Let E and F be two events of an experiment with sample space S. Suppose P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.1. Compute the values below. (a) P(E ∪ F) =

(b) P(Ec) =

(c) P(Fc ) =

(d) P(Ec ∩ F) =

(a) P(E∪F)= 0.8

(b) P(Ec)= 0.4

(c) P(Fc)= 0.7

(d) P(Ec∩F)= 0.8

Step-by-step explanation:

(a) It is called a union of two events A and B, and A ∪ B (read as "A union B") is designated to the event formed by all the elements of A and all of B. The event A∪B occurs when they do A or B or both.

If the events are not mutually exclusive, the union of A and B is the sum of the probabilities of the events together, from which the probability of the intersection of the events will be subtracted:

P(A∪B) = P(A) + P(B) - P(A∩B)

In this case:

P(E∪F)= P(E) + P(F) - P(E∩F)

Being P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.1

P(E∪F)= 0.6 + 0.3 - 0.1

P(E∪F)= 0.8

(b)  The complement of an event A is defined as the set that contains all the elements of the sample space that do not belong to A.  The Complementary Rule establishes that the sum of the probabilities of an event and its complement must be equal to 1. So, if P (A) is the probability that an event A occurs, then the probability that A does NOT occur is  P (Ac) = 1- P (A)

In this case: P(Ec)= 1 - P(E)

Then: P(Ec)= 1 - 0.6

P(Ec)= 0.4

(c) In this case: P(Fc)= 1 - P(F)

Then: P(Fc)= 1 - 0.3

P(Fc)= 0.7

(d)  The intersection of two events A and B, designated as A ∩ B (read as "A intersection B") is the event formed by the elements that belong simultaneously to A and B. The event A ∩ B occurs when A and B do at once.

As mentioned, the complementary rule states that the sum of the probabilities of an event and its complement must equal 1. Then:

P(Ec intersection F) + P(E intersection F) = P(F)

P(Ec intersection F) + 0.1 = 0.3

P(Ec intersection F)= 0.2

Being:

P(Ec∪F)= P(Ec) + P(F) - P(Ec∩F)

you get:

P(Ec∩F)= P(Ec) + P(F) - P(Ec∪F)

So:

P(Ec∩F)= 0.4 + 0.3 - 0.2

P(Ec∩F)= 0.8

auto insurance

step-by-step explanation:

b. 153 students do not participate in sports and seventy-five 12th graders were surveyed