AMC 10 · 2003 · #21
Grade 7 probabilityA bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A bag holds $2$ red beads and $2$ green beads. Three times in a row you pull out one random bead and drop in a red bead to replace it, no matter what color you pulled. Find the probability that after these three replacements every bead in the bag is red.
Givens: The bag starts with $2$ red beads and $2$ green beads, so $4$ beads total.; Each turn: pull one bead at random, then put a red bead back in its place.; This is done three times.; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{5}{32}$, (C) $\frac{9}{32}$, (D) $\frac{3}{8}$, (E) $\frac{7}{16}$.
Unknowns: The probability that all $4$ beads are red after the three replacements.
Understand
Restated: A bag holds $2$ red beads and $2$ green beads. Three times in a row you pull out one random bead and drop in a red bead to replace it, no matter what color you pulled. Find the probability that after these three replacements every bead in the bag is red.
Givens: The bag starts with $2$ red beads and $2$ green beads, so $4$ beads total.; Each turn: pull one bead at random, then put a red bead back in its place.; This is done three times.; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{5}{32}$, (C) $\frac{9}{32}$, (D) $\frac{3}{8}$, (E) $\frac{7}{16}$.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #15 Organize Information in More Ways, #1 Draw a Diagram, #7 Identify Subproblems
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.
Execute — Answer: C
7.SP.C.8 Step 1 Count only the green beads
- Forget which specific beads are which and just track how many green beads are in the bag.
- It starts at $2$.
- Every turn removes one bead and adds a red one, so the green count can only stay the same (if a red was pulled) or go down by one (if a green was pulled).
- To finish with zero greens after three turns, both greens must get pulled.
- That means across the three turns, a green bead is pulled on exactly two of them and a red bead on the other one.
💡 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.7 Step 2 Find the chance of a green pull at each state
- On any turn the bag has $4$ beads and each is equally likely to be pulled, so if there are $g$ greens the chance of pulling a green is $\frac{g}{4}$ and the chance of pulling a red is $\frac{4-g}{4}$.
- With $2$ greens: green $\tfrac{2}{4}=\tfrac12$, red $\tfrac12$.
- With $1$ green: green $\tfrac14$, red $\tfrac34$.
- With $0$ greens: red is certain, $\tfrac44=1$.
💡 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.8 Step 3 List the winning orders and multiply each branch
- A winning history is two green pulls and one red pull.
- The only choice is which turn is the red pull, giving three orders.
- For each, follow the green count and multiply the turn-by-turn chances.
- Red first (R, G, G): start $2$ greens, pull red $\tfrac12$ (still $2$), pull green $\tfrac12$ (now $1$), pull green $\tfrac14$ (now $0$): $\tfrac12\cdot\tfrac12\cdot\tfrac14=\tfrac{1}{16}$.
- Red second (G, R, G): green $\tfrac12$ (now $1$), red $\tfrac34$ (still $1$), green $\tfrac14$ (now $0$): $\tfrac12\cdot\tfrac34\cdot\tfrac14=\tfrac{3}{32}$.
- Red last (G, G, R): green $\tfrac12$ (now $1$), green $\tfrac14$ (now $0$), red $1$: $\tfrac12\cdot\tfrac14\cdot 1=\tfrac18$.
💡 Each ordered history is one path down the tree, and the chance of that whole path is the product of the chances along it.
5.NF.A.1 Step 4 Add the three winning probabilities
- The three orders are separate ways to win, so add their probabilities.
- Put them over the common denominator $32$: $\frac{1}{16}=\frac{2}{32}$, $\frac{3}{32}=\frac{3}{32}$, $\frac{1}{8}=\frac{4}{32}$.
- The total is $\frac{2+3+4}{32}=\frac{9}{32}$.
- That is choice $\textbf{(C)}$.
💡 Different winning stories can't happen together, so their chances simply add once written over the same denominator.
7.SP.C.8 Forget which specific beads are which and just track how many green beads are in 7.SP.C.7 On any turn the bag has $4$ beads and each is equally likely to be pulled, so if 7.SP.C.8 A winning history is two green pulls and one red pull. The only choice is which 5.NF.A.1 The three orders are separate ways to win, so add their probabilities. Put them Review
Reasonableness: The answer $\frac{9}{32}$ is between $0$ and $1$, as any probability must be, and it is under $\frac12$ — reasonable, since two of the three pulls must hit a green and greens grow scarce. A sanity check on the branches: the total probability of clearing both greens should not exceed the chance of getting at least two green pulls in three tries, and indeed $\frac{9}{32}\approx 0.28$ sits comfortably below that. The red-last order $\frac18$ is the biggest single contributor, which makes sense because pulling the greens early leaves an all-red bag whose final pull is guaranteed harmless.
Alternative: Instead of listing orders, work turn by turn with a state tally. Turn 1 leaves $1$ green with probability $\frac12$ (or stays at $2$ greens with probability $\frac12$). From $1$ green, turn 2 reaches $0$ greens with probability $\frac14$; from $2$ greens it can only reach $1$ green (probability $\frac12$) or stay at $2$. Carrying these state probabilities forward two more turns and summing the ways to land on $0$ greens after turn 3 gives the same $\frac{9}{32}$, a compact check that no winning path was missed.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Reframing the goal as "two green pulls and one red pull," listing the three winning orders, and multiplying the chances along each branch of the state tree.)7.SP.C.7Develop probability models and use them to find probabilities of events (Using the equally-likely model to get the chance of pulling a green as $\frac{g}{4}$ at each state of the bag.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining the three branch probabilities $\frac{1}{16}+\frac{3}{32}+\frac{1}{8}$ over the common denominator $32$ to get $\frac{9}{32}$.)
⭐ Track 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.
⭐ Track 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.
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