AMC 10 · 2016 · #17

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

probability-basicratio-proportionfloor-function bound-inequality-then-enumerateconvert-to-algebra ↑ Prerequisites: probability-basic

📏 Long solution 💡 4 insights

Problem

One red ball and N green balls (with N a multiple of 5) are placed in a line in random order. P(N) is the probability that at least

\frac{3}{5}

of the green balls land on the same side of the red ball. You are told P(5)=1 and that P(N)→

\frac{4}{5}

for large N. Find the smallest N with P(N) ＜

\frac{321}{400}

, then add up the digits of that N.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

The green balls are identical, so a random line-up is decided entirely by where the red ball slips in. Tool #16 (Change Focus) makes this the model: lay down the N green balls, and the red ball lands in one of the N+1 gaps, each gap equally likely. Now P(N) is just $\frac{favorable gaps}{N+1}$ . Tool #4 (Introduce a Variable) writes N=5k so the threshold $\frac{3}{5}$ N=3k is a clean whole number and the gaps are easy to count. Tool #13 (Convert to Algebra) turns P(N) ＜ $\frac{321}{400}$ into an inequality in k, which solves to a smallest k; from there N=5k and the digit sum follow.

1STEP 1

Reframe as dropping the red ball in a gap

Identical green balls make N+1 equally likely gaps for the red ball; with L green balls left, N-L sit right as L runs 0…N.

outcomes=N+1, L∈{0,1,…,N}, right pile=N-L

2STEP 2

Count the gaps that work

With N=5k the threshold is 3k; L ≥ 3k gives 2k+1 gaps and L ≤ 2k gives 2k+1 more, non-overlapping, so 4k+2 favorable gaps.

L ≥ 3k: 2k+1 gaps; L ≤ 2k: 2k+1 gaps; favorable=4k+2

3STEP 3

Write the probability and sanity-check it

Divide to get $\frac{4k+2}{5k+1}$ ; it checks out: k=1 gives $\frac{6}{6}$ =1 and large k gives ≈ $\frac{4}{5}$ , both matching the clues.

P(N)=

\frac{4k+2}{5k+1}

; k=1→

\frac{6}{6}

=1; k→∞→

\frac{4}{5}

4STEP 4

Solve the inequality for k

Solve $\frac{4k+2}{5k+1}$ ＜ $\frac{321}{400}$ by cross-multiplying: 479 ＜ 5k, so k ＞ 95.8 and the smallest whole k is 96.

400(4k+2) ＜ 321(5k+1)→ 479 ＜ 5k→ k ＞ 95.8→ k=96

5STEP 5

Find N and add its digits

So the least N=5k=480, whose digits 4, 8, 0 add to 12, giving answer (A).

N=5· 96=480, 4+8+0=12→(A)

Answer

The formula P(N)=

\frac{4k+2}{5k+1}

matches both anchors the problem hands you: P(5)=1 and the limit

\frac{4}{5}

, so the count of favorable gaps is trustworthy. Since P(N) decreases toward

\frac{4}{5}

=0.8 and

\frac{321}{400}

=0.8025 sits just above 0.8, the crossing happens at a large N, which fits N=480. Checking the boundary: at k=95,

\frac{382}{476}

=0.8025… which is not below

\frac{321}{400}

, while at k=96,

\frac{386}{481}

≈ 0.8025 ＜ 0.8025 holds, so N=480 is genuinely the first one to dip under.

💡Key takeaway

Since the green balls are identical, only the red ball's gap matters; count the gaps that leave a big enough pile, write the probability as a fraction in k, and solve the inequality to land on N=480 and digit sum 12.

Reframe as dropping the red ball in a gap
Count the gaps that work
Write the probability and sanity-check it
Solve the inequality for k
Find N and add its digits

Reframe as dropping the red ball in a gap

Count the gaps that work

Write the probability and sanity-check it

Solve the inequality for k

Find N and add its digits

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