# Math

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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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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2. Do u mean 9/39

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3. no, Reiny is correct.

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5. 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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6. thankyou!!!!!!!!!!!!!!

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