the problem states that 26 letters A,B…Z are arranged in a random order.
1) what is the probability that A comes before B in the random order?
2) what is the probability that A comes before Z in the random order?
so for the first one i know that total number of possible arrangements is 26!, then for A to be before B we would start with A being in the first position then we would have 25 ways for it to be before B, then move A one position then we would only be left with 24 ways, keep going like that, which means there is 25! ways for A to be before B, so probability is 25!/26! or 1/26, same method and answer would go for question 2. am i right or did i make mistake somewhere?
Best Answer
Yes, your reasoning and your answer are correct.
Another way to see this is to see that the probability that A comes before B is the same as the probability that A comes before some other letter. There are 25 letters that are not A. But there is the case where A is the last letter. There are 26 cases then, each having the same number of outcomes. The probability of each is 1/26.