AMC 10 · 2008 · #22
Grade 7 probabilityPick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only 3 flips ever happen, so there are just 8 possible paths. A branching tree diagram lays out every path and the fourth term it produces, so we can simply count how many of those land on a whole number.
Count the paths
Three flips separate term 1 from term 4, each with two outcomes, so there are 8 equally likely paths.
Three independent heads-or-tails choices always split into 8 equally likely paths.
Three independent two-way choices always split into eight equally likely paths.
▸ Why?
Each flip is made without regard to the others, so the option counts multiply.
▸ Why?
A fair coin makes every path just as likely, so the chance is a plain count over eight.
Write the two rules
Let x be the current term: heads sends it to and tails sends it to .
Naming the current value x turns each coin rule into one reusable formula.
6.EE.A.2Introduce A VariableGrow the tree
Apply both rules at every level: 6 branches to 11 and 2, then to 21, 4.5, 3, 0, then to eight fourth terms.
Applying both rules at every branch guarantees no possible outcome is missed.
7.NS.A.3Draw A DiagramCount integer endings
In flip order the fourth terms are 41, 9.5, 8, 1.25, 5, 0.5, -1, -1, and 5 of them are integers.
Reading the leaves of the tree turns the probability question into plain counting.
7.SP.C.8Make A Systematic ListForm the probability
Divide integer paths by all paths: the probability is , which is choice (D).
With equally likely paths, probability is just favorable paths over all paths.
7.SP.C.7Draw A DiagramWhen only a few coin flips can happen, draw the whole tree and just count the branches you want.
- Count the paths
- Write the two rules
- Grow the tree
- Count integer endings
- Form the probability