Three integers are chosen at random from the first 20 integers. The probability that their product is even, is

The total number of ways in which 3 integers can be chosen from first 20 integers is${20}_{C_3}$ .

The product of three integers will be even if at least one of the integers is even. Therefore, the required probability

 $=1-$ probability that none of three integers is even$=1-\frac{{10}_{C_3}}{{20}_{C_3}}=1-\frac{2}{19}=\frac{17}{19}$ .

[Three odd integers can be chosen in ${10}_{C_3}$ ways as there are 10 even and 10 odd integers.