Difference between revisions of "2008 AMC 10B Problems/Problem 17"
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==Solution 3 (combinatorics)== | ==Solution 3 (combinatorics)== | ||
− | The probability of getting the first voter to approve is <math>frac{7}{10} * frac{3}{10} * frac{3}{10}</math>. | + | The probability of getting the first voter to approve is <math>\frac{7}{10} * \frac{3}{10} * \frac{3}{10}</math>. |
This first voter, using combinations, can be arranged in 3 choose 1 ways, which simplifies into 3 ways. | This first voter, using combinations, can be arranged in 3 choose 1 ways, which simplifies into 3 ways. | ||
− | Multiplying 3 by <math>frac{7}{10} * frac{3}{10} * frac{3}{10}</math> gives <math>frac{189}{10000}</math> or $\mathrm{(B)}\$. | + | Multiplying 3 by <math>\frac{7}{10} * \frac{3}{10} * \frac{3}{10}</math> gives <math>\frac{189}{10000}</math> or $\mathrm{(B)}\$. |
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Revision as of 03:45, 8 August 2024
Contents
Problem
A poll shows that of all voters approve of the mayor's work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor's work?
Solution 1
Letting Y stand for a voter who approved of the work, and N stand for a person who didn't approve of the work, the pollster could select responses in different ways:
. The probability of each of these is
. Thus, the answer is
Solution 2
In more concise terms, this problem is an extension of the binomial distribution. We find the number of ways only 1 person approves of the mayor multiplied by the probability 1 person approves and 2 people disapprove:
Solution 3 (combinatorics)
The probability of getting the first voter to approve is .
This first voter, using combinations, can be arranged in 3 choose 1 ways, which simplifies into 3 ways.
Multiplying 3 by gives
or $\mathrm{(B)}$.
Video Solution by TheBeautyofMath
With explanation of how it helps on future problems, emphasizing "Don't Memorize, Understand" https://youtu.be/PO3XZaSchJc
~IceMatrix
See also
2008 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.