Question

Three natural numbers are taken at random from the set $A=\{x/1\le x\le 100,x\in N\}$. The probability that the $AM$ of the numbers taken is $25$ is

• A. $\frac{{}^{77}{C}_2}{{}^{100}{C}_3}$
• B. $\frac{{}^{25}{C}_2}{{}^{100}{C}_3}$
• C. $\frac{{}^{74}{C}_{72}}{{}^{100}{C}_{97}}$
• D. $\frac{{}^{75}{C}_3}{{}^{100}{C}_3}$

Answer 1

The correct option is C $\frac{{}^{74}{C}_{72}}{{}^{100}{C}_{97}}$

Let the three numbers be $a,b,c$.
Thus, $a+b+c=75$ where $a,b,c$ are positive integers.

This can be done in ${}_{(3-1)}^{(75-1)}C{=}_2^{74}C{=}_{72}^{74}C$ ways.

Thus, required probability = $\frac{{}_{72}^{74}C}{{}_3^{100}C}=\frac{{}_{72}^{74}C}{{}_{97}^{100}C}$