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    Hi there im massively confused on this question and is not as simple as it first looks.

    "A Personal Identification number (PIN) consists of 4 digits in order, each of which is one of the digits 0,1,2...9. Susie has difficulty remembering her PIN. She tries to remember it and writes it down what she thinks it is. The probability the first digit is correct is 0.8 and the prob the second digit is correct is 0.86. The probability that the first two digits are correct is 0.72. Find...

    a) The Probability that the second digit is correct given the first is correct.
    b) The prob that the first is correct and the second is incorrect.
    c) The prob that the first digit is incorrect and the second correct.
    d) The prob that the second is incorrect given the first is incorrect."


    Clearly the probabilities arent independent by the looks of it.

    I havent a clue where to start though so help please! Thanks.
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    This is nothing to do with binomial and geometric distributions, but fair enough.

    You need to apply the rule \mathbb{P} ( A | B ) \mathbb{P} (B) = \mathbb{P} (A \cap B) = \mathbb{P} (B|A) \mathbb{P}(A). You're given what \mathbb{P}(A), \mathbb{P}(B) and \mathbb{P}(A \cap B) are (where A is the event that the first digit is correct and B is the event that the second digit is correct), so you just have to work out the rest.

    EDIT: You also need to use the fact that \mathbb{P} ( \text{not}\, A ) = 1 - \mathbb{P} (A) and use similar rules to the above, replacing A and B with 'not A' and 'not B' here and there.
    • Thread Starter
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    Lol well weirdly this question is in the Binomial/Geometric Probability section.
 
 
 

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