2015 AMC 12B Problems/Problem 17

Revision as of 12:07, 5 March 2015 by Pi over two (talk | contribs) (Solution)


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 $n$ ?

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


Out of $n$ tosses, the probability of having 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 do this by multiplying by $\dbinom{n}{2}$.

For $3$ heads, the corresponding probability 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 $n$:

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

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

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

\[\frac{3}{n-2} = \frac{1}{3}\]

\[n-2 = 9\]

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

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

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

Invalid username
Login to AoPS