2015 AMC 12B Problem 17

Question

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

Set

\binom{n}{2}\left(\tfrac14\right)^2\left(\tfrac34\right)^{n-2}

= \binom{n}{3}\left(\tfrac14\right)^3\left(\tfrac34\right)^{n-3}

Cancel common factors to reach

\binom{n}{2}\cdot 3 = \binom{n}{3}

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.

Problem 17 in other years