2015 AMC 12B Problems/Problem 17: Difference between revisions
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==Problem== | ==Problem== | ||
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> ? | 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> | ||
==Solution== | ==Solution== | ||
When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails | When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails is | ||
Similarly, the probability of getting exactly 3 heads is < | <cmath>\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.</cmath> | ||
Similarly, the probability of getting exactly 3 heads is | |||
<cmath>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.</cmath> | |||
Now set the two probabilities equal to each other and solve for <math>n</math>: | Now set the two probabilities equal to each other and solve for <math>n</math>: | ||
| Line 20: | Line 24: | ||
<cmath>n = \fbox{\textbf{(D)}\; 11}</cmath> | <cmath>n = \fbox{\textbf{(D)}\; 11}</cmath> | ||
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) | |||
==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. | |||
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 
. When tossed 
times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of 
?
Solution
When tossed times, the probability of getting exactly 2 heads and the rest tails is
![\[\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.\]](http://latex.artofproblemsolving.com/f/c/6/fc6405c602dee5d12cb017541fb674a4f3bbb8e8.png)
Similarly, the probability of getting exactly 3 heads is
![\[\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.\]](http://latex.artofproblemsolving.com/9/2/c/92c95422814935e2a1744e40bb8ff525c4db41a3.png)
Now set the two probabilities equal to each other and solve for :
![\[\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}\]](http://latex.artofproblemsolving.com/d/e/e/deed9cb1861f787abde4c577294d240a24de5890.png)
![\[\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}\]](http://latex.artofproblemsolving.com/0/7/f/07f1b2a26ad277631bd46c1314db7092bcd944c6.png)
![\[3 = \frac{n-2}{3}\]](http://latex.artofproblemsolving.com/5/d/4/5d414de3ecbe06f23e64517629c32e1a0023e6d3.png)
![\[n-2 = 9\]](http://latex.artofproblemsolving.com/1/c/d/1cd6ebfe15d5b338b44028394dc9b9543fe488f7.png)
![\[n = \fbox{\textbf{(D)}\; 11}\]](http://latex.artofproblemsolving.com/c/2/7/c27f5da250dd157f54a2a7e3c2e4eb2eb7e4ed48.png)
Note: the original problem did not specify , so 
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 flips yielding two heads is of the form:
Similarly, the form for the probability of three heads is:
The probability of getting three heads (comapred to the probability of getting two) from flips is missing a factor of 
in the numerator. Thus, we need 
to add a factor of to the numerator of the probability of getting three heads.
Our testing equation becomes
![\[\dbinom{n}{2} \times 3 = \dbinom{n}{3}\]](http://latex.artofproblemsolving.com/5/1/4/514dbaae76ec78e6d85fbe7314205aa254e62f61.png)
since after factoring out the from , the remaining factorizations should be equal.
The only answer choice satisfying this condition is 
.
-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 | |
These problems are copyrighted © by the Mathematical Association of America. 
