Difference between revisions of "2008 AMC 10B Problems/Problem 17"
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==Solution 1== | ==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 <math>3</math> different ways: <math>\text{YNN, NYN, and NNY}</math>. The probability of each of these is <math>(0.7)(0.3)^2=0.063</math>. Thus, the answer is <math>3\cdot0.063=\boxed{ | + | 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 <math>3</math> different ways: <math>\text{YNN, NYN, and NNY}</math>. The probability of each of these is <math>(0.7)(0.3)^2=0.063</math>. Thus, the answer is <math>3\cdot0.063=0.189</math>, or <math>\boxed{B}</math> |
==Solution 2== | ==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: <math>{3\choose } \cdot(0.7)^1\cdot(1-0.7)^{(3-1)}=3\cdot0.7\cdot0.09=\boxed{\mathrm{(B)}\ {{{0.189}}}}</math> | + | 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: <math>{3\choose 1} \cdot(0.7)^1\cdot(1-0.7)^{(3-1)}=3\cdot0.7\cdot0.09=\boxed{\mathrm{(B)}\ {{{0.189}}}}</math> |
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| + | ==Solution 3 (combinatorics)== | ||
| + | The probability of getting the first voter to approve is <math>\frac{7}{10} * \frac{3}{10} * \frac{3}{10}</math>. | ||
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| + | This first voter, using combinations, can be arranged in 3 choose 1 ways, which simplifies into 3 ways. | ||
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| + | Multiplying <math>3</math> by <math>\frac{7}{10} * \frac{3}{10} * \frac{3}{10}</math> gives (B). | ||
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| + | ~PeterDoesPhysics | ||
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| + | PS: For a more challenging practice problem akin to this one, look to 2012 AMC 12A Problem 11. | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
Latest revision as of 18:39, 3 November 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 
, or 
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 by gives (B).
~PeterDoesPhysics
PS: For a more challenging practice problem akin to this one, look to 2012 AMC 12A Problem 11.
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. 
