Probability - four random integers between 0-9, that not more than two are the same 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?
 A: 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 $\frac{963}{1000}$.
A: You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.
The probability of getting four equal numbers is $\frac{10}{10000},$ because there are ten numbers available to be all equal.
The probability of having a triplet is $\frac{4\times10\times 9}{10000}$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.
Summing these two probabilities, you obtain $\frac{370}{10000}$ which is exactly $1-\frac{963}{1000}.$
A: You may first calculate the complementary probability.


*

*There are $10$ 4-digit groups with 4 identical digits

*There are $4 \choose 3$ choices of places to put 1 out of 10 digits into these 3 places and 9 other digits to fill the remaining place: ${4 \choose 3} \cdot 10 \cdot 9$


All together, the probability of getting 3 or 4 equal digits is
$$P(\mbox{3 or 4 equal digits}) =\frac{10 + {4 \choose 3} \cdot 10 \cdot 9}{10^4} = \frac{37}{10^3} \Rightarrow $$
$$P(\mbox{at most 2 equal}) = 1-\frac{37}{10^3} = \frac{963}{1000} $$
