Probability - selection with replacement and without replacement, At present I am revising for my upcoming Statistics exam on Friday. One topic in particular that I have difficulty grasping is selection without and with replacement. I was wondering if anybody knew a quick but effectual short-hand methodology to work out questions of the same format as the ones below? 1. A man has $7$ keys, of which only one will open the door of his home. (i) Suppose he tries the keys at random, discarding (i.e. setting aside) those that do not work. What is the probability that he will open the door on the $5$th try? (ii) Suppose he tries the keys at random without discarding those that do not work. What is the probability he will open the door on the $5$th try?

(i) Since the number of keys available keeps decreasing, the probability would be $${6\over 7}{5\over 6}\cdots {3\over 4}{1\over 3}=\frac{6*5*4*3}{7*6*5*4*3}=\frac{1}{7}$$

(ii) Now it would have to fail four times and then work one time: $$P=\left({6\over 7}\right)^4\left({1\over 7}\right)=\frac{6^4}{7^5}$$ Note that (i) is about twice as large as (ii) - it makes sense that it's bigger because the "bad" keys are being thrown out as they go.