To solve the problem of determining the probability that Stef rolls a 3 on the first cube and a number greater than 3 on the second cube, we need to follow a detailed step-by-step approach.
### Step 1: Identifying the Total Number of Possible Outcomes
Since Stef rolls two six-sided number cubes (each with faces numbered from 1 to 6), the total number of possible outcomes when rolling both cubes is calculated by multiplying the number of outcomes for each individual die:
[tex]\[
\text{Total outcomes} = 6 \times 6 = 36
\][/tex]
### Step 2: Identifying the Favorable Outcomes
Next, we identify the favorable outcomes, which are those outcomes where the first die shows a 3 and the second die shows a number greater than 3.
To break this down:
- Rolling a 3 on the first die has 1 specific way to occur (only when the first die shows a 3).
- Rolling a number greater than 3 on the second die includes the numbers 4, 5, and 6. This gives us 3 possible outcomes for the second die.
Therefore, the number of favorable outcomes can be calculated by multiplying the number of ways to achieve each condition together:
[tex]\[
\text{Favorable outcomes} = 1 \times 3 = 3
\][/tex]
### Step 3: Calculating the Probability
The probability [tex]\( P \)[/tex] of an event is given by the ratio of the number of favorable outcomes to the total number of outcomes:
[tex]\[
P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
\][/tex]
Using the values we have:
[tex]\[
P = \frac{3}{36}
\][/tex]
To simplify this fraction:
[tex]\[
P = \frac{1}{12}
\][/tex]
### Answer
Based on the calculation, the probability that Stef rolls a 3 on the first cube and a number greater than 3 on the second cube is:
[tex]\[ \frac{3}{36} = \frac{1}{12} \][/tex]
Therefore, the correct answer is:
B. [tex]\( \frac{3}{36} = \frac{1}{12} \)[/tex]
