# Drawing balls with replacement

I have the following question in an exam paper and I don't know exactly how to treat it. I would appreciate a hint with regards to what formula to use or such:

A bag contains 5 red balls and 1 black ball. Balls are randomly drawn from the bag
until the black ball is found, and the number of draws require is noted.

How much greater is the expected number of draws required if any red balls from
the bag is replaced rather than put to one side?


Now, I know that if there was no replacement, the black ball has a chance of 1/6+1/5+1/4+1/3+1/2 to be extracted. Please correct me if I'm wrong.

What happens when there's replacement involved?

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With no replacement, the black ball has probability $1$ to be (ultimately) extracted. The expression $\frac{1}{6}+\frac{1}{5}+\cdots$ has no clear connection to the problem. – André Nicolas Jul 24 '12 at 20:17

a) In general, if we have independent trials with probability of success $p$, then the expected number of trials until the first success is $\frac{1}{p}$. (This is a fact about the expectation of a random variable that has geometric distribution.) With replacement, we have independence, and $p=\frac{1}{6}$.
b) For without replacement, there is general theory, but it may be simplest to just calculate. Let $X$ be the number of trials until we get our black. Then $\Pr(X=1)=\frac{1}{6}$. Also, $\Pr(X=2)=\frac{5}{6}\cdot \frac{1}{5}$ (failure then success). And $\Pr(X=3)=\frac{5}{6}\cdot \frac{4}{5}\cdot \frac{1}{4}$. And so on. Notice anything? If you think about it, is it obvious without calculation?