Difference between revisions of "2008 AMC 10A Problems/Problem 22"

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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?
 
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?
  
<math>\textbf{(A)}\ \frac{1}{6} \qquad</math> <br /> <math>\textbf{(B)}\ \frac{1}{3} \qquad </math><br /> <math>\textbf{(C)}\ \frac{1}{2} \qquad </math><br /> <math>\textbf{(D)}\ \frac{5}{8} \qquad </math><br /> <math>\textbf{(E)}\ \frac{3}{4}</math>
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<math>\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{5}{8} \qquad \textbf{(E)}\ \frac{3}{4}</math>

Revision as of 18:33, 24 February 2008

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?

$\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{5}{8} \qquad \textbf{(E)}\ \frac{3}{4}$