The probability of selecting a $1 bill on the first draw is $\frac{5}{18}$ since there are 5 $1 bills out of a total of 18 bills. If a $1 bill is selected on the first draw and not replaced, there are 17 bills remaining with 6 of them being $10 bills. Therefore, the probability of selecting a $10 bill on the second draw given that a $1 bill was selected on the first draw is $\frac{6}{17}$.
Therefore, the overall probability of selecting a $1 bill, then a $10 bill without replacement is:
$$P(\text{\$1, then \$10}) = \frac{5}{18} \cdot \frac{6}{17} = \frac{5}{51}$$
The answer is $\boxed{\frac{5}{51}}$.
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)?
11/35
5/51
5/54
193/306
5 answers
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)?
11/35
5/51
5/54
193/306
11/35
5/51
5/54
193/306
The probability of selecting a $1 bill on the first draw is $\frac{5}{18}$ since there are 5 $1 bills$ out of a total of 18 bills. If a $1 bill$ is selected on the first draw and not replaced, there are 17 bills remaining with $6$ of them being $10$ bills. Therefore, the probability of selecting a $10$ bill on the second draw given that a $1$ bill was selected on the first draw is $\frac{6}{17}$.
Therefore, the overall probability of selecting a $1$ bill, then a $10$ bill without replacement is:
$$P(\text{\$1, then \$10}) = \frac{5}{18} \cdot \frac{6}{17} = \frac{5}{51}$$
The answer is $\boxed{\frac{5}{51}}$.
Therefore, the overall probability of selecting a $1$ bill, then a $10$ bill without replacement is:
$$P(\text{\$1, then \$10}) = \frac{5}{18} \cdot \frac{6}{17} = \frac{5}{51}$$
The answer is $\boxed{\frac{5}{51}}$.
$\boxed{\frac{5}{51}}$.
That is correct.