AMC 8 · 2019 · #15

Grade 6 probability

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

probability-basicratio-proportionset-partition identify-subproblems ↑ Prerequisites: probability-basicratio-proportion

📏 Short solution 💡 2 insights

Problem

On a beach, 50 people wear sunglasses and 35 people wear caps. Some wear both. If you pick a random cap-wearer, the chance they also have sunglasses is

\frac{2}{5}

. Now flip the question: if you pick a random sunglasses-wearer instead, what is the chance they are also wearing a cap?

Pick an answer.

(A)

$\frac{14}{85}$

(B)

$\frac{7}{25}$

(C)

$\frac{2}{5}$

(D)

$\frac{4}{7}$

(E)

$\frac{7}{10}$

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Venn Diagram

The trigger words 'both', 'also', and 'wearing X and Y' shout Venn diagram (Tool #12). Two overlapping circles for S (sunglasses) and C (caps) make the situation visible, and the key insight is that the overlap (people wearing BOTH) is one fixed number. Tool #16 (Change Focus) is the conceptual twist: the same overlap of 14 people is $\frac{14}{35}$ when viewed from the cap circle but $\frac{14}{50}$ when viewed from the sunglasses circle. We are just switching which group we treat as 'the whole' — same numerator, different denominator.

1STEP 1

Draw two overlapping circles: S (sunglasses, 50) and C (caps, 35). The overlap — people in both — is the unknown to fill first.

|S| = 50, |C| = 35, |S ∩ C| = ?

2STEP 2

The cap fraction fills the overlap: $\frac{2}{5}$ of the 35 cap-wearers wear sunglasses too, so the overlap holds 14 people.

|S ∩ C| =

\frac{2}{5}

× 35 =

\frac{2 × 35}{5}

\frac{70}{5}

= 14

3STEP 3

Switch perspective: those same 14 people sit in the 50-sunglasses circle, so the new probability is $\frac{14}{50}$ .

P(cap ∣ sunglasses) =

\frac{|S ∩ C|}{|S|}

\frac{14}{50}

4STEP 4

Reduce by the GCF of 2: $\frac{14}{50}$ becomes $\frac{7}{25}$ , which is choice (B).

\frac{14}{50}

\frac{14 ÷ 2}{50 ÷ 2}

\frac{7}{25}

→ (B)

Answer

\frac{7}{25}

The two fractions should be related by the ratio of group sizes. Since the sunglasses group (50) is larger than the cap group (35), the overlap is a smaller slice of sunglasses than of caps, so the new probability should be smaller than

\frac{2}{5}

. Check:

\frac{7}{25}

= 0.28, and

\frac{2}{5}

= 0.40. Yes, 0.28 ＜ 0.40, exactly as expected. The cross-check

\frac{2}{5}

× 35 =

\frac{7}{25}

× 50 = 14 also confirms both fractions point to the same 14 people.

💡Key takeaway

This AMC 8 problem only needs Grade 6 ratio reasoning — switching which group is 'the whole' while the overlap stays the same — that you already know!

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