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)?

77/190
3/100
3/95
2/5

Question

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)?

77/190
3/100
3/95
2/5

anonymous · Answer

i'm trying to get this math test done and i need help

jtvatsim · Answer

alright, what is your thought process on the question so far? how are you thinking to approach the problem? :)

anonymous · Answer

the probability the first one is a $1 is the number of $1 bills over the total number of bills

anonymous · Answer

i.e. $\frac{6}{20}=\frac{3}{10}$

anonymous · Answer

the probability the second is a $10 given that the first one is a $1 is $\frac{2}{19}$ the $19$ because you have already chosen a $1 bill and not replaced it

anonymous · Answer

to find your answer, multiply those numbers i.e. compute 
$$\frac{3}{10}\times \frac{2}{19}$$

jtvatsim · Answer

yup, that's the way it's done! :)