Three natural numbers are taken at random from the set of first 100 natural numbers. The probability that their A.M. is 25, is

$\frac{^{74}C_{72}}{^{100}C_{97}}$

Three natural numbers can be chosen out of 100 natural numbers in ${^{100}C}_3$ ways.

∴ Total number of elementary events =${^{100}C}_3$

The A.M. of three numbers is 25.

∴ Their sum =75

∴ Favourable number of elementary events

= Number of ways of selecting three numbers whose sum is 75

= Number of solution of x + y + z = 75 in N.

$={^{75-1}C}_{3-1}= {^{74}C}_2$

Hence, required probability $=\frac{^{74}C_{2}}{^{100}C_{3}}=\frac{^{74}C_{72}}{^{100}C_{97}}$