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Two different digits are selected at random from the digits $1$ through $9$. What is the probability that the sum of two digits is odd?

We have a sample space of $9 \times 8$ ways to choose two digits. To get an odd sum the digits should have different parities, so I have a doubt about the number of ways $5 \times 4$ or $5 \times 4 \times 2$ (because the first number can be odd or even) which one is correct?

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  • $\begingroup$ By stating that the sample space has $9 \cdot 8 = 72$ elements, you have implicitly considered the order of the digits, which is why there are $5 \cdot 4 \cdot 2 = 40$ favorable cases. If we ignore the order of the digits, there are $\binom{9}{2}$ ways to select two different digits and $\binom{5}{1}\binom{4}{1}$ ways to select one odd digit and one even digit, which yields the same probability. $\endgroup$ Apr 9, 2022 at 10:15

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One way to reason is to consider separate cases. Suppose the first integer selected is odd; then the second must be even. Since there are $5$ odd numbers and $4$ even numbers from $1$ through $9$ inclusive, it follows that there are $5 \times 4 = 20$ such choices.

Conversely, suppose the first integer is even; then the second must be odd. This again leads to $4 \times 5 = 20$ choices, for a total of $40$ in all.

Now if two distinct numbers are chosen at random without restriction, there are $9 \times 8 = 72$ such outcomes. So the desired probability is $40/72 = 5/9$.

Another way to perform the counting is to look at the complementary probability. What are the outcomes for which the sum is even? This occurs if the parity of each digit is the same. In the odd-odd case, this is simply $5 \times 4 = 20$. And in the even-even case, this is $4 \times 3 = 12$. So the total number of choices that result in an even sum is $20 + 12 = 32$ and the complementary probability is $32/72 = 4/9$.

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