Question

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be $6.$ To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts $1.$ If it comes up tails, he takes half of the previous term and subtracts $1.$ What is the probability that the fourth term in Jacob's sequence is an integer?

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{3}{4}$

Solution:

Starting from $6,$ the second terms are $11$ (heads) and $2$ (tails).

Continuing the tree, the eight equally likely fourth terms are $41, 9.5, 8, 1.25, 5, 0.5, -1, -1.$

Of these, $41, 8, 5, -1, -1$ are integers, so the probability is $\dfrac{5}{8}.$

Thus, the correct answer is D.

2008 AMC 10A Problem 22

Solution:

Problem 22 in Other Years