# Difference between revisions of "2015 AMC 12B Problems/Problem 17"

## Problem

An unfair coin lands on heads with a probability of $\tfrac{1}{4}$. When tossed $n>1$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the valu of $n$?

$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13$

## Solution

When tossed $n$ times, the probability of getting exactly 2 heads and the rest tails is

$$\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.$$

Similarly, the probability of getting exactly 3 heads is

$$\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.$$

Now set the two probabilities equal to each other and solve for $n$:

$$\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}$$

$$\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}$$

$$3 = \frac{n-2}{3}$$

$$n-2 = 9$$

$$n = \fbox{\textbf{(D)}\; 11}$$

Note: the original problem did not specify $n>1$, so $n=1$ 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)