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If X is chosen at random from ${1, 2, 3, 4}$ and Y is chosen at random from ${5, 6, 7}$ the probability that XY will be even is

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$\frac{1}{6}$

$\frac{5}{6}$

$\frac{2}{3}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$x \in {1, 2, 3, 4} y \in {5, 6, 7}$

A: xy is even $E ⟶ even, O ⟶ odd$

$P (A) = \begin{matrix} E & E & O & E & E & O \\ \frac{2}{4} \times & \frac{1}{3} + & \frac{2}{4} & \frac{1}{3} + & \frac{2}{4} \times & \frac{2}{3} \end{matrix}$

$= \frac{2 + 2 + 4}{2} = \frac{8}{12} = \frac{2}{3}$

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