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 value of $n$?
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13$
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](https://artofproblemsolving.com/wiki/index.php/User:Adihaya) ([talk](https://artofproblemsolving.com/wiki/index.php/User_talk:Adihaya)) 15:23, 19 February 2016 (EST)