2018 AMC 10B Problem 6

A box contains

5

chips, numbered

1,

2,

3,

4,

and

5.

Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds

4.

What is the probability that

3

draws are required?

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Solution

We need a third draw exactly when the first two chips still sum to

4

or less. The only such unordered pairs are

\{1,2\}

and

\{1,3\}.

Each can be drawn in either order, giving

4

favorable ordered prefixes. Imagine that a complete random ordering of all five chips is chosen in advance. Then all

5\cdot4=20

ordered first-two-chip prefixes are equally likely, even when the actual process would stop after the first chip. Thus the probability is

\frac{4}{20}=\frac{1}{5}.

Therefore, the answer is D.