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You have six $1 bills, eight $5 bills, two $10 bills, and four $20 bills in your wallet. You select a bill at random. Without replacing the bill, you choose a second bill. What is P($1,then $10)?

A. 77/190
B. 3/100
C.3/95
D.2/5

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Anonymous

  1. Very similar to the other I just answered for you.
    Difference in this one: the item is not replaced
    Prob
    = (6/20)(2/19)
    = 12/380
    = 3/95

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    Reiny

  2. Do u mean 9/39

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    Logan

  4. no, Reiny is correct.

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    prisoner

  5. The answer IS 9/39.

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    MissAshlei

  6. Unfortunately, this problem fails to address the probability that the first bill drawn was one of the two $10 bills, given that the first bill was not replaced (i.e., put back into the wallet). There was a 2/20 chance that you withdrew a $10 bill during the random draw, leaving a (n) possibility that the second draw had a 1/19 probability for drawing a $10 bill. You would have to solve for (n). Technically, there are two correct answers (a, & b) based upon the probability of the first draw.
    a = first draw not equal to $10 bill.
    b = first draw equal to $10 bill.

    a = 3/95
    b = 3/190

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    Daniel

  8. thankyou!!!!!!!!!!!!!!

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    boy

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