Difference between revisions of "2008 AMC 10A Problems/Problem 22"
Mathcrazed (talk | contribs) |
I like pie (talk | contribs) |
||
| Line 2: | Line 2: | ||
<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> | <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> | ||
| + | |||
| + | === Solution 2 === | ||
| + | {{solution}} | ||
| + | |||
| + | == See also == | ||
| + | {{AMC10 box|year=2008|ab=A|num-b=21|num-a=23}} | ||
Revision as of 19:20, 3 April 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?
Solution 2
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
