91Ó°ÊÓ

The problem states that events E and F are mutually exclusive, meaning that they can't occur at the same time. Additionally, we have been given the probabilities of E occurring \(P(E)=0.2\) and F occurring \(P(F)=0.5\).

Remember, mutually exclusive events cannot occur at the same time; hence, the intersection of E and F, denoted as \(E \cap F\), is an impossible event. Throughout probability theory, the probability of an impossible event is 0. So, \(P(E \cap F)=0\).

The union of E and F, denoted as \(E \cup F\), represents either event E occurring, event F occurring or both occurring. But since these events are mutually exclusive, we don't have to worry about them occurring simultaneously. The formula for computing the union of two mutually exclusive events is: \(P(E \cup F) = P(E) + P(F)\). Plugging in our values, our calculations will be: \(P(E \cup F) = 0.2 + 0.5 = 0.7\).

The notation \(E^{c}\) denotes the complement of E, meaning the event of E not occurring. In any probability space, the probability of an event plus the probability of its complement equals 1. Hence, \(P(E) + P(E^{c}) = 1\). Solving for \(P(E^{c})\), we get: \(P(E^{c}) = 1 - P(E) = 1 - 0.2 = 0.8\).

The expression \(E^{c} \cap F^{c}\) represents the probability of both E and F not occurring, which is the intersection of their individual complementary events. As in Step 4, we can use the complements to calculate this. However, these two complementary events aren't mutually exclusive so we cannot simply add their probabilities. Fortunately, the complement of \(E \cup F\) is exactly \(E^{c} \cap F^{c}\) (an intersection distributes over a union). So, we calculate the probability of \(E^{c} \cap F^{c}\) as follows: \(P(E^{c} \cap F^{c}) = P((E \cup F)^{c}) = 1 - P(E \cup F) = 1 - 0.7 = 0.3\). So, the answers to the given prompts are: a. \(P(E \cap F)=0\) b. \(P(E \cup F)=0.7\) c. \(P(E^{c})=0.8\) d. \(P(E^{c} \cap F^{c})=0.3\)