Computing probability for a printer?

Question

For a print machine, it's computed that the probability of printing a page with some misprints is 0.01. Using this print machine a book of 400 pages is going to be printed, compute (approxi- mately) the probability that there will be 8 pages with misprints.

Answer 1

Try to simplify the example first!

Say you are flipping a coin. The chance that you will get tails is 0.5, right? That's for one flip.

Alright say you are flipping the coin 2 times. The chance that you will get 2 tails is 0.5*0.5=0.25

What about if you are flipping the coin 3 times for a higher chance to get 2 tails? This means that you either want 2 tails and 1 heads, or 3 tails.

The probability of getting 2 tails and 1 head in a row is 0.5*0.5*0.5=0.125 (That's just (Tails)(Tails)(Heads), the probability of getting heads is also 0.5). The probability of getting 3 tails is 0.5*0.5*0.5=0.125 as well! (Tails)(Tails)(Tails) = 0.5*0.5*0.5. It just works out that way.

You can add both of these probabilities since you're okay with either of these results! At least 2 tails can be reworded as "I want either 2 tails or 3 tails". 0.125+0.125=0.25 chance of getting AT LEAST 2 tails in 3 flips

Note that the "at least" is important. If you wanted EXACTLY 2 tails, then they're asking for the probability of 2 tails and 1 head, which is only 0.125.

Your question didn't make that part really clear! But! Every time a page is printed, there's a 0.01 chance that it will be misprinted. If only one page is printed, the probability that it will be a misprinted one is 0.01. That's from the question. If two pages are printed, and you want exactly one misprint, the probability is (0.01)*(0.99) (The second number is the probability that the page will NOT be a misprint, since if you want exactly one misprint, the second page has to be printed normally) (I got the second number by taking 100% probability that you're printing a page, minus the 1% chance that it will be a misprint = the probability that it will not be a misprint).

Anyways, if you wanted the probability that AT LEAST one out of 2 pages would be misprinted, then it would be (0.01)(0.99)+(0.01)(0.01) (The probability that one or more pages will be misprinted out of 2 is the probability that only one will be misprinted + the probability that both would be misprinted).

Okay so you have 400 pages. If you wanted exactly 8 of them to be misprinted, that's a pretty small chance because it's very specific! In particular, you would have to do (0.01) multiplied by itself 8 times, times (0.99) multiplied by itself 392 times (the rest of the pages) So that's ($0.01^{8}$)*($0.99^{392}$) = approximately 1.94 * $10^{-18}$

that's a tiny number because obviously that's a tiny chance. If you're trying to find the probability that AT LEAST 8 sheets will be misprinted, the probability will be bigger just like in the other examples. But that's the probability that 8 will be misprinted + the probability that 9 will be misprinted + .... + the probability that all 400 will be misprinted.

SO

($0.01^{8}$)($0.99^{392}$) + ($0.01^{9}$)($0.99^{391}$) + ($0.01^{10}$)($0.99^{390}$) + ..... + ($0.01^{399}$)($0.99^{1}$) + ($0.01^{400}$)

That's a massive sum. But note that the last one is such a small number, because the probability that ALL of the 400 pages will be misprinted is really tiny!!! So if you wanted an approximation, you could just add up the first few and it would be good enough.