AMC 10 · 2003 · #21
Grade 7 probabilityPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The colors of individual beads don't matter — only how many green beads remain. So first reorganize the problem around that single number (Tool #15): the bag always has 4 beads, and each turn the green count either stays put (a red was pulled) or drops by one (a green was pulled). Starting from 2 greens, ending all red means both greens must be pulled at some point, so green must be pulled on exactly two of the three turns and red on the other one. That reframing turns a messy color story into a short counting problem. The chance of pulling green depends on how many greens are left, so a tree of states (Tool #1) shows the probability on each turn. There are only a few winning orders for which turn is the red pull, so a systematic list (Tool #2) of those orders catches every one, and finding each order's probability is a small subproblem (Tool #7) solved by multiplying along its branch.
Count only the green beads
Track only the green count: it starts at 2 and never rises, so reaching zero in three turns needs two green pulls and one red.
Since replacements are always red, the only way to clear the greens is to pull each green out — so two of the three pulls must be green.
7.SP.C.8Organize Information In More WaysFind the chance of a green pull at each state
Any turn has 4 equally likely beads, so with g greens a green pull has chance g/4: 1/2 at 2 greens, 1/4 at 1.
All four beads are equally likely, so the chance of grabbing a green is just how many greens there are out of four.
7.SP.C.7Draw A DiagramList the winning orders and multiply each branch
Only the red turn's position varies, so multiply down each of the three orders: RGG=1/16, GRG=3/32, GGR=1/8.
Each ordered history is one path down the tree, and the chance of that whole path is the product of the chances along it.
Each ordered history is one path, and the chance of that whole path is the product of the chances along it.
▸ Why?
Each draw's chance is set by the state it starts from, so the steps multiply along the path.
▸ Why?
Different winning histories can never happen together, so their chances simply add.
Add the three winning probabilities
The three orders are exclusive, so add over denominator 32: 2/32+3/32+4/32 = 9/32 — choice (C).
Different winning stories can't happen together, so their chances simply add once written over the same denominator.
5.NF.A.1Identify SubproblemsTrack just the green beads: since every replacement is red, you win only by pulling both greens, so list the few orders that do it and add up their chances.
- Count only the green beads
- Find the chance of a green pull at each state
- List the winning orders and multiply each branch
- Add the three winning probabilities