Question

A red die is tossed and then a green dieis tossed. What is the probability thatthe red die shows a six or the green dieshows a six?Hint: The two events are not mutually exclusive. So to the find theprobability of the union, use:P(A or B) = P(A) + P(B) - P(A and B)[?]

Answer 1

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:

$P(A)=P(B)=\frac{1}{6}$

Each roll is independent from each other, then, the probability of both events happening simultaneously is given by their product:

$P(A\:and\:B)=P(A)P(B)$

Using the additive rule of probability, we have the following equation for our problem:

$\begin{gathered} P(A\:or\:B)=P(A)+P(B)-P(A\:and\:B) \\ =P(A)+P(B)-P(A)P(B) \\ =\frac{1}{6}+\frac{1}{6}-\frac{1}{6^2} \\ =\frac{2}{6}-\frac{1}{36} \\ =\frac{12}{36}-\frac{1}{36} \\ =\frac{12-1}{36} \\ =\frac{11}{36} \end{gathered}$

the probability that the red die shows a six or the green die shows a six is 11/36.