Question

A money envelope contains five $10 bills, ten $20 bills, and fifteen $50 bills. Sandra randomly selects two bills without returning the first bill. What is the probability of getting a $10 bill then a $50 bill? Write your answer in the simplest form of fraction.(1 point)

Answer 1

To find the probability of Sandra randomly selecting a $10 bill and then a $50 bill from the envelope, we'll first determine the total number of bills in the envelope and then calculate the probabilities step by step.

1. **Count the total number of bills:**
- Number of $10 bills = 5
- Number of $20 bills = 10
- Number of $50 bills = 15

Total number of bills = 5 + 10 + 15 = 30 bills.

2. **Calculate the probability of selecting a $10 bill first:**
The probability of selecting a $10 bill on the first draw is the number of $10 bills divided by the total number of bills:
\[
P(\text{first bill is } \$10) = \frac{5}{30} = \frac{1}{6}
\]

3. **Calculate the probability of selecting a $50 bill second:**
After drawing a $10 bill, there will be:
- Total bills left = 30 - 1 = 29 bills,
- Number of $50 bills left = 15 (no $50 bill has been drawn yet).

Thus, the probability of selecting a $50 bill on the second draw is:
\[
P(\text{second bill is } \$50 | \text{first bill was } \$10) = \frac{15}{29}
\]

4. **Calculate the combined probability:**
The combined probability of both events happening (first selecting a $10 bill and then selecting a $50 bill) is:
\[
P(\text{first is } \$10 \text{ and second is } \$50) = P(\text{first is } \$10) \times P(\text{second is } \$50 | \text{first is } \$10)
\]
\[
= \frac{1}{6} \times \frac{15}{29} = \frac{15}{174}
\]

5. **Simplify the fraction:**
Now we simplify \(\frac{15}{174}\). The greatest common divisor of 15 and 174 is 3:
\[
\frac{15 \div 3}{174 \div 3} = \frac{5}{58}
\]

Therefore, the probability of Sandra selecting a $10 bill followed by a $50 bill is \(\boxed{\frac{5}{58}}\).