Let's call the event of the red die to show a six as event A, and the event of the green die to show a six as event B.
The theoretical probability is defined as the ratio of the number of favourable outcomes to the number of possible outcomes. On both dices, we have 6 possible outcomes(the numbers from 1 to 6), with one favourable outcome(the number 6), therefore, the probabilities of those events are:
![P(A)=P(B)=(1)/(6)](https://img.qammunity.org/2023/formulas/mathematics/college/5pvorpf5u5sfnc2hmqiafkf5vy3bn4inw3.png)
Each roll is independent from each other, then, the probability of both events happening simultaneously is given by their product:
![P(A\:and\:B)=P(A)P(B)](https://img.qammunity.org/2023/formulas/mathematics/college/yjwnoctbawumsegl4kgys705uk0c10fnvl.png)
Using the additive rule of probability, we have the following equation for our problem:
![\begin{gathered} P(A\:or\:B)=P(A)+P(B)-P(A\:and\:B) \\ =P(A)+P(B)-P(A)P(B) \\ =(1)/(6)+(1)/(6)-(1)/(6^2) \\ =(2)/(6)-(1)/(36) \\ =(12)/(36)-(1)/(36) \\ =(12-1)/(36) \\ =(11)/(36) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lxt7wzzd6z15mnz260pcl4d83apfdw7t3f.png)
the probability that the red die shows a six or the green die shows a six is 11/36.