AMC 10 · 2015 · #17
Grade 7 probabilityPick an answer.
Because every coin is fair and independent, all 2⁸ = 256 stand/sit patterns are equally likely, so the probability is just (number of safe patterns)/256. The phrase "no two adjacent standing" is hard to count all at once, so Tool #7 (Identify Subproblems) splits the work by how many people stand: 0, 1, 2, 3, or 4 (five or more cannot fit without touching). The easy counts come from Tool #2 (Make a Systematic List), and the one stubborn case — exactly 3 standing — bends to Tool #16 (Count the Complement): count all triples, then throw out the ones that put two standers side by side.
Set up probability as a fraction
All 256 outcomes are equally likely.
When every outcome is equally likely, probability is just a counting job: good outcomes over all outcomes.
7.SP.C.7Identify SubproblemsSplit by how many stand
Splitting by how many stand gives five cases.
Breaking outcomes into non-overlapping cases by a clear feature turns one hard count into a few easy ones.
7.SP.C.8Identify SubproblemsCount 0, 1, and 2 standers
The small cases count directly.
List the outcomes in an orderly way and small cases count themselves.
7.SP.C.8Make A Systematic ListCount 3 standers by complement
The next uses a complement.
When the good cases are tangled, count everything and peel off the bad ones.
When the good cases are tangled, count everything and peel off the bad ones.
▸ Why?
Every outcome either qualifies or fails, so the two counts add up to the whole.
▸ Why?
The cases are told apart by a clear feature and never overlap, so the counts can simply be added.
Count 4 standers
The largest case allows only two patterns.
Packed to the limit, the only room left is a strict alternation, and a circle gives just two of those.
7.SP.C.8Make A Systematic ListAdd up and divide
Adding and dividing gives 47/256, choice (A).
Sum the non-overlapping case counts, then divide once by the total.
7.NS.A.3Identify SubproblemsEvery coin flip is equally likely, so count the safe seatings by how many people stand (0, 1, 2, 3, 4), add them to 47, and divide by 256.
- Set up probability as a fraction
- Split by how many stand
- Count 0, 1, and 2 standers
- Count 3 standers by complement
- Count 4 standers
- Add up and divide