Question

@Salty (I luv u, just this last question, idk if u can solve it though :p)

1. On average, @dumbcow corrects 1 word per 800 typed words.
2. A page has 200 words on average.
3. Probability of at least 1 correction per typed page is 0.0265 (we solved this in the previous question).
4. If at least1 correction per page is made, then it needs to be retyped.

What is the probability that a page has at least 3 attempts?
(I think we use binomial? :o)

Answer 1

i luv you too<3

Answer 2

Hmm, well the answer key says 0.0007

Answer 3

wait
in this sentence-"What is the probability that a page has at least 3 attempts?"
does attempt means corrections?

Answer 4

ohh wait it said "atleast 3 attempts"

Answer 5

"atleast"

Answer 6

yeah, attempt means there's a 'correction' to be made.

Answer 7

So basically we're trying to find the complement of that.
The probability of 1 attempt and 2 attempts.

Answer 8

Too hard? :(

Answer 9

almost done ig

Answer 10

so we gotta subtract the probability of getting 0, 1and 2 attempts from 1

so
we already know that 1-P(0)-P(1) is 0.0265
and P(2) is what we need to calculate \(P(2)=\large\frac{(0.25)^2e^{-0.25}}{2!}=0.024337\)

soo P(needed)=1-P(0)-P(1)-P(2) =

Answer 11

but the answer according to this is 0.002162

Answer 12

aaaahhh....

Answer 13

whateverrrrr

Answer 14

lol its annoying when you're doing stuff right(correct according to yourself) and still the answer doesn't match

Answer 15

ima recheck
i almost made a ton of mistakes while solving this..

Answer 16

i rechecked but my answers still the same

i think i'm missing something here
" If at least1 correction per page is made, then it needs to be retyped.
What is the probability that a page has at least 3 attempts?"