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August 17th, 2012, 12:03 PM   #1
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Binomial distribution and the typewritten text

The probability that a typewritten page is free from error is 1/2;
the probability that it contains exactly n errors is inversely proportional to n for n =1,2,3,4;
and the probability that it contains more than 4 errors is 1/12.

Calculate the probability that
i) a page will contain exactly one error [Ans: 1/5]
ii) six pages will consist of four pages containing no error and the remaining two pages containing exactly one error each [Ans:3/80]

I think I must involve the binomial distribution here but i dont know to apply the given information into in to get the answers. Please help.
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August 18th, 2012, 05:41 AM   #2
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Re: Binomial distribution and the typewritten text

Hey UnscaryMidnight.

You can't just rely on the binomial theorem: you need to use the statements and make a deduction. You are given:

P(X = 0) = 1/2
P(X > 4) = 1/12
P(X = a) = c/a for some c where a = 1,2,3.4

And also P(X = 0) + P(X = 1 or 2 or 3 or 4) + P(X > 4) = 1/2 + c/a + 1/12 = 1.

Remember that in general, you can't just default to an assumption like binomial or something else: if you are given things like this you need to double check them just in case they might be different.

Based on the above hint, can you then solve for P(X = 1) (the first question)?
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August 18th, 2012, 06:42 AM   #3
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Re: Binomial distribution and the typewritten text

Thanks for replying!

Based P(X = 0) + P(X = 1 or 2 or 3 or 4) + P(X > 4) = 1/2 + c/a + 1/12 = 1.
then P(X=1 or 2 or 3 or 4) = 5/12
my logic then says 5/12 - P(2 or 3 or 3) should give me the answer...? but then how would i work out the probability of 2 or 3 or 4 errors? and now i dont think i can use the binomial theorem as i dont seem to have enough parameters for it...
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August 18th, 2012, 08:21 PM   #4
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Re: Binomial distribution and the typewritten text

Sorry I made a mistake.

P(X = 1 OR 2 OR 3 OR 4) = c/1 + c/2 + c/3 + c/4 not just c/a: the a was meant to be a place-holder for the actual realization. You know that all probabiltiies add up to 1 and you only have a c value that you need to solve for. This c will give you the probabilities for X = 1,2,3,4 which you will be able to use.
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August 19th, 2012, 09:34 AM   #5
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Re: Binomial distribution and the typewritten text

Thanks!
I got the first part now!
i get c= 1/5
and since P(x=1) = c/1 it will be 1/5

what would ur advice be for the second part?
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August 19th, 2012, 09:41 AM   #6
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Re: Binomial distribution and the typewritten text

What i did was (1/2)^4 for the 4 pages with no errors and then added (1/5)^2 but that doesnt get me the right answer...
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August 19th, 2012, 05:42 PM   #7
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Re: Binomial distribution and the typewritten text

Hint: Remember that you can many different combinations of errors. How many combinations can you get in 6 pages if you have 4 that have no errors and 2 that have one error: (Hint think of the binomial distribution)?
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