Question

Please go through with me how to solve this problem..
You have three $1 bills, four $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).

A. 1/12

B. 5/81

C. 5/72

D. 2/27

Answer 1

Could you help me with this? @agent0smith

Answer 2

What is the probability that will you get $10 on the first draw?

Answer 3

Well $10 out of $43 ... would I just make that a fraction? Or .... I don't even know >.<

Answer 4

There are 9 bills in your wallet. 2 of them are tens. The probability you will get a 10 is 2/9

Answer 5

So you draw out the 10. How many bills are in your wallet now?

Answer 6

If you draw out one of the tens, then there would be 8 bills left, both tens, would be 7..

Answer 7

You had 9 bills in your wallet. You took out 1 bill. 9-1=8. You now have 8 bills in your wallet. How many of them are $1 bills?

Answer 8

3 of them are $! bills

Answer 9

$1 bills, lol my fault

Answer 10

So what is the probability you will draw out a $1 bill on your next draw?

Answer 11

3/8 ?

Answer 12

Yes. So now you should multiply those individual probabilities to get your answer:
\[\frac{2}{9}\times\frac{3}{8}\]

Answer 13

Do i cross multiply or multiply the numerator x numerator and denominator x denominator?

Answer 14

Cancel the two and then multiply numerators and multiply denominators.

Answer 15

What do you mean by cancel the two? o.o