2015 AMC 12B Problems/Problem 17

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 valu 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)

2015 AMC 12B (Problems • Answer Key • Resources)
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2015 AMC 12B Problems/Problem 17

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Problem

Solution

Solution 2

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