Two distinct numbers are chosen at random from the first 15 natural numbers. The probability that the sum will be divisible by 3 is

$\frac{31}{91}$

Out of first 15 natural numbers, 3 natural numbers can be chosen is ${^{15}C}_3$ ways.

When we divide a natural number by 3, it leaves either 0 or 1 or 2 as the remainder, so, let us divide 15 natural numbers into the following groups:

$G_1 : 1 \,\,\,\, 4 \,\,\,\, 7 \,\,\,\, 10 \,\,\,\, 13$

$G_2 : 2 \,\,\,\, 5 \,\,\,\, 8 \,\,\,\, 11 \,\,\,\, 14$

$G_3 : 3 \,\,\,\, 6 \,\,\,\, 9 \,\,\,\, 12 \,\,\,\, 15$

The sum of three natural numbers chosen from these 15 natural numbers will be divisible by 3 if either all the three natural numbers are chosen from the same group or one natural number is chosen from each group.

∴ Favourable number of elementary events 

$= ({^5C}_3 + {^5C}_3+ {^5C}_3) + ({^5C}_1×{^5C}_1×{^5C}_1)= 155$

Hence, required probability $=\frac{155}{^{15}C_3}=\frac{155×6}{15×14×13}=\frac{31}{91}$