AMC 10 · 2005 · #15
Grade 7 probabilityAn envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $20 or more?
Pick an answer.
AMC 10 2005 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: An envelope holds eight bills: two $1$s, two $5$s, two $10$s, and two $20$s. You draw two of them at once (without replacement). Find the probability that the two bills you draw add up to $20$ dollars or more.
Givens: Eight bills total: $\$1,\$1,\$5,\$5,\$10,\$10,\$20,\$20$; Two bills are drawn at random, without replacement; The order of the two bills does not matter; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{2}{5}$, (C) $\tfrac{3}{7}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{2}{3}$
Unknowns: The probability that the sum of the two drawn bills is $\$20$ or greater
Understand
Restated: An envelope holds eight bills: two $1$s, two $5$s, two $10$s, and two $20$s. You draw two of them at once (without replacement). Find the probability that the two bills you draw add up to $20$ dollars or more.
Givens: Eight bills total: $\$1,\$1,\$5,\$5,\$10,\$10,\$20,\$20$; Two bills are drawn at random, without replacement; The order of the two bills does not matter; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{2}{5}$, (C) $\tfrac{3}{7}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{2}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List
Every outcome (a pair of bills) is equally likely, so the probability is just (number of good pairs) / (total pairs). Tool #7 (Identify Subproblems) splits the good pairs into two clean cases: pairs that include a $\$20$, and pairs that do not. The first case is easiest counted with Tool #16 (Count the Complement) — count all pairs, then remove the pairs with no $\$20$. The second case is a short, finite check handled by Tool #2 (Make a Systematic List): among the remaining bills, which pairs still reach $\$20$?
Execute — Answer: D
7.SP.C.8 Step 1 Count all possible pairs
- Treat the eight bills as eight distinct objects.
- Choosing $2$ of them without regard to order is a combination: $\binom{8}{2}=\frac{8\cdot 7}{2}=28$.
- So there are $28$ equally likely pairs, which will be the denominator of the probability.
💡 Since every pair of bills is just as likely as any other, counting the pairs is the whole game.
7.SP.C.8 Step 2 Any pair with a $20 already wins
- If one of the two bills is a $\$20$, the smallest the other bill can be is $\$1$, giving $\$21$ — already at least $\$20$.
- So every pair that contains at least one $\$20$ is a success. Count these by the complement: from all $28$ pairs, remove the pairs that use none of the two $\$20$s.
- Those avoid-the-twenties pairs are chosen from the other $6$ bills: $\binom{6}{2}=15$.
- So pairs with at least one $\$20$ number $28-15=13$.
💡 It is easier to count the pairs that dodge both twenties and subtract than to list every pair that grabs one.
7.SP.C.8 Step 3 Check the pairs with no $20
Now look only at the $15$ pairs drawn from the six smaller bills ($\$1,\$1,\$5,\$5,\$10,\$10$) and ask which still reach $\$20$. List the possible sums: $10+10=20$ (success), while $10+5=15$, $10+1=11$, $5+5=10$, $5+1=6$, and $1+1=2$ all fall short. Only the two-tens pair works, and there is exactly one such pair, $\binom{2}{2}=1$.
💡 Without a twenty, the only way to hit $\$20$ is to pair the two biggest remaining bills, the tens.
7.SP.C.7 Step 4 Add the cases and form the probability
- The successful pairs are the $13$ that contain a $\$20$ plus the $1$ two-tens pair, for $13+1=14$ good pairs out of $28$. The probability is $\frac{14}{28}=\frac{1}{2}$.
- That is choice (D).
💡 Good outcomes over all equally likely outcomes gives the probability, and $14$ out of $28$ is exactly half.
7.SP.C.8 Treat the eight bills as eight distinct objects. Choosing $2$ of them without re 7.SP.C.8 If one of the two bills is a $\$20$, the smallest the other bill can be is $\$1$ 7.SP.C.8 Now look only at the $15$ pairs drawn from the six smaller bills ($\$1,\$1,\$5,\ 7.SP.C.7 The successful pairs are the $13$ that contain a $\$20$ plus the $1$ two-tens pa Review
Reasonableness: The answer $\tfrac{1}{2}$ is believable: a $\$20$ appears in a healthy share of pairs and guarantees a win, so a fifty-fifty chance is in the right ballpark — not tiny, not near-certain. A cross-check by the complement confirms it. The pairs that fall short of $\$20$ are exactly the pairs with no $\$20$ except the two-tens pair: $15-1=14$ failing pairs, matching the $28-14=14$ successes. Since successes and failures are both $14$, the probability is $\tfrac{1}{2}$ either way. The trap choice (C) $\tfrac{3}{7}=\tfrac{12}{28}$ comes from forgetting to count the $10+10$ pair (or a twenty-with-twenty pair); losing two good pairs would drop $14$ to $12$.
Alternative: Count the complement directly: the probability the sum is *less* than $\$20$. A pair sums to less than $\$20$ exactly when it contains no $\$20$ and is not the two-tens pair. Pairs with no $\$20$: $\binom{6}{2}=15$; remove the one $10+10$ pair to get $15-1=14$ failing pairs. So $P(\text{sum}<20)=\tfrac{14}{28}=\tfrac{1}{2}$, and the answer is $1-\tfrac{1}{2}=\tfrac{1}{2}$.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $\binom{8}{2}=28$ total pairs, the $13$ pairs containing a $\$20$ (via the complement $28-\binom{6}{2}$), and the single $10+10$ pair among the rest.)7.SP.C.7Develop probability models and use them to find probabilities of events (Treating all $28$ pairs as equally likely and computing the probability as $\frac{14}{28}=\frac{1}{2}$.)
⭐ When every pick is equally likely, count the winning pairs over all pairs — and split the count into easy cases, like 'has a twenty' versus 'no twenty', so nothing gets missed or double-counted.
⭐ When every pick is equally likely, count the winning pairs over all pairs — and split the count into easy cases, like 'has a twenty' versus 'no twenty', so nothing gets missed or double-counted.
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