AMC 10 · 2004 · #4

Grade 7 probability
Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)
probability-basiccomplementary-countingdigit-counting complementary-countingsystematic-enumeration ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
One integer is picked at random from 10 through 99, every one equally likely. Find the probability that at least one of its two digits is a 7.

Pick an answer.

(A)
$\dfrac{1}{9}$
(B)
$\dfrac{1}{5}$
(C)
$\dfrac{19}{90}$
(D)
$\dfrac{2}{9}$
(E)
$\dfrac{1}{3}$
How to solve
Strategy Change Focus / Count the Complement

The phrase "at least one" is the standard trigger for Tool #16 (Change Focus / Count the Complement): the wanted event splits into three overlapping cases, while its opposite — "neither digit is a 7" — is one clean case with no overlap to police. Tool #2 (Make a Systematic List) supplies the digit-by-digit count that makes the complement easy. Tool #12 (Draw a Venn Diagram) then re-counts the same set the other way, as two overlapping groups minus their shared member, so the two methods can be checked against each other.

1STEP 1

Count every possible pick

The picks number 90, all equally likely.

99-10+1=90 equally likely outcomes
2STEP 2

Flip to "no 7 anywhere"

Flipping to the complement turns three overlapping cases into one clean condition.

#(at least one 7)=90-#(no 7 at all)
3STEP 3

Count the numbers with no 7

Eight tens digits times nine ones digits gives 72 with no seven.

8 × 9=72 numbers with no digit equal to 7
4STEP 4

Subtract and simplify

Subtracting leaves 18, and reducing gives 1/5.

90-72=18, 18/90=(18 · 1)/(18 · 5)=1/5
5STEP 5

Re-count the other way to confirm

A direct count that removes the double-counted 77 confirms 1/5, choice (B).

10+9-1=18, 18/90=1/5 (B)
Answer
1/5
The result 1/5=0.2 lies between 0 and 1, as any probability must. A rough estimate agrees: a digit is a 7 roughly one time in ten, and there are two digit slots, so something near 0.2 is expected. The wrong choices line up with specific slips, which is a good sign the right one is being read correctly: 1/9=10/90 counts only the block 70–79 and forgets sevens in the ones place; 19/90 is 10+9 with the double-counted 77 never removed; 2/9=20/90 adds 10+10, both overcounting the ones-digit sevens (there are 9, not 10, because 07 is not in range) and ignoring the overlap.
💡Key takeaway

When a question says "at least one", count the ones with none and subtract — a single clean count beats three overlapping ones.

  • Count every possible pick
  • Flip to "no 7 anywhere"
  • Count the numbers with no 7
  • Subtract and simplify
  • Re-count the other way to confirm

More like this

Same archetype — closest grade level first.