2008 AMC 10B Problems/Problem 17

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 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 $3$ different ways: $\text{YNN, NYN, and NNY}$ . The probability of each of these is $(0.7)(0.3)^2=0.063$ . Thus, the answer is $3\cdot0.063=\boxed{\mathrm{(B)}\ {{{0.189}}}}$

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: ${3\choose 1} \cdot(0.7)^1\cdot(1-0.7)^{(3-1)}=3\cdot0.7\cdot0.09=\boxed{\mathrm{(B)}\ {{{0.189}}}}$

Solution 3 (combinatorics)

The probability of getting the first voter to approve is $\frac{7}{10} * \frac{3}{10} * \frac{3}{10}$ .

This first voter, using combinations, can be arranged in 3 choose 1 ways, which simplifies into 3 ways.

Multiplying 3 by $\frac{7}{10} * \frac{3}{10} * \frac{3}{10}$ gives $\frac{189}{10000}$ 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

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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