2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 6 Easy

A box contains 5 chips, numbered 1, 2, 3, 4, and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required? (2018 AMC 10B Problem, Question#6)

  • A.

    \dfrac{1}{15}

  • B.

    \frac{1}{10}

  • C.

    \frac{1}{6}

  • D.

    \frac{1}{5}

  • E.

    \frac{1}{4}

Answer:D

Notice that the only four ways such that no less than 2 draws are requred are 1, 2; 1, 3; 2, 1;  and 3, 1. Notice that each of those cases has a  \frac{1}{5}\cdot \frac{1}{4} chance, so the answer is \frac{1}{5}\cdot \frac{1}{4}\cdot 4= \frac{1}{5}, or \rm D.

Notice that only the first two draws are important, it doesn’t matter what number we get third because no matter what combination of 3 numbers is picked, the sum will always be greater than 5. Also, note that it is necessary to draw a 1 in order to have 3 draws, otherwise 5 will be attainable in two or less draws. So the probabilty of getting a 1 is \frac{1}{5}. It is necessary to pull either a 2 or 3 on the next draw and the probabilty of that is \frac{1}{2}.But,  the order of the draws can be switched so we get \frac{1}{5}\cdot \frac{1}{2}\cdot 2= \frac{1}{5}, or \rm D.