Difference between revisions of "2015 AMC 12B Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | An unfair coin lands on heads with a probability of <math>\tfrac{1}{4}</math>. When tossed <math>n>1</math> times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the | + | An unfair coin lands on heads with a probability of <math>\tfrac{1}{4}</math>. When tossed <math>n>1</math> times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the valu of <math>n</math>? |
<math>\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13</math> | <math>\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13</math> |
Revision as of 22:23, 25 December 2018
Contents
Problem
An unfair coin lands on heads with a probability of . When tossed times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the valu of ?
Solution
When tossed times, the probability of getting exactly 2 heads and the rest tails is
Similarly, the probability of getting exactly 3 heads is
Now set the two probabilities equal to each other and solve for :
Note: the original problem did not specify , so was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. — @adihaya (talk) 15:23, 19 February 2016 (EST)
Solution 2
Bash it out with the answer choices! (not really a rigorous solution)
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.