2017 AMC 8

Complete problem set with solutions and individual problem pages

Problem 10 Easy

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?

  • A.

    \frac {1}{10}

  • B.

    \frac 15

  • C.

    \frac {3}{10}

  • D.

    \frac 25

  • E.

    \frac 12

Answer:C

Solution 1 

There are \binom{5}{3} possible groups of cards that can be selected. If 4 is the largest card selected, then the other two cards must be either 1, 2, or 3, for a total \binom{3}{2} groups of cards. Then, the probability is just {\frac{{\dbinom{3}{2}}}{{\dbinom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}.

 

Solution 2 

P (no 5)= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{2}{5}. This is the fraction of total cases with no fives. p (no 4 and no 5)= \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} = \frac{6}{60} = \frac{1}{10}. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. \frac{2}{5} - \frac{1}{10} = \frac{3}{10} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}.

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