Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #15
Grade 6 probabilityPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 14/35 when viewed from the cap circle but 14/50 when viewed from the sunglasses circle. We are just switching which group we treat as 'the whole' — same numerator, different denominator.
Draw the two groups
Draw two overlapping circles: S (sunglasses, 50) and C (caps, 35). The overlap — people in both — is the unknown to fill first.
Sorting people into categories (sunglasses-only, cap-only, both) is the kindergarten skill of putting objects into groups and counting each group.
K.MD.B.3Draw A Venn DiagramFind the overlap
The cap fraction fills the overlap: of the 35 cap-wearers wear sunglasses too, so the overlap holds 14 people.
Multiplying a fraction by a whole number to find 'part of a group' is the Grade 4 fraction skill of finding 2/5 of something.
4.NF.B.4Draw A Venn DiagramSwitch to the sunglasses side
Switch perspective: those same 14 people sit in the 50-sunglasses circle, so the new probability is .
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.
If instead a sunglasses-wearer is picked at random, the chance the person also wears a cap is the 14 who wear both taken out of the 50 who wear sunglasses.
▸ Why?
The person is drawn at random from the sunglasses-wearers, so the chance they also wear a cap is simply how many of the 50 wear both — which is the same 14 — measured against the 50 in the group.
▸ Why?
The 14 who wear both are one fixed set of people; pointing to them from inside the cap group or from inside the sunglasses group pairs each person with themselves, so the favorable count stays 14 either way.
▸ Why?
Picking 'at random' means every sunglasses-wearer is equally likely to be the one chosen, so the probability of landing on someone who also wears a cap is their favorable count over the total of 50 sunglasses-wearers.
Simplify the fraction
Reduce by the GCF of 2: becomes , which is choice (B).
Reducing 14/50 to 7/25 is the Grade 4 equivalent-fractions skill — same value, smaller numbers.
4.NF.A.1Draw A Venn DiagramThis 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!
- Draw the two groups
- Find the overlap
- Switch to the sunglasses side
- Simplify the fraction
A parent dashboard for the family lives at sensimlab.com.