2015 AMC 12B Problems/Problem 17: Difference between revisions
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==Solution== | ==Solution== | ||
When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails | When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails is | ||
Similarly, the probability of getting exactly 3 heads is < | <cmath>\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.</cmath> | ||
Similarly, the probability of getting exactly 3 heads is | |||
<cmath>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.</cmath> | |||
Now set the two probabilities equal to each other and solve for <math>n</math>: | Now set the two probabilities equal to each other and solve for <math>n</math>: | ||
Revision as of 00:27, 12 March 2015
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 value 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
:
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 | |
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