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.

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10.

A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 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
Difficulty rating: 1020
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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