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=\mathrm{(B)}\ {{{0.189}}}</math> | + | 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{\mathrm{(B)}\ {{{0.189}}}}</math> |
| − | == | + | ==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 } \cdot(0.7)^1\cdot(1-0.7)^{(3-1)}=3\cdot0.7\cdot0.09=\boxed{\mathrm{(B)}\ {{{0.189}}}}</math> | ||
Revision as of 19:16, 23 March 2023
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: 
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. 
