Three natural numbers are taken at random from the set of first 100 natural numbers. The probability that their A.M. is 25, is
Answer & explanation
Correct answer: option 3
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 $=\frac{^{74}C_{2}}{^{100}C_{3}}=\frac{^{74}C_{72}}{^{100}C_{97}}$