Three numbers are chosen at random from numbers 1 to 30. The probability that the minimum of the chosen numbers is 9 and maximum is 25, is

$\frac{3}{812}$

Out of first 30 natural numbers, three natural numbers can be chosen in ${^{30}C}_3$ ways.

If the minimum and maximum of the numbers chosen are 9 and 25 respectively, then we have to select a number from the remaining 15 numbers i.e. 10, 11,...,...,. 24.

.. Favourable number of elementary events = ${^{15}C}_1$.

Hence, required probability $=\frac{^{15}C_1}{^{30}C_3}=\frac{3}{812}$