AMC 10 · 2010 · #18

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccombinations-basiccaseworksymmetry-argument complementary-countingcasework ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

Bernardo draws 3 different numbers from {1,2,3,4,5,6,7,8,9} and writes them in descending order to form a 3-digit number. Silvia does the same, but she draws 3 different numbers from {1,2,3,4,5,6,7,8}. Every choice of 3 numbers is equally likely. Find the probability that Bernardo's number is larger than Silvia's number.

Pick an answer.

(A)

$\frac{47}{72}$

(B)

$\frac{37}{56}$

(C)

$\frac{2}{3}$

(D)

$\frac{49}{72}$

(E)

$\frac{39}{56}$

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

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.

1STEP 1

A chosen set is one number

Descending order fixes the number once the 3 values are chosen, so compare sets: Bernardo has C(9, 3)=84, Silvia has C(8, 3)=56.

C(9, 3)=84, C(8, 3)=56

2STEP 2

Case 1: Bernardo picks a 9

A leading 9 beats Silvia's best, 876, so he wins outright; fixing it leaves C(8, 2)=28 of 84 picks, probability 1/3.

P(9 chosen)=(C(8, 2))/(C(9, 3))=28/84=1/3

3STEP 3

Case 2: no 9 means a fair fight

With probability 2/3 he has no 9 and draws from Silvia's own pool, 1 through 8, so each is equally likely to be larger.

P(no 9)=1-1/3=2/3; P(B ＞ S)=P(S ＞ B)

4STEP 4

Remove the tie, split the rest

Case 2 also allows a tie: identical sets occur with probability 1/56, so the rest splits in half, giving 55/112.

P(tie)=1/56; P(B ＞ S ∣ no 9)=1/2(1-1/56)=55/112

5STEP 5

Weight each case

Weight each case: Case 1 gives 1/3·1=56/168, and Case 2 gives 2/3·55/112=55/168.

1/3 · 1=56/168, 2/3·55/112=55/168

6STEP 6

Add the cases

Adding them, 56/168+55/168=111/168=37/56, which is choice (B).

56/168+55/168=111/168=37/56 → (B)

Answer

37/56

The answer 37/56≈ 0.661 should be a bit more than 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 ＜ 37/56 ＜ 1. Sanity-check the pieces — the guaranteed-win case alone gives 1/3≈ 0.33, and the fair-fight case adds another 55/168≈ 0.33, totaling about 0.66, matching. The distractors 47/72 and 49/72 come from mishandling the tie, and 2/3 ignores the tie entirely; our count keeps the tie explicit.

💡Key takeaway

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.

A chosen set is one number
Case 1: Bernardo picks a 9
Case 2: no 9 means a fair fight
Remove the tie, split the rest
Weight each case
Add the cases