AMC 10 · 2005 · #14
Grade 7 probabilityPick an answer.
The experiment has two stages, and the second stage depends on the first. Splitting it into subproblems by asking only one thing about stage one -- was the damaged face odd or even? -- is enough, because that single fact determines the whole answer for stage two. The important care point is that the damaged face is chosen in proportion to how many dots it carries, so the two cases are not equally likely and must be weighted by dot counts, not by face counts.
Weight the cases by dots, not faces
The dot is uniform, not the face, so each face carries its own weight.
The dot is picked, not the face, so a face with more dots is a bigger target.
7.SP.C.7Identify SubproblemsRemoving a dot flips one parity
Removing one dot flips exactly one face's parity.
Taking away one dot always switches a face between odd and even, so only the count of odd faces moves, by one.
Removing one dot always switches a single face between odd and even, so the count of odd faces moves by exactly one.
▸ Why?
Taking away one from a number flips it from odd to even or from even to odd, never leaving it as it was.
▸ Why?
The removed dot came from an odd face or from an even one and never from both, so the two cases simply add.
Probability inside each case
So the odd-face count becomes either 2 or 4.
Once the die is fixed, an odd top just means landing on one of the odd faces.
7.SP.C.7Identify SubproblemsCombine the two cases
Weighting the two cases and adding gives 11/21.
Each case contributes its own chance, scaled by how often that case happens.
5.NF.A.1Identify SubproblemsCheck by counting all outcomes
A flat count of all outcomes confirms 11/21, choice (D).
Listing every equally likely dot-and-roll pair turns the probability into a plain fraction of counts.
7.SP.C.8Organize Information In More WaysThe dot is chosen, not the face, so faces with more dots are more likely to be hit; weight each case by its dot count instead of treating the six faces as equally likely.
- Weight the cases by dots, not faces
- Removing a dot flips one parity
- Probability inside each case
- Combine the two cases
- Check by counting all outcomes