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"In vase A, there are 3 red balls and 5 white balls. In vase B, there are 6 red balls and 4 white balls. Two balls are to be drawn at random, without replacement, from each vase. What is the probability of selecting 4 red balls?"

(Answer: 1/28)

So I thought about doing 4 cases and then adding them up (don't pick red ball in A, don't pick red ball in B...pick red ball in A, pick red ball in B...etc...).

I did this:
(3/8)(3/7)(6/10)(6/9) +
(3/8)(2/7)(6/10)(5/9) +

But it doesn't work. What am I doing wrong?

  • Math -

    Since you pick 2 from each vase, and you want 4 reds, there is only one way, that is, you pick 2 from A and 2 from B.

    which is C(3,2)/(C(8,2) * C(6,2)/C(10,2)
    = 3/28 * 15/45 = 1/28

  • Math -

    Oh yeah! Thank you!

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