We are asked to determine the probability of rolling a 6 and then rolling a number less than 2. To do that we will use the product rule probabilities since we want to find the probability of two independent events happening:
![P(AandB)=P(A)P(B)](https://img.qammunity.org/2023/formulas/mathematics/college/u0y5sb3ps38xaw3fjxujarpywsv1g9325z.png)
Where:
![\begin{gathered} A=\text{ rolling a 6} \\ B=\text{ rolling a number less than 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cbcxbcscj78myxo1twhfwxzsjgixg035l9.png)
To determine the probability of rolling a 6 we need to have into account that there are 6 possible outcomes out of which only one is a 6. Therefore, the probability is:
![P(A)=(1)/(6)](https://img.qammunity.org/2023/formulas/mathematics/college/g06h4iuclk4jjkbly6onhnapc1gdh15le4.png)
To determine the probability of B we need to have into account that in a 6-sided die the numbers that are less than 2 are (1), this means that there is only one number less than 2 out of 6 possible numbers. Therefore, the probability is:
![P(B)=(1)/(6)](https://img.qammunity.org/2023/formulas/mathematics/college/bmvkuo95bn25r6er8c9anmwoxasvucrww8.png)
Now, we substitute in the product rule:
![P(AandB)=((1)/(6))((1)/(6))](https://img.qammunity.org/2023/formulas/mathematics/college/ava3ts48i32furt80z1p1bhdur7oio4pm7.png)
Solving the product:
![P(AandB)=(1)/(36)](https://img.qammunity.org/2023/formulas/mathematics/college/t0r0nmzoihnb4inhte0dfl0ow4x8cn9n5q.png)
Therefore, the probability is 1/36.