To find the probability of Sandra first selecting a $10 bill and then a $50 bill without replacement, we first need to determine the total number of bills and the distribution of bills.
Let's denote the number of each type of bill:
- Let \( x \) be the number of $10 bills.
- Let \( y \) be the number of $20 bills.
- We know there are 15 $50 bills.
So, the total number of bills in the envelope is:
\[ \text{Total Bills} = x + y + 15 \]
Step 1: Calculate the probability of selecting a $10 bill first.
The probability of selecting a $10 bill first is given by the ratio of the number of $10 bills to the total number of bills:
\[ P(\text{10 first}) = \frac{x}{x + y + 15} \]
Step 2: Calculate the probability of selecting a $50 bill second.
After selecting the first bill (which was a $10 bill), we do not return it. Thus, we now have one less total bill and the same number of $50 bills.
The new total number of bills is:
\[ \text{New Total} = x + y + 15 - 1 = x + y + 14 \]
The probability of selecting a $50 bill after selecting a $10 bill is given by the ratio of the number of $50 bills to the new total number of bills:
\[ P(\text{50 second | 10 first}) = \frac{15}{x + y + 14} \]
Step 3: Calculate the joint probability.
The joint probability of both events occurring (selecting a $10 bill first and then a $50 bill) is the product of the probabilities of the individual events:
\[ P(\text{10 then 50}) = P(\text{10 first}) \times P(\text{50 second | 10 first}) = \frac{x}{x + y + 15} \times \frac{15}{x + y + 14} \]
So, the final expression for the probability of getting a $10 bill and then a $50 bill is:
\[ P(\text{10 then 50}) = \frac{x \cdot 15}{(x + y + 15)(x + y + 14)} \]
However, we cannot simplify this further without knowing the values of \( x \) and \( y \). Therefore, the answer is:
\[ \text{Probability} = \frac{15x}{(x + y + 15)(x + y + 14)} \]
If \( x \) and \( y \) values are known, we can compute a numerical answer. In absence of explicit values, we have the theoretical framework.