2015 AMC 12B Problem 17

Below is the professionally curated solution for Problem 17 of the 2015 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12B solutions, or check the answer key.

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Concepts:binomial probability

Difficulty rating: 1830

17.

An unfair coin lands on heads with a probability of $\dfrac14.$ When tossed $n$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of $n?$

$5$

$8$

$10$

$11$

$13$

Solution:

Setting the two probabilities equal and cancelling the common powers of $\tfrac14$ and $\tfrac34$ gives $\binom{n}{2}\cdot\dfrac34 = \binom{n}{3}\cdot\dfrac14.$

This becomes $\dfrac{n(n-1)}{2}\cdot 3 = \dfrac{n(n-1)(n-2)}{6},$ so $\dfrac32 = \dfrac{n-2}{6},$ giving $n - 2 = 9$ and $n = 11.$

Thus, the correct answer is D.

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