Difference between revisions of "2015 AMC 12B Problems/Problem 17"
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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> | <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> | ||
| − | <cmath>\frac{ | + | <cmath>\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}</cmath> |
| − | <cmath> | + | <cmath>3 = \frac{n-2}{3}</cmath> |
| − | |||
| − | |||
<cmath>n-2 = 9</cmath> | <cmath>n-2 = 9</cmath> | ||
Revision as of 00:22, 12 March 2015
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 could be written as 
. However, we must account for the different orders that this could occur in (for example like HTT or THT or TTH). We can account for this by multiplying by 
.
Similarly, the probability of getting exactly 3 heads is 
.
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)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
