# 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 $n$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value 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 could be written as ${\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}$. However, we mus account for the different orders that this could occur in (for example like HTT or THT or TTH). We can account for this by multiplying by $\dbinom{n}{2}$.

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 $n$:

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

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

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

$$\frac{3}{n-2} = \frac{1}{3}$$

$$n-2 = 9$$

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