Call a household prosperous if its income exceeds $100,000.

Question

Call a household prosperous if its income exceeds $100,000. Call the household educated if the householder completed college. Select an American household at random, and let AA be the event that the selected household is prosperous and BB the event that it is educated. According to the Current Population Survey, P(A)=0.13P(A)=0.13, P(B)=0.251P(B)=0.251, and the probability that a household is both prosperous and educated is 0.0640.064. Round to 3 decimal places, where necessary.

a) What is the probability that the household selected is neither prosperous nor educated? Hint: Draw a venn diagram and apply the complement rule. 

b) Given that the household is either educated or prosperous, what is the probability that they are prosperous? 

c) Are events AA and BB independent? Justify. 

 

d) Are the events AA and BB mutually exclusive? Justify. 

Answer 1

I'll start with an explanation for part a). If you can understand the following explanation, it should get you through most of the rest of the exercise.

Your probability space features the events A and B, where the probability of each event (P(A) and P(B)) and the probability of their intersection, P(A ∩ B), is known. You want to know the probability that neither A or B occurs, P(A^C ∩ B^C), where "E^C" is notation for the complement of any event E.

A useful shortcut for solving this is De Morgan's laws.

De Morgan's laws are as follows:

1) The union of the complements of two events is the complement of their intersection:

(A^C ∪ B^C) = (A ∩ B)^C

2) The intersection of the complements of two events is the complement of their union:

(A^C ∩ B^C) = (A ∪ B)^C

I would recommend getting some scratch paper and drawing the Venn Diagrams for each of these

In this case you'll want the second of De Morgan's laws. Of course you'll probably recall the complement rule:

P(E^C) = 1 - P(E)

Combining these, we get the following calculation:

P(A^C ∩ B^C) = 1 - P(A ∪ B)

So now all we need to find is the probability of the union of A and B. If you know the probability of each event and the probability of their intersection, you can find the union's probability:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

As you might intuit from your Venn diagram illustration, adding P(A) and P(B) together without subtracting the intersection means counting P(A ∩ B) twice, because the intersection is both a subset of A and a subset of B, and in this case, since it has nonzero probability, it's not an empty set.

So to wrap things up, we have all three of these principles in effect (just be careful with distributing that minus sign multiplication):

P(A^C ∩ B^C) = 1 - P(A ∪ B) = 1 - [P(A) + P(B) - P(A ∩ B)]

Now for part b). This part's a bit more straightforward, as the only additional principle in play is the concept of conditional probability. The idea behind a conditional probability is that you're narrowing the probability space from what it was before, to only including the given event.

So suppose you want to know the probability of some event E, conditional on another event F.

Then the corresponding probability is the intersection of E and F, taken in proportion to all of F, as if the event space were narrowed to only include F and its subsets:

P (E | F) = P(E ∩ F) / P(F).

In this case you're asked to find the probability of A, given either event A or B.

This gives us P(A | A ∪ B) = P(A ∩ (A ∪ B)) / P(A ∪ B).

It so happens in this case that A is a subset of its union with B, so the intersection of A and its union with B is just A:

A ∩ (A ∪ B) = A

So we have the special case where we just divide the subset's probability by the given event's probability:

P(A | A ∪ B) = P(A) / P(A ∪ B).

And of course we already found P(A ∪ B) in part a), so the solution is trivial from there.

c) Recall that A and B are mutually independent only if P(A | B) = P(A), and if P(B | A) = P(B).

d) Your Venn diagram should make it clear whether A and B are mutually exclusive. Recall that the condition for mutually exclusive events is P(A ∪ B) = P(A) + P(B), the special case where (A ∩ B) is an empty set.