Difference between revisions of "2010 AMC 12A Problems/Problem 15"
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− | A coin is altered so that the probability that it lands on heads is less than <math>\frac{1}{2}</math> and when the coin is flipped four times, the | + | A coin is altered so that the probability that it lands on heads is less than <math>\frac{1}{2}</math> and when the coin is flipped four times, the probability of an equal number of heads and tails is <math>\frac{1}{6}</math>. What is the probability that the coin lands on heads? |
<math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}</math> | <math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}</math> | ||
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[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:47, 3 July 2013
Problem
A coin is altered so that the probability that it lands on heads is less than and when the coin is flipped four times, the probability of an equal number of heads and tails is . What is the probability that the coin lands on heads?
Solution
Let be the probability of flipping heads. It follows that the probability of flipping tails is .
The probability of flipping heads and tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.
As for the desired probability both and are nonnegative, we only need to consider the positive root, hence
Applying the quadratic formula we get that the roots of this equation are . As the probability of heads is less than , we get that the answer is .
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.