2015 AMC 12B Problems/Problem 17
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 in the numerator. Thus, we need to add a factor of to the numerator of the probability of getting three heads. Our testing equation becomes
since after factoring out the 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 |
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All AMC 12 Problems and Solutions |
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