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

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==Problem==
 
==Problem==
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An unfair coin lands on heads with a probability of <math>\tfrac{1}{4}</math>. When tossed <math>n</math> times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of <math>n</math> ?
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An unfair coin lands on heads with a probability of <math>\tfrac{1}{4}</math>. When tossed <math>n>1</math> times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of <math>n</math>?
  
 
<math>\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13</math>
 
<math>\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13</math>
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Note: <math>n=1</math> also works, but it is assumed in the above that the problem writers meant where <math>n>1</math>.
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Note: the original problem did not specify <math>n>1</math>, so <math>n=1</math> was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:23, 19 February 2016 (EST)
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==Solution 2==
 +
 
 +
Bash it out with the answer choices! (not really a rigorous solution)
 +
 
 +
==Solution 2.5==
 +
 
 +
In order to test the answer choices efficiently, realize that the probability <math>n</math> flips yielding two heads is of the form:
 +
 
 +
<math>\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}</math>
 +
 
 +
Similarly, the form for the probability of three heads is:
 +
 
 +
<math>\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}</math>
 +
 
 +
The probability of getting three heads (comapred to the probability of getting two) from <math>n</math> flips is missing a factor of <math>3</math> in the numerator. Thus, we need <math>\dbinom{n}{3}</math> to add a factor of <math>3</math> to the numerator of the probability of getting three heads.
 +
Our testing equation becomes
 +
 
 +
<cmath>\dbinom{n}{2} \times 3 = \dbinom{n}{3}</cmath>
 +
 
 +
since after factoring out the <math>3</math> from <math>\dbinom{n}{3}</math>, the remaining factorizations should be equal.
 +
 
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The only answer choice satisfying this condition is <math>\fbox{\textbf{(D)}\;11}</math>.
 +
 
 +
-Solution by Joeya
  
 
==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}}

Latest revision as of 20:24, 13 February 2021

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: 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)

Solution 2.5

In order to test the answer choices efficiently, realize that the probability $n$ flips yielding two heads is of the form:

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

Similarly, the form for the probability of three heads is:

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

The probability of getting three heads (comapred to the probability of getting two) from $n$ flips is missing a factor of $3$ in the numerator. Thus, we need $\dbinom{n}{3}$ to add a factor of $3$ to the numerator of the probability of getting three heads. Our testing equation becomes

\[\dbinom{n}{2} \times 3 = \dbinom{n}{3}\]

since after factoring out the $3$ from $\dbinom{n}{3}$, the remaining factorizations should be equal.

The only answer choice satisfying this condition is $\fbox{\textbf{(D)}\;11}$.

-Solution by Joeya

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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