1995 AHSME Problems/Problem 20

Problem

If $a,b$ 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


$\mathrm{(A) \ \frac {2}{5} } \qquad \mathrm{(B) \ \frac {59}{125} } \qquad \mathrm{(C) \ \frac {1}{2} } \qquad \mathrm{(D) \ \frac {64}{125} } \qquad \mathrm{(E) \ \frac {3}{5} }$

Solution

The probability of $ab$ being odd is $\left(\frac 35\right)^2 = \frac{9}{25}$, so the probability of $ab$ being even is $1 - \frac{9}{25} = \frac {16}{25}$.

The probability of $c$ being odd is $3/5$ and being even is $2/5$.

$ab+c$ is even if $ab$ and $c$ are both odd, with probability $\frac{9}{25} \cdot \frac{3}{5} = \frac{27}{125}$; or $ab$ and $c$ are both even, with probability $\frac{16}{25} \cdot \frac{2}{5} = \frac{32}{125}$. Thus the total probability is $\frac{59}{125} \Rightarrow \mathrm{(B)}$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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