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2008 AMC 10A Problem 22

Problem 22 of 25HarderNumber TheoryCounting & Probability

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6.6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1.1. If it comes up tails, he takes half of the previous term and subtracts 1.1. What is the probability that the fourth term in Jacob’s sequence is an integer?

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Solution

Starting from 6,6, the second terms are 1111 (heads) and 22 (tails). Continuing the tree, the eight equally likely fourth terms are 41,9.5,8,1.25,5,0.5,1,1.41, 9.5, 8, 1.25, 5, 0.5, -1, -1. Of these, 41,8,5,1,141, 8, 5, -1, -1 are integers, so the probability is 58.\dfrac{5}{8}. Thus, the correct answer is D.

More practice

Concepts: basic probability · tree diagram · parity

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.