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Difference between revisions of "2015 AMC 12B Problems/Problem 17"

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==Solution==
 
==Solution==
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Out of <math>n</math> tosses, the probability of having exactly <math>2</math> heads and the rest tails could be written as <math>{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}</math>. 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 <math>\dbinom{n}{2}</math>.
  
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For <math>3</math> heads, the corresponding probability is <math>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}</math>.
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Now set the two probabilities equal to each other and solve for <math>n</math>:
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<cmath>\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}</cmath>
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<cmath>\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}}</cmath>
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<cmath>\frac{n(n-1)}{2!} \cdot \frac{3!}{n(n-1)(n-2)} = \frac{1}{4} \cdot {\left( \frac{3}{4} \right)}^{-1}</cmath>
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<cmath>\frac{3}{n-2} = \frac{1}{3}</cmath>
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<cmath>n-2 = 9</cmath>
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<cmath>n = \fbox{\textbf{(D)}\; 11}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}
 
{{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:07, 5 March 2015

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

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 (Problems • Answer Key • Resources)
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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