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Parity probability of ab+c

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November 21, 2005
probability

Problem Statement

If a,ba,b and cc are three (not necessarily different) numbers chosen randomly and with replacement from the set {1,2,3,4,5}\{1,2,3,4,5 \}, the probability that ab+cab+c is even is
<spanclass=′latex−bold′>(A)</span> 25<spanclass=′latex−bold′>(B)</span> 59125<spanclass=′latex−bold′>(C)</span> 12<spanclass=′latex−bold′>(D)</span> 64125<spanclass=′latex−bold′>(E)</span> 35<span class='latex-bold'>(A)</span>\ \dfrac{2}{5} \qquad <span class='latex-bold'>(B)</span>\ \dfrac{59}{125} \qquad <span class='latex-bold'>(C)</span>\ \dfrac{1}{2} \qquad <span class='latex-bold'>(D)</span>\ \dfrac{64}{125} \qquad <span class='latex-bold'>(E)</span>\ \dfrac{3}{5}
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