Question

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

Answer 1

To find the probability of Sandra selecting a $10 bill first and then a $50 bill, we start by determining the total number of bills in the envelope.

Here's the breakdown of the bills:
- Number of $10 bills: 5
- Number of $20 bills: 10
- Number of $50 bills: 15

Total number of bills:
\[
5 + 10 + 15 = 30 \text{ bills}
\]

Now, we calculate the probability of each event:

1. **Probability of selecting a $10 bill first**:
The number of favorable outcomes (selecting a $10 bill) is 5, and the total number of bills is 30. Thus, the probability \( P(\text{$10 first}) \) is:
\[
P(\text{$10 first}) = \frac{5}{30} = \frac{1}{6}
\]

2. **Probability of selecting a $50 bill second**:
After selecting a $10 bill, there are now 29 bills left in total (30 - 1 = 29). The number of $50 bills remains the same at 15. So, the probability \( P(\text{$50 second | $10 first}) \) is:
\[
P(\text{$50 second | $10 first}) = \frac{15}{29}
\]

3. **Combined probability**:
We multiply the two probabilities together to find the combined probability of both events occurring (selecting a $10 bill first and then a $50 bill):
\[
P(\text{$10 first and $50 second}) = P(\text{$10 first}) \times P(\text{$50 second | $10 first}) = \frac{1}{6} \times \frac{15}{29} = \frac{15}{174}
\]

4. **Simplifying the fraction**:
The fraction \( \frac{15}{174} \) can be simplified. The greatest common divisor (GCD) of 15 and 174 is 3:
\[
\frac{15 \div 3}{174 \div 3} = \frac{5}{58}
\]

So, the probability of pulling a $10 bill first and then a $50 bill is:
\[
\boxed{\frac{5}{58}}
\]