# 2015 AMC 12B Problems/Problem 17

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## 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 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 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)

## Solution 2.5

In order to test the answer choices efficiently, realize that the probability $n$ flips yielding two heads is of the form: $\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}$

Similarly, the form for the probability of three heads is: $\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}$

The probability of getting three heads (comapred to the probability of getting two) from $n$ flips is missing a factor of $3$ in the numerator. Thus, we need $\dbinom{n}{3}$ to add a factor of $3$ to the numerator of the probability of getting three heads. Our testing equation becomes $$\dbinom{n}{2} \times 3 = \dbinom{n}{3}$$

since after factoring out the $3$ from $\dbinom{n}{3}$, the remaining factorizations should be equal.

The only answer choice satisfying this condition is $\fbox{\textbf{(D)}\;11}$.

-Solution by Joeya

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 