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: $n=1$ also works, but it is assumed in the above that the problem writers meant where $n>1$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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