Three different numbers are selected at random from the set `A={1, 2, 3,…, 10}`. The probability that the product of two of the numbers is equal to third, is

Options

A. `(3)/(4)`
B. `(1)/(40)`
C. `(1)/(8)`
D. `(39)/(40)`

Answer

B

Explanation

Out of 10 numbers, three numbers can be chosen in `.^(10)C_(3)` ways.
`therefore` Total number of elementary events `=.^(10)C_(3)`
The product of two numbers, out of the chosen numbers, will be equal to the third number if the numbers are chosen in one of the following ways :
(2,3, 6), (2, 4, 8), (2, 5, 10)
`therefore` Favourable number of elementary events =3
Hence, required probability `=(3)/(.^(10)C_(3))=(1)/(40)`