Probability that among 3 random digits two different one

Question

I have been trying to solve the following problem:

What is the probability that among 3 random digits, there appear exactly 2 different ones?

The formula for no repititions is:

(n*(n-1)...(n-r+1))/n^r

So, for the first digit there are 10 possibilities, for the second 9. Next, the third digit should be the same as previous two: it is binomial(2,1) And there are total 10^3 possibilities of ordering with replacement: My formula is:

(10*9*binomial(2,1))/10^3

But it leads to the wrong answer. Could you please help me understand where is my mistake and the logic of solving such problems? Thank you

Answer 1

There are $10$ choices for the digit that appears twice and then $9$ choices for the digit that appears once. There are $\binom{3}{1}=3$ ways to arrange the digits. This gives $10\cdot9\cdot3=270$ choices out of $1000$. That is, a $0.27$ probability.

Answer 2

Your mistake is that you don't consider the case where the first digit repeats. You have the following exhaustive and exclusive cases:

• The first digit repeats: There are $10$ options for the first digit, which can appear in any of the $2$ remaining positions, and the other digit is one in $9$. There are $10 \cdot 2 \cdot 9 = 180$ cases here
• The first digit doesn't repeat: You can select the first digit in $10$ ways, and the other two (which repeat) in $9$ ways, for a total of $10 \cdot 9 = 90$

Adding both gives $180 + 90 = 270$ cases.