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If There Is Three $1, 4 $5 Bills And Two $10 Bills In Your Wallet. You select a bill at random Without replacing the bill you choose a second bill at random. Find p ($10 then $1).

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    The probability p(10 then 1) = p(10)*p(1|10) = (2/(3+4+2))(3/(3+4+1))
      = 2/9*3/8 = 2*3/(9*8) = 6/72 = 1/12

    P(1|10) is the probability of selecting a $1 bill, given that you have already selected a $10 bill. Under that condition, there is one $10 bill remaining, and the total number of bills is 8.

    1 0

    Oddman 

    answered 1 year ago

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