Two numbers a and b are selected at random from 1,2,. 100 and are multiplied. The probability that the product ab is divisible by 3, is

$\frac{83}{150}$

Two numbers can be chosen out of first  100 and natural numbers on ${^{100}C}_2$ ways.

∴ Total number of elementary events= ${^{100}C}_2$

The product ab is divisible by 3 if at least one of the two numbers is divisible by 3.

∴ Required probability

=1 - Probability that none of the two numbers chosen is divisible by 3.

$= 1 - \frac{^{67}C_2}{^{100}C_2}=1- \frac{67×33}{50×99}= 1 - \frac{67}{150} =\frac{83}{150}$