Difference between revisions of "2008 AMC 10B Problems/Problem 17"

(Alternative Solution)
m
Line 8: Line 8:
 
==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 \Rightarrow \boxed{B}</cmath>
 
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>
 +
 +
==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==
 
==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 20:33, 20 October 2021

Problem

A poll shows that $70\%$ 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?

$\mathrm{(A)}\ {{{0.063}}} \qquad \mathrm{(B)}\ {{{0.189}}} \qquad \mathrm{(C)}\ {{{0.233}}} \qquad \mathrm{(D)}\ {{{0.333}}} \qquad \mathrm{(E)}\ {{{0.441}}}$

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 $(0.7)(0.3)^2=0.063.$ Thus, the answer is $3 \cdot 0.063=0.189\Rightarrow \boxed{B}$

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: \[{3\choose 1} \cdot (0.7)^1\cdot(1-0.7)^{(3-1)} = 3 \cdot 0.7 \cdot 0.09 = 0.189 \Rightarrow \boxed{B}\]

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 (ProblemsAnswer KeyResources)
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. AMC logo.png