Difference between revisions of "2015 AMC 12B Problems/Problem 17"
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The probability of getting three heads (comapred to the probability of getting two) from <math>n</math> flips is missing a factor of three in the numerator. Thus, we need <math>\dbinom{n}{3}</math> to add a factor of 3 to the numerator of the probability of getting three heads. | The probability of getting three heads (comapred to the probability of getting two) from <math>n</math> flips is missing a factor of three in the numerator. Thus, we need <math>\dbinom{n}{3}</math> to add a factor of 3 to the numerator of the probability of getting three heads. | ||
− | Our testing equation becomes | + | Our testing equation becomes |
− | < | + | <cmath>\dbinom{n}{2} \times 3 = \dbinom{n}{3}</cmath>. |
− | + | since after factoring out the three from <math>\dbinom{n}{3}</math>, the remaining factorizations should be equal. | |
The only answer choice satisfying this condition is <math>\fbox{\textbf{(D)}\;11}</math>. | The only answer choice satisfying this condition is <math>\fbox{\textbf{(D)}\;11}</math>. |
Revision as of 13:11, 11 January 2021
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 :
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)
Solution 2.5
In order to test the answer choices efficiently, realize that the probability flips yielding two heads is of the form:
Similarly, the form for the probability of three heads is:
.
The probability of getting three heads (comapred to the probability of getting two) from flips is missing a factor of three in the numerator. Thus, we need to add a factor of 3 to the numerator of the probability of getting three heads. Our testing equation becomes
.
since after factoring out the three from , the remaining factorizations should be equal.
The only answer choice satisfying this condition is .
-Solution by Joeya
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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