Three numbers are selected simultaneously from the set {1, 2, 3, …., 25}. The probability that the product of selected numbers is divisible by 4, is equal to

$\frac{773}{1150}$

Product can’t be divisibly by 4 in the following mutually exclusive cases;

(i) All the selected numbers are odd.

corresponding ways = ${ }^{13} C_3=286$

(ii) Two of the selected numbers are odd and another is a multiple of 2 only.

Corresponding ways = ${ }^{13} C_2 . { }^6 C_1=468$

Thus, required probability

$=1-\frac{(286+468)}{{ }^{25} C_3}=\frac{773}{1150}$