1995 AMC 12 Problem 20

Attempt Problem 20 of the 1995 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AMC 12 solutions, or check the answer key.

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20.

a,

b

and

c

are three (not necessarily different) numbers chosen randomly and with replacement from the set

\{1,2,3,4,5\},

the probability that

ab+c

is even is

\frac25

\frac{59}{125}

\frac12

\frac{64}{125}

\frac35

The sum is even when

ab

and

c

have the same parity

The product

ab

is odd only when both selected factors are odd

There are

3

odd and

2

even choices. The product

ab

is odd with probability

\frac{9}{25}

and even with probability

\frac{16}{25}.

Therefore

\begin{aligned} P(ab+c\text{ even}) &=\frac9{25}\cdot\frac35\\ &\quad{}+\frac{16}{25}\cdot\frac25\\ &=\frac{59}{125}. \end{aligned}

Thus the correct answer is B.

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