Problem 19 · 1998 AJHSME
Hard
Counting & Probability
enumerate-cases
Tamika selects two different numbers at random from the set {8, 9, 10} and adds them. Carlos takes two different numbers at random from the set {3, 5, 6} and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?
Show answer
Answer: A — 4/9.
Show hints
Hint 1 of 2
The two people act independently, so the real sample space is small: every one of Tamika's 3 possible results paired with every one of Carlos's 3 — only 9 equally likely matchups. List each person's possible outcomes first.
Still stuck? Show hint 2 →
Hint 2 of 2
Organize by Carlos's value: against his low number Tamika always wins, against his high number she always loses — only his middle value needs a closer look.
Show solution
Approach: list each person's outcomes, count winning matchups out of 9
- Pinning down outcomes: Tamika picks two of {8,9,10} and adds → sums 17, 18, 19. Carlos picks two of {3,5,6} and multiplies → products 15, 18, 30. Each value is equally likely, and the two choices are independent, so there are 3 × 3 = 9 equally likely matchups.
- Count Tamika's wins by Carlos's value: vs Carlos = 15, all 3 of her sums (17,18,19) win → 3; vs Carlos = 18, only 19 wins → 1; vs Carlos = 30, none win → 0.
- That's 3 + 1 + 0 = 4 wins out of 9, so the probability is 4/9.
- Why this transfers: for two independent random choices, the outcomes multiply (3 × 3 = 9 equally likely pairs), and a tidy table or 'sort by one person's value' keeps you from miscounting. Sanity check: 4 of 9 is just under half, which feels right since Carlos's products spread higher than Tamika's sums.
Mark:
· log in to save