Question 535145: Suppose 200 misprints ar distributed randomly throughout a book of 600 pages. Find the probability that a given page contains.
(a) exactly two misprints
(b) two or more misprints
Answer by SwiftAlbatross(13)   (Show Source):
You can put this solution on YOUR website! Assuming the chance of a misprint being on any page is uniform and that each
misprint is an independent event, we can think of a misprint being randomly
assorted into 600 pages and this occurrence happening 200 times. Therefore, the
probability a misprint will go to any page is 1/600 since there are 600 pages and
a misprint may go into any of them evenly. Then, like flipping coins, the next
misprint event happens independently of the previous one and so its probability of
going to any page is also 1/600, and so on and so on for all 200 misprints. Then,
just like coins, you can multiply their probabilites together. For example, what
is the probability of flipping 3 Heads in 3 coin tosses: Simply (1/2)*(1/2)*(1/2).
The same logic can be used here: For instance, the probability that one page will
have all 200 misprints is (1/600)^200.
Part (a) asks that there are exactly 2 misprints on one page and the other 198
misprints must be on the other 599 pages. So you can do
(1/600) * (1/600) * (599/600)^198 * 200C2
where 200C2 is 200 Choose 2 or C(200,2) [Combinations: I'll explain later]
Part (b) asks that one page contains 2 or more misprints. We could calculate the
probability of a page getting exactly 2 misprints and then add to it the
probability of getting exactly 3 misprints and then add the probability of getting
exactly 4 misprints and so on, but that is time consuming so we can just subtract
the probability of getting exactly one misprint on that page and exactly no misprints on
that page from 1, since this is the same probability. So
1 - (1/600) * (599/600)^199 * 200C1 - (599/600)^200 * 200C0
To convince yourself of this logic, use smaller numbers and work with actual
paper. For example, suppose you have only 6 pages and 2 misprints going among
them. A misprint may go to the 6 pages evenly. So the chance page 1 will have 2
misprints is :
(1/6) * (1/6)
The first misprint must go to page 1 AND the second misprint must go to page 1.
exactly one misprint:
(1/6) * (5/6) + (5/6) * (1/6)
The first misprint must go to page 1 AND the second misprint must go to pages 2-6
OR
The first misprint must go to pages 2-6 AND the second misprint must go to page 1
no misprints:
(5/6) * (5/6)
The first misprint must go to pages 2-6 AND the second misprint must go to pages 2-6.
one or more misprints:
(1/6) * (5/6) + (5/6) * (1/6) + (1/6) * (1/6) = 1- (5/6) * (5/6)
Either write the case for exactly one misprint plus the case for exactly two
misprints or subtract the case for exactly no misprints from one.
Even though I write first and second misprint, they are independent events so the order of
the misprints is inconsequential but to keep track of which misprint goes where, I
order them.
Now imagine you have four misprints in 6 pages.
P(Exactly 4 misprints on a given page) =
(1/6) * (1/6) * (1/6) * (1/6)
P(Exactly 3 misprints on a give page) =
(1/6)*(1/6)*(1/6)*(5/6) + (1/6)*(1/6)*(5/6)*(1/6) + (1/6)*(5/6)*(1/6)*(1/6) +
(5/6)*(1/6)*(1/6)*(1/6) = (1/6)*(1/6)*(1/6)*(5/6)*4
P(Exactly 2 misprints on a given page) =
(1/6)*(1/6)*(5/6)*(5/6) + (1/6)*(5/6)*(1/6)*(5/6) + (1/6)*(5/6)*(5/6)*(1/6) +
(5/6)*(1/6)*(1/6)*(5/6) + (5/6)*(1/6)*(5/6)*(1/6) + (5/6)*(5/6)*(1/6)*(1/6) =
6 * (1/6)*(1/6)*(5/6)*(5/6)
As you can see, the number being multiplied is just the differnt combinations of
misprints appearing from the total number of misprints. So nCr = (# of total
misprints) C (# of misprints appearing on target page). For the case of exactly 2
misprints on a given page that preceded, this is 4C2 or 6.
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