Let's break down the solution step-by-step to determine the probability of rolling a 2 first and then a 1 with a six-sided number cube.
1. Understanding the Cube:
- A standard number cube has 6 faces, numbered from 1 to 6.
2. Probability of Rolling a 2 First:
- There is only one face with the number 2 on the cube.
- The probability (P) of rolling a 2 on the first roll is:
[tex]\[
P(\text{First roll is 2}) = \frac{1}{6}
\][/tex]
3. Probability of Rolling a 1 Second (Given That the First was a 2):
- Once you have rolled a 2, you still have a full number cube (all 6 faces to choose from) for the second roll.
- There is only one face with the number 1.
- The probability (P) of rolling a 1 on the second roll is:
[tex]\[
P(\text{Second roll is 1}) = \frac{1}{6}
\][/tex]
4. Combining the Probabilities:
- To find the overall probability of both events happening in sequence (rolling a 2 first and then a 1 second), we need to multiply the probabilities of each independent event.
- Mathematically, this is expressed as:
[tex]\[
P(\text{First roll is 2 and second roll is 1}) = P(\text{First roll is 2}) \times P(\text{Second roll is 1 given first roll is 2})
\][/tex]
- Substituting the known probabilities:
[tex]\[
P(\text{First roll is 2 and second roll is 1}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}
\][/tex]
Therefore, the probability of rolling a 2 first and then rolling a 1 is [tex]\(\frac{1}{36}\)[/tex].
None of the provided answer choices directly match [tex]\(\frac{1}{36}\)[/tex]. This might indicate a typo or misinterpretation in the question choices, but based on the calculation, [tex]\(\frac{1}{36}\)[/tex] is the correct answer if such a choice were to be available or re-evaluated in a multiple-choice context.
