AMC 10 Weekly Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 18 Medium

From five line segments of lengths 2, 4, 6, 8, 10, three are chosen at random. The probability that these three line segments can form a triangle is （   ）．

  • A.

    \frac{1}{5}

  • B.

    \frac{3}{10}

  • C.

    \frac{2}{5}

  • D.

    \frac{3}{5}

  • E.

    \frac{7}{10}

Answer:B

Choosing any 3 line segments, the possible sets are \{2,4,6\},\ \{2,4,8\},\ \{2,4,10\},\ \{2,6,8\},\ \{2,6,10\},\ \{2,8,10\},\ \{4,6,8\},\ \{4,6,10\},\ \{4,8,10\},\ \{6,8,10\}, a total of 10 basic events.

 

Among these, the sets of three line segments that can form a triangle are

\{4,6,8\},\ \{4,8,10\},\ \{6,8,10\}, a total of 3.

 

Therefore, the probability that three chosen line segments can form a triangle is

P = \frac{3}{10}.

 

Thus, the answer is B.

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