You have five $1 bills, four $5 bills six $10 bills and three $20 bills. You select a bills at random. Without replacing the bill, You choose a second bill. What is P($1, then $10)?
B. 5/51
I think it's B. The probabilities change every time a bill is pulled out. The probability of pulling out a 1 first is 5 out of 18 which is .227 then the probability of pulling out a 10 is 6 out of 17 because you don't replace the bill. since the events are independent of each other, you get .35, then you multiply them to get .098 which in fraction form is answer B.
Elena is correct because i just took my test! =)
it's B
wow, butt sir. I love your choice of animals. c;;;
The probability of selecting a $1 bill first is 6/20, since there are 6 $1 bills out of a total of 20 bills in the wallet. After not replacing the first bill, there will be 19 bills left in the wallet, among which only two are $10 bills. Therefore, the probability of selecting a $10 bill second, given that a $1 bill was selected first, is 2/19.
Since the two events are independent, we can multiply their probabilities to find the probability of both events happening:
P($1, then $10) = P(selecting a $1 bill first) x P(selecting a $10 bill second, given that a $1 bill was selected first)
P($1, then $10) = 6/20 x 2/19
P($1, then $10) = 3/95
Therefore, the answer is D, 3/95.