A person while dialing a telephone number, forgets the last three digits of the number but remembers that exactly two of them are same. He dials the number randomly. The probability that he dialed the correct number, is equal to.

$\frac{1}{270}$

Total number of ways of dialing the last three digits such that exactly 2 of them are same

$={ }^{10} C_2 . 2 . \frac{3 !}{2 !}=270$

Thus, required probability $=\frac{1}{270}$