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

Options

A. `(1)/(406)`
B. `(1)/(812)`
C. `(3)/(812)`
D. none of these

Answer

C

Explanation

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.
`therefore` Favourable number of elementary events `= .^(15)C_(1)`.
Hence, required probability `=(.^(15)C_(1))/(.^(30)C_(3))=(3)/(812)`