AMC 10 · 2010 · #18
Grade 7 probabilityBernardo randomly picks 3 distinct numbers from the set {1,2,3,...,7,8,9} and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1,2,3,...,6,7,8} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?
Pick an answer.
AMC 10 2010 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: Bernardo draws 3 different numbers from {1,2,3,4,5,6,7,8,9} and writes them in descending order as a 3-digit number. Silvia draws 3 different numbers from {1,2,3,4,5,6,7,8} and does the same. Find the probability that Bernardo's number is larger than Silvia's.
Givens: Bernardo picks 3 distinct numbers from {1,...,9} and arranges them in descending order; Silvia picks 3 distinct numbers from {1,...,8} and arranges them in descending order; Each 3-number choice is equally likely; Answer choices: (A) $\frac{47}{72}$, (B) $\frac{37}{56}$, (C) $\frac{2}{3}$, (D) $\frac{49}{72}$, (E) $\frac{39}{56}$
Unknowns: The probability that Bernardo's 3-digit number is greater than Silvia's
Understand
Restated: Bernardo draws 3 different numbers from {1,2,3,4,5,6,7,8,9} and writes them in descending order as a 3-digit number. Silvia draws 3 different numbers from {1,2,3,4,5,6,7,8} and does the same. Find the probability that Bernardo's number is larger than Silvia's.
Givens: Bernardo picks 3 distinct numbers from {1,...,9} and arranges them in descending order; Silvia picks 3 distinct numbers from {1,...,8} and arranges them in descending order; Each 3-number choice is equally likely; Answer choices: (A) $\frac{47}{72}$, (B) $\frac{37}{56}$, (C) $\frac{2}{3}$, (D) $\frac{49}{72}$, (E) $\frac{39}{56}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
Tool #16 (Change Focus / Count the Complement): the clever move is to notice that once Bernardo has no 9 he is drawing from the exact same pool as Silvia, so the chance he wins equals the chance she wins — and the only leftover is a tie, which I remove and split. Tool #7 (Identify Subproblems): I break the event into two clean cases, Bernardo picks a 9 versus he does not, because the 9 is the single thing that makes his set different from Silvia's. Tool #2 (Make a Systematic List): counting choices with combinations tells me how likely the 9 case is and how likely a tie is.
Execute — Answer: B
7.SP.C.8 Step 1 A chosen set is one number
- Because the digits are written in descending order, once you choose 3 numbers the 3-digit number is fixed.
- So comparing numbers is the same as comparing the two chosen sets.
- Bernardo chooses 3 of 9 numbers, giving $\binom{9}{3}=84$ equally likely numbers; Silvia chooses 3 of 8, giving $\binom{8}{3}=56$ equally likely numbers.
💡 Descending order removes all the arranging, so picking a set is the same as picking a number.
7.SP.C.7 Step 2 Case 1: Bernardo picks a 9
- If Bernardo's set contains a 9, his number starts with 9, and Silvia's largest possible number is 876, so Bernardo automatically wins.
- Count how often this happens: fix the 9 and choose the other two numbers in $\binom{8}{2}=28$ ways, out of $84$ total.
- So the probability of this case is $\frac{28}{84}=\frac{1}{3}$, and inside it Bernardo wins for sure.
💡 A leading 9 beats anything Silvia can build, so this whole slice is a guaranteed win.
7.SP.C.7 Step 3 Case 2: no 9 means a fair fight
- The remaining probability, $1-\frac{1}{3}=\frac{2}{3}$, is the chance Bernardo has no 9.
- Then he is choosing 3 numbers from {1,...,8}, exactly the same pool Silvia uses.
- Since they draw from identical pools the same way, Bernardo is just as likely to have the bigger number as Silvia is.
- The two outcomes are mirror images.
💡 When two people draw the same way from the same pool, neither is favored to be larger.
7.SP.C.8 Step 4 Remove the tie, split the rest
- In Case 2 there are three outcomes: Bernardo bigger, Silvia bigger, or a tie.
- A tie means they pick the identical set; with Silvia's set fixed, Bernardo matches it with probability $\frac{1}{56}$.
- The other $1-\frac{1}{56}=\frac{55}{56}$ splits evenly between the two mirror outcomes, so Bernardo wins Case 2 with probability $\frac{1}{2}\cdot\frac{55}{56}=\frac{55}{112}$.
💡 Take out the ties, then hand each player exactly half of what's left.
5.NF.B.4 Step 5 Weight each case
- Now weight each case by how likely it is.
- Case 1 happens with probability $\frac{1}{3}$ and gives a certain win, contributing $\frac{1}{3}\cdot 1=\frac{1}{3}$.
- Case 2 happens with probability $\frac{2}{3}$ and wins with probability $\frac{55}{112}$, contributing $\frac{2}{3}\cdot\frac{55}{112}=\frac{110}{336}=\frac{55}{168}$.
💡 Each case counts only as much as its chance of happening, so multiply before you add.
5.NF.A.1 Step 6 Add the cases
- Add the two contributions over the common denominator $168$: $\frac{56}{168}+\frac{55}{168}=\frac{111}{168}$.
- Dividing top and bottom by 3 gives $\frac{37}{56}$.
- So the probability that Bernardo's number is larger is $\frac{37}{56}$, which is (B).
💡 Two separate winning routes add up to the total chance of winning.
7.SP.C.8 Because the digits are written in descending order, once you choose 3 numbers th 7.SP.C.7 If Bernardo's set contains a 9, his number starts with 9, and Silvia's largest p 7.SP.C.7 The remaining probability, $1-\frac{1}{3}=\frac{2}{3}$, is the chance Bernardo h 7.SP.C.8 In Case 2 there are three outcomes: Bernardo bigger, Silvia bigger, or a tie. A 5.NF.B.4 Now weight each case by how likely it is. Case 1 happens with probability $\frac 5.NF.A.1 Add the two contributions over the common denominator $168$: $\frac{56}{168}+\fr Review
Reasonableness: The answer $\frac{37}{56}\approx 0.661$ should be a bit more than $\frac{1}{2}$, and it is: Bernardo's advantage is only the extra number 9, so he should win a little more than half the time, not overwhelmingly. Check the fraction is valid: $0<\frac{37}{56}<1$. Sanity-check the pieces — the guaranteed-win case alone gives $\frac{1}{3}\approx 0.33$, and the fair-fight case adds another $\frac{55}{168}\approx 0.33$, totaling about $0.66$, matching. The distractors $\frac{47}{72}$ and $\frac{49}{72}$ come from mishandling the tie, and $\frac{2}{3}$ ignores the tie entirely; our count keeps the tie explicit.
Alternative: Count ordered pairs directly. There are $84\times 56 = 4704$ equally likely (Bernardo, Silvia) pairs. The $28\times 56=1568$ pairs where Bernardo holds a 9 are all wins. Among the other $56\times 56=3136$ pairs (both from {1,...,8}), $56$ are ties and the remaining $3080$ split evenly, giving $1540$ Bernardo wins. Total wins $=1568+1540=3108$, and $\frac{3108}{4704}=\frac{37}{56}$, the same answer.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the equally likely picks with combinations and counting the tie outcomes when both draw from {1,...,8}.)7.SP.C.7Develop probability models and use them to find probabilities of events (Building the case model — probability of Bernardo drawing a 9, and using symmetry to say each player is equally likely to be larger.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Weighting each case by its probability, such as multiplying 2/3 by 55/112.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding the two case contributions over a common denominator to reach 37/56.)
⭐ When both players draw the same way from the same pool, neither is favored — so remove the ties, split the rest in half, and the only real edge comes from Bernardo's extra 9.
⭐ When both players draw the same way from the same pool, neither is favored — so remove the ties, split the rest in half, and the only real edge comes from Bernardo's extra 9.
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