Formula used: Selection of r things out of total n things:\(^nC_r = \frac{n!}{r!\times (n-r)!}\)Probability of occurrence of the event:P(E) = \(\frac{n(E)}{n(S)}\)Where,n(E) = Number of favorable outcomen(S) = Number of possible outcomeCalculation:Given digits are 1, 2, 3, 4, and 5.As discussed above we know, that two digits can be chosen out of 5 in 5C2 ways.? n(S) = 5C2 = \(\frac{5!}{2!\times 3!}\) ? n(S) = \(\frac{5 \times 4 \times 3!}{2 \times 1\times 3!} = 10\)? n(S) = 10Since the last digit in the product appears as 0 which is only possible when any of 2 or 4 is multiplied by 5 so possible required result,(E) = {(2, 5), (5, 2), (4, 5), (5, 4)}? n(E) = 4Hence, required probabilityP(E) = n(E)/n(S) = 4/10? P(E) = 2/5? Required probability is \(\frac{2}{5}\).