AMC 10 Step by Step

2008 AMC 10B Problem 17

P16-20~2 minPrint

A poll shows that 70%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?

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Key insight

One fixed order, like yes-no-no, has chance 0.7 times 0.3 times 0.3. The single yes can come on any of 3 occasions, so multiply by 3.

Solution

Each occasion is a separate random pick: "approves" with chance 0.70.7, "does not" with chance 0.30.3. List every way to get exactly one "approves".

  1. There are three orders with exactly one approval:
OrderChance
yes, no, no0.7×0.3×0.3=0.0630.7 \times 0.3 \times 0.3 = 0.063
no, yes, no0.3×0.7×0.3=0.0630.3 \times 0.7 \times 0.3 = 0.063
no, no, yes0.3×0.3×0.7=0.0630.3 \times 0.3 \times 0.7 = 0.063

2. Each order has the same chance, 0.0630.063, because it multiplies the same three numbers.
3. These orders cannot happen at the same time, so add them:
0.063+0.063+0.063=3×0.063=0.189.0.063 + 0.063 + 0.063 = 3 \times 0.063 = 0.189.

The answer is (B) 0.189\boxed{\textbf{(B)}\ 0.189}.

Why this works

For separate random picks, the chance of one exact order is the product of the individual chances. "Exactly one" does not say which occasion, so count the orders that fit and multiply. Choice (A) is the chance of just one order; choice (E), 0.4410.441, is the chance of exactly two approvals. Listing the orders keeps you from mixing these up.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Each occasion independently yields an approval with probability 0.70.7 and a disapproval with probability 0.30.3.

A specific pattern with one approval and two disapprovals, say approve-disapprove-disapprove, has probability
0.7⋅0.3⋅0.3=0.063.0.7 \cdot 0.3 \cdot 0.3 = 0.063.
The single approval could occur on any of the 33 occasions, and these three patterns are disjoint, so
3⋅0.063=0.189.3 \cdot 0.063 = 0.189.

The answer is (B) 0.189\boxed{\textbf{(B)}\ 0.189}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.