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    probability distribution Q



    A player rolls two dice and is awarded 1 point if the same number appears on the upface of both dice. If different numbers appear, he gets no points. Let X be the number of points a player could earn in three attempts. What is the probability of X?

    X {0, 1, 2, 3}
    P(X) {0.579, 0.347, 0.069, 0.005}
    ______________________________________________________________

    Dice 1
    1
    2
    3
    4
    5
    6

    Dice 2
    1
    2
    3
    4
    5
    6

    so the way i see it is that there is a 1/6 chance of getting one number from each die so the probability of getting two numbers the same should be 1/6 x 1/6 = 0.0277 but thats different from the answers given above and why does the X-value only go up to 0, 1, 2, and 3 where there are 6 numbers on a dice?

    anyone know how to do this problem?
    thanks in advance

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    X is the number of times you get a match between the two dice in three attempts. Since there are only 3 attempts, then you can't get any more than 3 matches.

    P(match on any attempt) = 1/6
    P(no match on any attempt) = 5/6

    This is now just like a binomial experiment, with n=3 trials, p(success)=1/6, p(fail)=5/6.

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    Thanks, i got it!

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    Hi John:

    I have another probability distribution question that I am hoping you can explain to me like the one you did before so I can do it on my own. Thanks in advance.

    The "READY CASH" machine at a local bank required that the user correctly enter a 3 digit access code prior to making a transaction with a bank card. The 3 digit code is generated by selecting any of the digits from 0 to 9 but each digit may be used only once in each code. The machine is also very unforgiving in what when any single digit is entered incorrectly the machine terminates the transaction and confiscates the users' card. A thief is trying to make a transaction with a stolen card. The their randomly selects digits before the card is onfiscated or access is obtianed. Show the probability distribution of X.

    X {1, 2, 3}
    P(X) {0.90. 0.089, 0.011}

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    Is there something missing from the problem description - it's not clear what X represents.

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    sorry
    here it is again:

    The "READY CASH" machine at a local bank required that the user correctly enter a 3 digit access code prior to making a transaction with a bank card. The 3 digit code is generated by selecting any of the digits from 0 to 9 but each digit may be used only once in each code. The machine is also very unforgiving in what when any single digit is entered incorrectly the machine terminates the transaction and confiscates the users' card. A thief is trying to make a transaction with a stolen card. The theif randomly selects digits to enter but is aware that each digit may be used only once. Let X br the number of digits entered before the card is confiscated or access is obtained. Show the probability distribution of X.

    X {1, 2, 3}
    P(X) {0.90. 0.089, 0.011}

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    I get P(X) {0.90. 0.089, 0.0097}

    X=1 --> (9/10)
    X=2 --> (1/10)*(8/9)
    X=3 --> (1/10)*(1/9)*(7/8)

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