Given that x is to be chosen at random from the set {1,2,3}, y is to be chosen at random from the set {4,5,6}, and z is to be chosen at random from the set {7,8,9,10} and we need to find what is the probability that xyz will be even

We know that Total Probability is always equal to 1

=> P(Odd) + P(Even) = 1
=> P(Even) = 1 - P(Odd)

To get xyz as odd we need to have all of them as odd.
=> P(xyz = Odd) = P(x = odd) * P(y = odd) * P(z = odd) = \(\frac{2}{3}\) * \(\frac{1}{3}\) * \(\frac{2}{4}\)

(Because for x we have two choices (1,3) to get odd out of 3, for y we have one choice (5) to get odd out of 3 and for z we have two choices (7,9) to get odd out of 4)

= \(\frac{1}{9}\)

=> P(Even) = 1 - P(Odd) = 1 - \(\frac{1}{9}\) = \(\frac{9 - 1}{9}\) = \(\frac{8}{9}\)

So, Answer will be E.
Hope it helps!