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
Answer & explanation
Correct answer: option 2
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.
$= 1 - \frac{^{67}C_2}{^{100}C_2}=1- \frac{67×33}{50×99}= 1 - \frac{67}{150} =\frac{83}{150}$