We have tried 2 methods to solve a problem previously.
We have tried Method#1 100 times and it has been successful 70 times and for Method#2, we have tried it 10 times and 6 times it has been successful.
Now I want to know if I want to decide which method to use for the next experiment, which method has the higher possibility of success.
If we assume that methods always have the same probability of success $p_1$ and $p_2$ respectively, then we can make some estimate of these (unknown to us) parameters based on empirical evidence. Maximum likelihood estimate will give us $p_1 = 0.7, \, p_2 = 0.6$ and that will be that.
Now comes the tricky part (spoiler: answer for your question as it is won't change). Suppose that second method had $8/10$ success rate. The probability of this happening with $p_2 = 2/3$ would be $45*2^8/3^{10} \approx 0.2$. In other words, while on average second method would be better, there would be a significant chance of it being worse.
If all you care about is having your success rate as high as possible, simply choosing method with best success rate will give you what you want. However, if you wish to, say, get average success rate $0.75$ with least "chance of surprise", actual answer in this altered example would be a mixed strategy (using first strategy with probability $r_1$ and second with probability $r_2$). The higher you set your intended average rate, the greater becomes the dispersion (probability of getting significantly higher or lower than average).
Alternatively, you can try to get the best possible average result for the given deviation (in general case, you will get a mixed strategy as well). Exact means of computing the best mix in these cases are studied in modern portfolio theory.