You have five $1 bills, four $5 bills, six $10 bills, and three $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. 11/35
B. 5/51
C. 5/54
D. 193/306
I'm not really sure what the answer is I just need some help to solve this problem. Thank you :)
Multiply the probabilities after finding them. The first will be 5/(5+4+6+3). Remember that you are not replacing the bills.
To solve this problem, we need to calculate the probability of selecting a $1 bill first and then a $10 bill without replacement.
First, let's calculate the total number of bills in the wallet. We have:
Total number of bills = 5 ($1 bills) + 4 ($5 bills) + 6 ($10 bills) + 3 ($20 bills) = 5 + 4 + 6 + 3 = 18 bills
Next, let's calculate the probability of selecting a $1 bill first. There are 5 $1 bills out of a total of 18 bills. Therefore, the probability of selecting a $1 bill first is:
Probability of selecting a $1 bill = Number of $1 bills / Total number of bills = 5 / 18
After selecting the $1 bill, we have 17 bills remaining in the wallet. Now, we want to calculate the probability of selecting a $10 bill from the remaining bills. We have:
Number of $10 bills = 6
Total number of bills remaining = 17
Therefore, the probability of selecting a $10 bill second, given that a $1 bill was selected first, is:
Probability of selecting a $10 bill second = Number of $10 bills / Total number of bills remaining = 6 / 17
To find the probability of selecting a $1 bill first and then a $10 bill, we multiply the two probabilities:
P($1, then $10) = (5 / 18) * (6 / 17)
Now, let's simplify this expression:
P($1, then $10) = (5 * 6) / (18 * 17) = 30 / 306
Finally, let's simplify the fraction:
P($1, then $10) = 1 / 10.2
Since this does not match any of the answer choices, let's convert it to a decimal:
P($1, then $10) ≈ 0.098
So, the closest answer choice is B. 5/51.
Answer: B. 5/51