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
Answer & explanation
Correct answer: option 3
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
$=1-\frac{(286+468)}{{ }^{25} C_3}=\frac{773}{1150}$