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Probability - Error in lines Watch

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    I am really stumped on how do do this question:

    A page of typescript contains 40 lines, with 80 characters per line. Each character has probability p=0.001 (independently of the others) of being erroneous.
    Given that the page contains exactly two errors, what is the probability that they occur on separate lines?

    Well to find that the page contains exactly two errors:

    Let random variable X = no. of errors in the page
    We can use the binomial distribution to find the probability:

    P(X=2) = \binom{3200}{2} (0.001)^2 (0.999)^{3198} = 0.209

    I am not sure how to move on from here.
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    (Original post by aspiring_doge)
    I am really stumped on how do do this question:

    A page of typescript contains 40 lines, with 80 characters per line. Each character has probability p=0.001 (independently of the others) of being erroneous.
    Given that the page contains exactly two errors, what is the probability that they occur on separate lines?

    Well to find that the page contains exactly two errors:

    Let random variable X = no. of errors in the page
    We can use the binomial distribution to find the probability:

    P(X=2) = \binom{3200}{2} (0.001)^2 (0.999)^{3198} = 0.209

    I am not sure how to move on from here.
    what is the probability of having 1 error in a single line of 80 characters?
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    (Original post by aspiring_doge)
    I am really stumped on how do do this question:

    A page of typescript contains 40 lines, with 80 characters per line. Each character has probability p=0.001 (independently of the others) of being erroneous.
    Given that the page contains exactly two errors, what is the probability that they occur on separate lines?
    This is one of those naughty questions where you are given redundant information to lead you off onto dead ends.

    First, notice that you are given that there are two errors on the page; you don't need to work out the probability of there being two errors - the p=0.001 is redundant.

    Second, notice that each line contains the same number of characters. Therefore the probability of an error occurring in line A is the same as the probability of an error occurring in line B.

    Therefore the problem is reduced to the problem of allocating two balls (the errors) to 40 urns (the lines) so that no urn contains two balls.

    How many ways are there of allocating two balls to forty urns? How many of these configurations have two balls in an urn? Can you finish it off now?
 
 
 
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