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

(Solution 2.5)
m (Solution 2.5)
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<math>\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}</math>.
 
<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 three in the numerator. Thus, we need <math>\dbinom{n}{3}</math> to add a factor of 3 to the numerator of the probability of getting three heads.
+
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
 
Our testing equation becomes
  
 
<cmath>\dbinom{n}{2} \times 3 = \dbinom{n}{3}</cmath>.
 
<cmath>\dbinom{n}{2} \times 3 = \dbinom{n}{3}</cmath>.
  
since after factoring out the three from <math>\dbinom{n}{3}</math>, the remaining factorizations should be equal.  
+
since after factoring out the <math>3</math> from <math>\dbinom{n}{3}</math>, the remaining factorizations should be equal.  
  
 
The only answer choice satisfying this condition is <math>\fbox{\textbf{(D)}\;11}</math>.
 
The only answer choice satisfying this condition is <math>\fbox{\textbf{(D)}\;11}</math>.

Revision as of 14:20, 11 January 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 (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

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