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:
[tex]P(A)=P(B)=\frac{1}{6}[/tex]
Each roll is independent from each other, then, the probability of both events happening simultaneously is given by their product:
[tex]P(A\:and\:B)=P(A)P(B)[/tex]
Using the additive rule of probability, we have the following equation for our problem:
[tex]\begin{gathered} P(A\:or\:B)=P(A)+P(B)-P(A\:and\:B) \\ =P(A)+P(B)-P(A)P(B) \\ =\frac{1}{6}+\frac{1}{6}-\frac{1}{6^2} \\ =\frac{2}{6}-\frac{1}{36} \\ =\frac{12}{36}-\frac{1}{36} \\ =\frac{12-1}{36} \\ =\frac{11}{36} \end{gathered}[/tex]
the probability that the red die shows a six or the green die shows a six is 11/36.
