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    All the letters in a particular office are typed by either Pat, or by Lyn.
    The probability that a letter type by Pat will contain one or more errors is 0.3.
    The probability that a letter typed by Lyn will contain one or more errors is 0.05.

    On any one day, 6% of the letters typed in the office are typed by Pat. One letter is chosen at random from those typed on that day.
    Show that the probability that it will contain one or more errors is 0.065 (I managed to do this)

    Given that each of the 2 letters chosen at random from the day's typing contains one or more errors, find, to 4 decimal place, the probability that one was typed by Pat and the other by Lyn.

    what I thought was:

    P(Pat wrote the letter & made an error in a particular day) = 0.06*0.03
    P(Lyn wrote the letter & made an error in a particular day) = 0.94*0.05

    so 2*(0.94*0.05)(0.06*0.03) = 0.001692

    this isn't the correct answer and in the solutions they seem to have divided what I got by 0.065^2

    could anyone explain why (do so as if I am a baby).

    thanks.
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    (Original post by DH.)
    All the letters in a particular office are typed by either Pat, or by Lyn.
    The probability that a letter type by Pat will contain one or more errors is 0.3.
    The probability that a letter typed by Lyn will contain one or more errors is 0.05.

    On any one day, 6% of the letters typed in the office are typed by Pat. One letter is chosen at random from those typed on that day.
    Show that the probability that it will contain one or more errors is 0.065 (I managed to do this)

    Given that each of the 2 letters chosen at random from the day's typing contains one or more errors, find, to 4 decimal place, the probability that one was typed by Pat and the other by Lyn.

    what I thought was:

    P(Pat wrote the letter & made an error in a particular day) = 0.06*0.03
    P(Lyn wrote the letter & made an error in a particular day) = 0.94*0.05

    so 2*(0.94*0.05)(0.06*0.03) = 0.001692

    this isn't the correct answer and in the solutions they seem to have divided what I got by 0.065^2

    could anyone explain why (do so as if I am a baby).

    thanks.
    This is what we call conditional probability. Let me explain.

    If X and Y are defined to be events such that

    X: "both letters contain an error"
    Y: "they typed one letter each"

    what you've worked out is the probability that both letters contain and error and they typed one of these letters each, i.e

    P(X \cap Y)

    Can you see why?

    So now you want to answer

    "given both letters contain an error, find the probability that they typed one letter each".

    Hint: Recall and fill in the following in for events A and B

    P(A|B) =...

    Does that help?

    P.S I almost fell asleep typing this!! Stats is so uninteresting!
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    (Original post by Indeterminate)
    This is what we call conditional probability. Let me explain.

    If X and Y are defined to be events such that

    X: "both letters contain an error"
    Y: "they typed one letter each"

    what you've worked out is the probability that both letters contain and error and they typed one of these letters each, i.e

    P(X \cap Y)

    Can you see why?

    So now you want to answer

    "given both letters contain an error, find the probability that they typed one letter each".

    Hint: Recall and fill in the following in for events A and B

    P(A|B) =...

    Does that help?

    P.S I almost fell asleep typing this!! Stats is so uninteresting!
    thanks!
 
 
 
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