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

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

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

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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