Difference between revisions of "2008 AMC 10B Problems/Problem 17"
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The pollster could select responses in 3 different ways: YNN, NYN, and NNY, where Y stands for a voter who approved of the work, and N stands for a person who didn't approve of the work. The probability of each of these is <math>(0.7)(0.3)^2=0.063.</math> Thus, the answer is <math>3 \cdot 0.063=0.189\Rightarrow \boxed{B}</math> | The pollster could select responses in 3 different ways: YNN, NYN, and NNY, where Y stands for a voter who approved of the work, and N stands for a person who didn't approve of the work. The probability of each of these is <math>(0.7)(0.3)^2=0.063.</math> Thus, the answer is <math>3 \cdot 0.063=0.189\Rightarrow \boxed{B}</math> | ||
==Alternative Solution== | ==Alternative Solution== | ||
| − | 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: <cmath>{3\choose 1} \cdot (0.7)^1\cdot(1-0.7)^{(3-1)} = 3 \cdot 0.7 \cdot 0.09 = 0.189 | + | 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: <cmath>{3\choose 1} \cdot (0.7)^1\cdot(1-0.7)^{(3-1)} = 3 \cdot 0.7 \cdot 0.09 = 0.189 \Rightarrow \boxed{B}</cmath> |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2008|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:12, 7 June 2021
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
The pollster could select responses in 3 different ways: YNN, NYN, and NNY, where Y stands for a voter who approved of the work, and N stands for a person who didn't approve of the work. The probability of each of these is 
Thus, the answer is 
Alternative Solution
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: ![\[{3\choose 1} \cdot (0.7)^1\cdot(1-0.7)^{(3-1)} = 3 \cdot 0.7 \cdot 0.09 = 0.189 \Rightarrow \boxed{B}\]](http://latex.artofproblemsolving.com/d/e/9/de972688074e320dfd31f4142dd724144cab4c9a.png)
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. 
