Pre-Cal

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a red number cube and a white number cube are rolled. Find the probability that the red number cube shows a 2, given that the sum showing on the two number cubes is less than or equal to 5.

My work: (red cube = [1,2,3,4]
white cube = [4,3,2,1]

Probability of a 2 on the red cube is 1/4. Is this correct?

  • Pre-Cal -

    Block the whole matrix out with read along the top and white down
    write all the sums in the boxes of the matrix, which contains 36 numbers.
    Of those 36,
    4+3+2+1 = 10
    have sums less than or equal to 5
    Of those 10, 3 show the number 2 on red
    w1-r2 , w2-r2 , w3-r2
    so 3/10

  • Pre-Cal -

    If the sum is five, then red,white can be

    1,4
    2,3
    3,2
    4,1
    if the sum is four, then
    1,3
    2,2
    3,1
    if the sum is three, then
    1,2
    2,1
    if the sum is two, then
    1,1

    I see 10 possible combinations, and the red is 2 in 3 of them.

    Pr(red=2, sum is =<5) is 3/10

  • Pre-Cal -

    If the sum of the two cubes is 2, 3, 4, or 5 (all the possibilities that are allowed), then the red cube can only be 1, 2, 3 or 4.
    The possibile combinations are:
    W R
    1 4
    1 3
    1 2
    1 1
    2 3
    2 2
    2 1
    3 2
    3 1
    4 1
    There are three combinations with a red 2 out of ten allowed. That makes the probability 3/10

  • Pre-Cal -

    LOL - tomorrow we do Blackjack :)

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