Anonymous,HowGan

An urn contains 4 red balls numbered from 1 to 4 and 6 white balls numbered from 5 to 10.we choose simultaneously 2 balls fron the urn.?
calculate the probability of the following events:
-the drawn balls are red
-the drawn balls have different colors

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3 months ago 2

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  1. Puzzling

    PART 1 - The two drawn balls are red

    On the first draw, there are 4 balls out of 10 that are red:
    P(first is red) = 4/10 = 2/5

    On the second draw, there are 3 balls out of 9 that are red:
    P(second is red) = 3/9 = 1/3

    Multiply to get the probability of both events:
    P(both are red) = 2/5 x 1/3 = 2/15

    Answer:
    2/15

    PART 2 - The two drawn balls are different colors.

    Let's first figure out the probability of drawing a red ball and then a white ball:

    On the first draw, there are 4 balls out of 10 that are red:
    P(first is red) = 4/10 = 2/5

    On the second draw, there are 6 balls out of 9 that are white:
    P(second is white) = 6/9 = 2/3

    Multiply to get the probability of both events:
    P(first is red, second is white) = 2/5 x 2/3 = 4/15

    But we have a second, equivalent case where the first ball could have been white and the second ball could have been red. It has the same probability of 4/15.

    If you aren't convinced we can figure the separate probability as:
    On the first draw, there are 6 balls out of 10 that are white:
    P(first is white) = 6/10 = 3/5

    On the second draw, there are 4 balls out of 9 that are red:
    P(second is red) = 4/9

    Multiply to get the probability of both events:
    P(first is white, second is red) = 3/5 x 4/9 = 12/45 = 4/15

    Add both cases:
    4/15 + 4/15
    = 8/15

    Answer:
    8/15

    P.S. If you are familiar with the "n choose k" formula, you can do this even quicker.

    PART 1:
    From the 4 red, choose 2 --> 4C2 = (4 x 3) / (2 x 1) = 6 ways
    From the 10 balls, choose *any* 2 --> 10C2 = (10 x 9) / (2 x 1) = 45 ways

    Divide to get the probability:
    6/45 = 2/15

    PART 2:
    From the 4 red, choose 1 --> 4C1 = 4 ways
    From the 6 white, choose 1 --> 6C1 = 6 ways
    Total ways to pick 1 red and 1 white = 4 x 6 = 24 ways
    From the 10 balls, choose *any* 2 --> 10C2 = 45 ways

    Divide to get the probability:
    24/45 = 8/15

  2. ?

    draw R p = 4/10
    then draw R p = 3/9
    p = (4/10)(3/9) = 2/15
    answer is 2/15
    draw R p= 4/10
    then draw W = 6/9
    p = (4/10)(6/9) = 4/15
    Or
    draw W p= 6/10
    then draw R p = 4/9
    p = (6/10)(4/9) = 4/15

    p (different colors) = 4/15+4/15 = 8/15

    draw W p = 6/10
    then draw W p = 5/9
    p = (6/10)(5/9) = 5/15 = 1/3

    RR = 2/15
    RW = 4/15
    WR = 4/15
    WW = 5/15

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