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)?
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.
B. 5/51
To find the probability of selecting a $1 bill followed by a $10 bill without replacement, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's calculate the total number of bills we have:
Total bills = Number of $1 bills + Number of $5 bills + Number of $10 bills + Number of $20 bills
= 5 + 4 + 6 + 3
= 18
Now, let's calculate the number of favorable outcomes, which is selecting a $1 bill followed by a $10 bill:
Number of $1 bills = 5
Number of $10 bills = 6
Since the bills are not replaced after each selection, the number of $1 bills decreases to 4 when we select the second bill, and the number of $10 bills decreases to 5.
To calculate the probability, we multiply the probabilities of selecting a $1 bill and then a $10 bill:
P($1, then $10) = (Number of $1 bills / Total bills) * (Number of $10 bills / (Total bills - 1))
P($1, then $10) = (5 / 18) * (6 / 17)
Simplifying this expression:
P($1, then $10) = 30 / 306
P($1, then $10) ≈ 0.098
Therefore, the probability of selecting a $1 bill first and then a $10 bill without replacement is approximately 0.098 or 9.8%.