To calculate the probability of at least one win when there is a 1/10 probability of winning and ten trials directly is a bit of a challenge. There are many different ways that you can satisfy this criteria. You can win exactly once, exactly twice, exactly three times, and so on. To complicate things further there are 10 ways you can win once and even more ways you can win multiple times. The exact number of ways you can win a certain number n times is equal to the number of ways n
 items can be selected out of a set of 10, which can be calculated using the combination function. For a set of 10 the number of combinations of sizes from 0 to 10 are 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, and 1. So there are 10 ways to win once, 45 ways to win twice, 120 ways to win 3 times, and so on.
So now to actually calculate some probability. the probability of winning exactly once is equal to the product of the probability of winning one game (1/10) times the probability of losing 9 games. (9/10^9) times the number of possible ways to win exactly one game (10). Similarly we can look at the probability of winning exactly two games as the probability of winning 2 games (1/10^2) times the probability of losing 8 games (9/10^8) times the number of ways to win exactly 2 games (45). The same math can be extended to every possible exact number of wins.
The last step in calculating this probability is to remember that we are not interested in any particular number of wins - we are only interested in the probability of winning at least once. In other words, the probability of winning exactly once or exactly twice or exactly three times or so on. So to calculate the total probability of winning at least once we just have to add together the probabilities of winning exactly once, exactly twice, exactly three times and so on, and then we will finally have the probability of winning at least once out of ten games.

If on the other hand you wanted to add the probabilities, the problem here is that the probabilities are not mutually exclusive. It is certainly possible to win both the first game and the second game, so if you add the probabilities you are actually double counting any outcomes in which you win both games. In this case you would need to take the sum of the probability of winning either game minus the probability of winning both games to account for the double counting - an idea known as the inclusion-exclusion principle.
Expanding this out to 10 games, we can look at the probability of winning 1 game as 1/10, the probability of winning 2 games as 1/100, the probability of winning 3 games as 1/1000, and so on. Since there are 10 ways to win once we add the probabilities together to get 10*1/10, or 1. We then subtract the double counted sets of two wins or 45*1/100. This actually double subtracts the sets of three wins, so we again add 120*1/1000, then subtract 210*1/10,000, and go on alternating adding and subtracting until we’ve counted all of our sets - yet again we get the same exact answer - the probability of winning at least once is .651322 or just under 2/3.