Let's address each scenario one by one and match them to the given probabilities.
### Scenario 1:
The probability that both numbers are odd numbers less than 6 if the same numbers cannot be chosen twice.
To find the probability for this scenario:
1. Identify the odd numbers less than 6: these are 1, 3, and 5.
2. Calculate the total probability of choosing two different odd numbers without replacement from the total set of numbers (1 through 9).
The given probability for this scenario is:
[tex]\[
\frac{1}{12}
\][/tex]
### Scenario 2:
The probability that both numbers are greater than 6 if the same number can be chosen twice.
To find the probability for this scenario:
1. Identify the numbers greater than 6: these are 7, 8, and 9.
2. Calculate the probability of choosing one of these greater numbers twice (with replacement) from the total set of numbers (1 through 9).
The given probability for this scenario is:
[tex]\[
\frac{1}{9}
\][/tex]
### Scenario 3:
The probability that both numbers are even numbers if the same numbers cannot be chosen twice.
To find the probability for this scenario:
1. Identify the even numbers: these are 2, 4, 6, and 8.
2. Calculate the total probability of choosing two different even numbers without replacement from the total set of numbers (1 through 9).
The given probability for this scenario is:
[tex]\[
\frac{1}{6}
\][/tex]
### Summary of Matching:
- The probability that both numbers are odd numbers less than 6 if the same numbers cannot be chosen twice. [tex]$\xrightarrow{\ } \frac{1}{12}$[/tex]
- The probability that both numbers are greater than 6 if the same number can be chosen twice. [tex]$\xrightarrow{\ } \frac{1}{9}$[/tex]
- The probability that both numbers are even numbers if the same numbers cannot be chosen twice. [tex]$\xrightarrow{\ } \frac{1}{6}$[/tex]
