Correct Answer - B
Out of 10 numbers, three numbers can be chosen in .10C3 ways.
∴ Total number of elementary events =.10C3
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)
∴ Favourable number of elementary events =3
Hence, required probability =
3
.10C3
=
1
40