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10 January, 16:43

Two fair dice are rolled. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; (b) X is the value on the first die and Y is the larger of the two values; (c) X is the smallest and Y is the largest value obtained on the dice.

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  1. 10 January, 17:11
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    a)

    P (X = x₀, Y = 2x₀) = 1/36

    P (X = x₀, Y = k) = 1/18 for k between x₀+1 and 2x₀-1 inclusive

    Every other event has probability 0. x₀ is any number between 1 and 6 inclusive.

    b)

    P (X = x₀, Y = x₀) = x₀/36

    P (X = x₀, Y = k) = 1/36 for k between x₀+1 and 6 inclusive.

    x₀ is between 1 and 6 inclusive. Every other event has probability 0.

    c)

    P (X = x₀, Y = x₀) = 1/36

    P (X = x₀, Y = k) = 1/18 with k between x₀+1 and 6 inclusive

    x₀ between 1 and 6 inclusive. Any other event has probability 0.

    Step-by-step explanation:

    Note that there are 36 possible results for the dice

    a)

    P (X = 1, Y = 2)

    This is obtained only when both dices are 1, hence its probability is 1/36

    P (X = 1, Y = k) = 0 (k > 1)

    because if the largest value of the dice is 1, then both dices are 1

    P (X = 2, Y = 3)

    one dice is 2, the other one is 3, hence there are 2 possibilities and the probability is 2/36 = 1/18

    P (X = 2, Y = 4)

    This happens only if both dices are 2, hence the probability is 1/36.

    P (X = 2, Y = k) = 0 (k > 2)

    same argument of above. If the largest dice is 2, then the sum is either 3 or 4.

    P (X = 3, Y = 4), P (X = 3, Y = 5)

    in both given events we need one dice to be 3 and the other dice to be 1 for the first event and 2 for the second event. In both cases, there are only 2 favourable cases, hence the probability of the event is 2/36 = 1/18

    P (X = 3, Y = 6)

    This event happens only when both dices are 3, hence the probability is 1/36

    This should show a pattern. As long as x₀ is between 1 and 6, if y₀ is between x+1 and 2x-1, then the probability P (X = x₀, Y = y₀) is 1/18 (either first dice is x₀, second dice is y₀-x₀ or first dice is x₀ and second dice is y₀ - x₀), also P (X = x₀, Y = 2x₀) = 1/36 (both dices are x₀). Every other event has probability 0.

    b) We can separate them using conditional probability and the fact that both dices results are independent with each other.

    P (X = x₀, Y = y₀) = P (X = x₀) * P (Y = y₀ | X = x₀)

    P (X = x₀) = 1/6 for any value x₀ between 1 and 6.

    If y₀ is x₀, this means that the first dice has the largest value, so the second dice is between 1 and x₀, and the probability of this event is x₀/6 (x₀ favourable cases over 6 possible ones).

    If y₀ is not x₀, then it should be higher (otherwise the event would be impossible and it would have probability 0). As long as y₀ is between 2 and 6, the probability of this event is 1/6.

    Thus

    P (X = x₀, Y = x₀) = 1/6 * x₀/6 = x₀/36

    P (X = x₀, Y = x₀ + k) = (1/6) ² = 1/36 (k > 0)

    Every other probability is 0

    c)

    P (X = x₀, Y = x₀) = 1/36 (because both dices are equal to x₀ in this event)

    P (X = x₀, Y = x₀+k) = 2/36 = 1/18 (here k > 0. One possibility is the first dice is x₀ and the second one is x₀+k, and the remaining possibility is the first dice is x₀+k and the second dice is x₀)

    Evert other event has probability 0.
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