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
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
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 $\boxed{\frac{5}{51}}$.
$\boxed{\frac{5}{51}}$.
That is correct.
To find the probability of selecting a $1 bill followed by a $10 bill without replacement, we need to calculate two things: the probability of selecting a $1 bill on the first draw and the probability of selecting a $10 bill on the second draw given that a $1 bill has already been selected.
First, let's calculate the probability of selecting a $1 bill on the first draw.
Total number of bills = 5 + 4 + 6 + 3 = 18
Number of $1 bills = 5
P($1 on the first draw) = Number of $1 bills / Total number of bills
P($1 on the first draw) = 5 / 18
Now, since we didn't replace the first bill, the total number of bills is reduced to 17 for the second draw. Also, the number of $1 bills is reduced to 4, and the number of $10 bills is 6.
P($10 on the second draw given $1 on the first draw) = Number of $10 bills / Total number of bills for the second draw
P($10 on the second draw given $1 on the first draw) = 6 / 17
To find the combined probability, we multiply the two probabilities:
P($1, then $10) = P($1 on the first draw) * P($10 on the second draw given $1 on the first draw)
P($1, then $10) = (5 / 18) * (6 / 17)
Simplifying this expression:
P($1, then $10) = 30 / (18 * 17)
P($1, then $10) = 30 / 306
Therefore, the correct answer is:
P($1, then $10) = 5/51