Binomial distribution problem This is a question in my statistics book:


I could answer perfectly till part (c) but honestly, I have no idea why the answer to (d) is $0.4005$.
Why is the top divided by $0.065^2$ and where did the multiply with $2$ come from?
 A: $$P(\text{Lisa typed exactly 1}|\text{ Both had errors})={P(\text{Lisa typed one, Pat typed one, and both had errors})\over P(\text{both had errors})}$$
The probability that Pat types one and makes an error is $(0.06)(0.3)$, while the probability that Lisa types one and makes an error is $(0.94)(0.05)$.  So the probability that Pat types the first, Lisa types the second, and both make errors is $(0.06)(0.3)(0.94)(0.05)$.  But it's also possible that Lisa types the first and Pat types the second, and so $$P(\text{Lisa typed one, Pat typed one, and both had errors})=2(0.06)(0.3)(0.94)(0.05)$$
and that's where the 2 on the top comes from.  $P(\text{both had errors})$ is found using similar logic.  Pat can type both and make errors on both $\rightarrow (0.06)^2(0.3)^2$, Lisa can type both and make errors on both $\rightarrow (0.94)^2(0.05)^2$, or each can type one of them and make errors on both $\rightarrow 2(0.06)(0.3)(0.94)(0.05)$.  So
$$P(\text{both had errors})=(0.06)^2(0.3)^2+(0.94)^2(0.05)^2+2(0.06)(0.3)(0.94)(0.05)=\left((0.06)(0.3)+(0.94)(0.05)\right)^2=(0.065)^2$$
And so you end up with
$${2(0.06)(0.3)(0.94)(0.05)\over (0.065)^2}\approx 0.4005$$
Edit: You also know from part c that a randomly chosen letter will contain errors with probability $0.065$, so it stands to reason that the probability of two randomly chosen letters both containing errors is $0.065^2$, which is considerably less complicated.
