2017 AMC 8 Problem 10

Attempt Problem 10 of the 2017 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

A box contains five cards, numbered 1,1, 2,2, 3,3, 4,4, and 5.5. Three cards are selected randomly without replacement from the box. What is the probability that 44 is the largest value selected?
110 \dfrac{1}{10}
15 \dfrac{1}{5}
310 \dfrac{3}{10}
25 \dfrac{2}{5}
12 \dfrac{1}{2}
Answer: C
Concepts:basic probabilitycombinations
Small hint:
Count all ways to choose 33 cards from the 55 cards.
Big hint:
If 44 is the largest selected card, the other two selected cards must come from 1,1, 2,2, and 33.
Video solution:
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Written solution:
The number of ways to choose 33 cards from 55 is (53)=10.{5 \choose 3} = 10. If 44 is the largest value selected, then the other two cards have to be chosen from {1,2,3}.\{1, 2, 3\}. There are (32)=3{3 \choose 2} = 3 ways to do this. The probability is then 310.\dfrac{3}{10}.
Thus, C is the correct answer.
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