 10 pairs.

Solution

Three draws are required exactly when the first two draws do not already exceed 44, but the third draw does.

Two distinct chips from {1,2,3,4,5}\{1, 2, 3, 4, 5\} sum to at most 44 only for the pairs {1,2}\{1, 2\} (sum 33) and {1,3}\{1, 3\} (sum 44). In either case the smallest remaining chip is at least 22, so the third draw pushes the total past 44 automatically. Hence "three draws" is the same event as "the first two chips are {1,2}\{1,2\} or {1,3}\{1,3\}."

There are (52)=10\binom{5}{2} = 10 equally likely unordered pairs for the first two chips, and 22 of them work:
210=15.\frac{2}{10} = \frac{1}{5}.

The answer is (D) 15\boxed{\textbf{(D)}\ \frac{1}{5}}.

Why this works

A stopping-time question is usually easiest to phrase as a condition on the first few draws: "stops at step 3" means "not stopped by step 2, stopped by step 3." Here the second half of the condition is automatic, so the whole problem collapses to counting pairs with a small sum.

Alternative approach

Ordered draws: the first two chips can be drawn in 5⋅4=205 \cdot 4 = 20 ordered ways, and the favorable orders are (1,2),(2,1),(1,3),(3,1)(1,2), (2,1), (1,3), (3,1), giving 420=15\frac{4}{20} = \frac{1}{5} again.

The trap

Forgetting that the third draw is then automatic, or counting only one order of the pair while using all 20 ordered pairs as the total.

Common mistakes

  • Forgetting that the third draw is then automatic, or counting only one order of the pair while using all 20 ordered pairs as the total.
  • Including the pair {1,4}\{1, 4\} or {2,3}\{2, 3\} (sum 55), which already exceeds 44 after two draws.

Techniques

Organized listing / direct enumeration