To find the probability of Sandra selecting a $10 bill first and then a $50 bill, we first need to determine the total number of bills in the envelope.
The total number of bills is:
- 5 (for $10 bills)
- 10 (for $20 bills)
- 15 (for $50 bills)
So, the total number of bills is: \[ 5 + 10 + 15 = 30 \]
Step 1: Calculate the probability of selecting a $10 bill first.
The probability of selecting a $10 bill is: \[ P(\text{$10 bill first}) = \frac{\text{Number of $10 bills}}{\text{Total number of bills}} = \frac{5}{30} = \frac{1}{6} \]
Step 2: Calculate the probability of selecting a $50 bill second after selecting a $10 bill.
After selecting a $10 bill, there are now 29 bills remaining (since one bill has been taken out) and the number of $50 bills is still 15.
The probability of then selecting a $50 bill is: \[ P(\text{$50 bill second | $10 bill first}) = \frac{\text{Number of $50 bills}}{\text{Total remaining bills}} = \frac{15}{29} \]
Step 3: Calculate the combined probability.
To find the overall probability of both events happening (selecting a $10 bill first and then a $50 bill), we multiply the two probabilities: \[ P(\text{$10 bill and then $50 bill}) = P(\text{$10 bill first}) \times P(\text{$50 bill second | $10 bill first}) = \frac{1}{6} \times \frac{15}{29} \]
Calculating this gives: \[ \frac{1}{6} \times \frac{15}{29} = \frac{15}{174} \]
Now, we simplify \(\frac{15}{174}\): Both 15 and 174 can be divided by 3: \[ \frac{15 \div 3}{174 \div 3} = \frac{5}{58} \]
Thus, the probability of getting a $10 bill first and then a $50 bill is: \[ \boxed{\frac{5}{58}} \]