Question

the outcomes of two variables are (low, medium, high) and (on, off), respectively. an experiment is conducted in which the outcomes of each of the two variables are observed. the probabilities associated with each of the six possible outcome - pairs are given in the accompanying two - way table. consider the events below. complete parts a through h.
a: {high}
b: {high or on}
c: {off and low}
d: {medium}
a. find p(a). p(a)=0.07 (type an exact answer in simplified form.)
b. find p(b). p(b)=0.71 (type an exact answer in simplified form.)
c. find p(c). p(c)=0.26 (type an exact answer in simplified form.)
d. find p(d). p(d)=0.15 (type an exact answer in simplified form.)
e. find p(a^c). p(a^c)=0.93 (type an exact answer in simplified form.)
f. find p(a∪b). p(a∪b)= (type an exact answer in simplified form.)

Explanation:

Step1: Recall probability rules

For a given event \(A\), the probability of the complement \(A^c\) is \(P(A^c)=1 - P(A)\). For the union of two events \(A\) and \(B\), if we know the probabilities of individual - events and their relationships, we can find \(P(A\cup B)\).

Step2: Calculate \(P(A)\)

Given \(P(A) = 0.07\)

Step3: Calculate \(P(B)\)

Given \(P(B)=0.71\)

Step4: Calculate \(P(C)\)

Given \(P(C)=0.26\)

Step5: Calculate \(P(D)\)

Given \(P(D)=0.15\)

Step6: Calculate \(P(A^c)\)

Using the formula \(P(A^c)=1 - P(A)\), substituting \(P(A) = 0.07\), we get \(P(A^c)=1 - 0.07=0.93\)

Step7: Calculate \(P(A\cup B)\)

Since no information about the relationship between \(A\) and \(B\) (such as if they are mutually - exclusive) is given other than their individual probabilities, if we assume they are mutually - exclusive (the problem does not state otherwise), then \(P(A\cup B)=P(A)+P(B)\). Substituting \(P(A) = 0.07\) and \(P(B)=0.71\), we get \(P(A\cup B)=0.07 + 0.71=0.78\)

Answer:

0.78