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, 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 $n$ :

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

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

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

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

$n-2 = 9$

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

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2015 AMC 12B Problems/Problem 17

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