Typing Probability The old printer in the computer room skips 1 character in 20 (chance of 1/20 of skipping a character)
From the beginning of a new line, how many characters can the old printer be expected to print correctly before the first character is skipped?
Assuming that half a line (40 characters) has been printed correctly, from this point how many correct characters can be expected before the first character is skipped?
So, my thinking is
$$E(w)=1/p$$
$$E(w)=20$$ 
It can be expected to print 20 characters before the first one is skipped.
As for the second part, using the result from the first part, if 40 were printed correctly then 0 more correct characters can be expected before the first character is correct?
Really unsure about this problem, especially the second part.
 A: We model the situation by assuming independence. Let $X$ be the number of characters  up to and including the first character skipped. Then $X$ has geometric distribution with parameter $p=\frac{1}{20}$, so $E(X)=20$.
However, the question asked for the expected number of correctly printed characters before the first skip. This is $X-1$, and therefore has mean $19$.
As to the second question, we use the memorylessness of the geometric distribution. 
The answer is again $19$. 
A: If $X$ denotes the number of characters correctly printed until the first is skipped, $C$ denotes the event that the first character is printed correctly and $S$ the (complementary) event that the first character is skipped then:$$\mathbb{E}X=\mathbb{E}\left(X\mid C\right)P\left(C\right)+\mathbb{E}\left(X\mid S\right)P\left(S\right)=\left(1+\mathbb{E}X\right)\frac{19}{20}+0.\frac{1}{20}$$
and consequently: $$\mathbb{E}X=19$$
Here $\mathbb{E}\left(X\mid C\right)=1+\mathbb{E}X$ because after typing a correct character the same process starts again keeping in mind that $1$ correct character has allready been printed.
