Four integers are chosen at random between 0 and 9, inclusive. Find the probability that (a) not more than 2 are the same.
What I tried: all unique numbers: 63/125,
two same numbers: 72/1000
And then add them both. But the answer in the book is 963/1000. I'm not getting such a high probability. Where have I made the mistake?
Best Answer
You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $\binom42$ ways. Then you get $\frac{936}{1000}$, as José wrote. Since the book says $\frac{963}{1000}$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $\frac{\binom42\binom{10}2}{10^4}=\frac{27}{1000}$, so the total is then $\frac{963}{1000}$.