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A bag contains 3 black balls, 4 green balls and 5 yellow balls.

a) if two balls are picked at random without replacement, find the probability that Botha balls are of the same color.
b) How many black balls must be added to the bag so that the probability of drawing a black ball is 1/2.

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4 answers
  1. P(black,black) = 3/12 * 2/11
    do the same steps for P(green,green) and P(red,red)
    then add them all up, since the only ways to get two of the same color is to get two blacks, greens, or reds.

    There are 12 balls in all
    There are 3 black balls
    so, you need to add x black balls to make them 50% --
    12+x = 2(3+x)

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  2. a) you got 3 cases: both black, both green, or both yellow
    without replacement, prob of both black = (3/12)(2/11) = 1/22
    without replacement, prob of both green = .....
    without replacement, prob of both yellow = ...
    add them up

    b) number of blacks to be added ---- x
    So now you have x+3 black balls out of 12+x in total
    prob (black) = (x+3)/(x+12) = 1/2
    solve for x

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  3. ways of picking two balls ... 12C2 ... 66
    ... two black balls ... 3C2 ... 3
    ... two green balls ... 4C2 ... 6
    ... two yellow balls ... 5C2 ... 10

    p(both same color) = (3 + 6 + 10) / 66

    b) (3 + x) / (12 + x) = 1/2

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  4. thank you

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