Question

two numbers are selected from a set of first 90 natural numbers.find the probability that the product of randomly selected number is divisible by3

Answer 1

Number of favourable outcomes are 30

Total no.of outcome are 90

Probability of no.divisible by 3 are=30/90

=1/3

Answer 2

Between first 90 natural numbers ,number of numbers divisible by 3 are =3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,......., 66,69,72,75,78,81,84,87,90.=30 in number

Now,two numbers are selected at random

Probability of an event $=\frac{\text{Total favorable outcome}}{\text{total possible outcome}}$

Probability of selecting 2 numbers out of 90 natural numbers which are divisible by 3

$=\frac{_{2}^{30}\textrm{C}}{_{2}^{90}\textrm{C}}\\\\=\frac{\frac{30!}{28!\times2!}}{\frac{90!}{88!\times2!}}\\\\=\frac{29\times15}{89 \times45}\\\\=\frac{29}{267}$