AMC 10 · 2015 · #11

Grade 7 probability
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-numbersprimality-testprobability-basic systematic-enumeration ↑ Prerequisites: prime-numbersprobability-basic
📏 Medium solution 💡 2 insights
Problem
Look at every positive integer below 100 in which each digit is one of the prime digits. Pick one of those numbers at random. Find the probability that the number you picked is itself prime.

Pick an answer.

(A)
$\dfrac{8}{99}$
(B)
$\dfrac{2}{5}$
(C)
$\dfrac{9}{20}$
(D)
$\dfrac{1}{2}$
(E)
$\dfrac{9}{16}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The pool of numbers is small, so tool #2 (Make a Systematic List) lets us write out every qualifying number without missing or repeating any. A probability is a fraction, so tool #7 (Identify Subproblems) splits the work into two independent counts: how many numbers qualify (the denominator) and how many of them are prime (the numerator). With both counts in hand, tool #3 (Eliminate Possibilities) matches the reduced fraction to exactly one answer choice.

1STEP 1

Find the allowed digits

A digit is prime only for 2, 3, 5, or 7, so every qualifying number is built solely from {2, 3, 5, 7}.

prime digits = {2, 3, 5, 7}
2STEP 2

Count all qualifying numbers

One-digit picks give 4 numbers and two-digit picks give 4 × 4 = 16, so the whole pool has 4 + 16 = 20 numbers.

4 + (4 × 4) = 4 + 16 = 20
3STEP 3

List and test for primality

The four one-digit numbers are all prime, and among the two-digit ones only 23, 37, 53, 73 survive, giving 8 primes in all.

primes: 2, 3, 5, 7, 23, 37, 53, 73
4STEP 4

Form the probability and choose

With 8 primes out of the pool of 20, the probability is 820\frac{8}{20}, which reduces to 25\frac{2}{5}, matching choice (B).

8/20 = 2/5 → (B)
Answer
2/5
About 8 of 20 numbers being prime is a roughly 40% rate, which is believable because the pool is deliberately stocked with prime digits and excludes even endings like 0,4,6,8 entirely. The fraction 25\frac{2}{5} also sits sensibly between the tiny 899\frac{8}{99} and the larger 12\frac{1}{2} choices. Recounting confirms exactly 8 primes and 20 total, so 25\frac{2}{5} holds.
💡Key takeaway

For an equally-likely probability, list the whole pool, count how many fit, and put the good count over the total.

  • Find the allowed digits
  • Count all qualifying numbers
  • List and test for primality
  • Form the probability and choose

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