AMC 8 · 2019 · #15

Grade 6 probability

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

📏 Short solution 💡 2 insights

Problem

On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is $\frac{2}{5}$ . If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?

$\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$

Pick an answer.

(A)

$\frac{14}{85}$

(B)

$\frac{7}{25}$

(C)

$\frac{2}{5}$

(D)

$\frac{4}{7}$

(E)

$\frac{7}{10}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: $50$ people wear sunglasses (set $S$, so $|S| = 50$); $35$ people wear caps (set $C$, so $|C| = 35$); Of the cap-wearers, $\frac{2}{5}$ also wear sunglasses; Answer choices: (A) $\frac{14}{85}$, (B) $\frac{7}{25}$, (C) $\frac{2}{5}$, (D) $\frac{4}{7}$, (E) $\frac{7}{10}$

Unknowns: The fraction of sunglasses-wearers who also wear a cap

Understand

Plan

Primary tool: #12 Draw a Venn Diagram

Secondary: #16 Change Focus / Complement

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.

Execute — Answer: B

#12 Draw a Venn Diagram K.MD.B.3 Step 1

Draw two overlapping circles labeled $S$ (sunglasses, $50$ people) and $C$ (caps, $35$ people).
The shaded overlap region is the unknown we need to fill in first — the number of people in BOTH circles.

$$|S| = 50, \quad |C| = 35, \quad |S \cap C| = ?$$

💡 Sorting people into categories (sunglasses-only, cap-only, both) is the kindergarten skill of putting objects into groups and counting each group.

#12 Draw a Venn Diagram 4.NF.B.4 Step 2

Use the cap-wearers' fraction to fill in the overlap.
The problem says $\frac{2}{5}$ of the $35$ cap-wearers also wear sunglasses, so the overlap is $\frac{2}{5}$ of $35$.

$$|S \cap C| = \tfrac{2}{5} \times 35 = \tfrac{2 \times 35}{5} = \tfrac{70}{5} = 14$$

💡 Multiplying a fraction by a whole number to find 'part of a group' is the Grade 4 fraction skill of finding $\frac{2}{5}$ of something.

#16 Change Focus / Complement 6.RP.A.3 Step 3

Now switch perspective.
The same $14$ people in the overlap are also part of the sunglasses circle.
Out of the $50$ sunglasses-wearers, $14$ of them also wear a cap, so the new probability is $\frac{14}{50}$.

$$P(\text{cap} \mid \text{sunglasses}) = \dfrac{|S \cap C|}{|S|} = \dfrac{14}{50}$$

💡 Switching the 'whole' from the cap group to the sunglasses group is ratio reasoning: the part stays $14$, the whole changes from $35$ to $50$.

#12 Draw a Venn Diagram 4.NF.A.1 Step 4

Simplify the fraction by dividing the top and the bottom by their greatest common factor, which is $2$.
The result matches answer choice (B).

$$\dfrac{14}{50} = \dfrac{14 \div 2}{50 \div 2} = \dfrac{7}{25} \;\Rightarrow\; \textbf{(B)}$$

💡 Reducing $\tfrac{14}{50}$ to $\tfrac{7}{25}$ is the Grade 4 equivalent-fractions skill — same value, smaller numbers.

[1] #12 K.MD.B.3 Draw two overlapping circles labeled $S$ (sunglasses, $50$ people) and $C$ (caps

[2] #12 4.NF.B.4 Use the cap-wearers' fraction to fill in the overlap. The problem says $\frac{2}

[3] #16 6.RP.A.3 Now switch perspective. The same $14$ people in the overlap are also part of the

[4] #12 4.NF.A.1 Simplify the fraction by dividing the top and the bottom by their greatest commo

Review

Reasonableness: 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 $\tfrac{2}{5} \times 35 = \tfrac{7}{25} \times 50 = 14$ also confirms both fractions point to the same $14$ people.

Alternative: Tool #13 (Convert to Algebra): set $|S \cap C| = x$. From the cap side, $\tfrac{x}{35} = \tfrac{2}{5}$, giving $x = 14$. Then the sunglasses-side probability is $\tfrac{x}{50} = \tfrac{14}{50} = \tfrac{7}{25}$. Same answer, slightly more formal. The Venn diagram path is more intuitive for an elementary student because the overlap is literally drawn.

CCSS standards used (min grade 6)

K.MD.B.3 Classify objects into given categories and count the numbers in each (Sorting the beachgoers into the categories sunglasses-only, cap-only, and both, which the Venn diagram makes visible.)
4.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $\tfrac{2}{5} \times 35 = 14$ to find how many cap-wearers also wear sunglasses (the Venn overlap).)
4.NF.A.1 Explain why a fraction is equivalent to another fraction (Reducing $\tfrac{14}{50}$ to $\tfrac{7}{25}$ by dividing numerator and denominator by $2$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Switching the reference whole from the $35$ cap-wearers to the $50$ sunglasses-wearers while the overlap of $14$ stays the same — a ratio-reasoning move.)

⭐ 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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