AMC 10 · 2015 · #11
Grade 7 probabilityPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
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}.
Knowing exactly which digits are allowed turns a vague set into a short, listable one.
4.OA.B.4Make A Systematic ListCount 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.
Each digit slot is an independent choice of 4, so multiply the choices to count the combinations.
3.OA.A.1Identify SubproblemsList 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.
A short divisibility check on the last digit kills most candidates before any real testing.
4.OA.B.4Make A Systematic ListForm the probability and choose
With 8 primes out of the pool of 20, the probability is 8/20, which reduces to 2/5, matching choice (B).
Probability of an equally-likely event is just favorable count over total count, reduced.
7.SP.C.7Eliminate PossibilitiesFor 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