Probability, Set Theory, Venn Diagrams

In summary: That is, the stuff outside A or B is the stuff not in A but also not in B. In particular, whatever is in A ∪ B must be in A or B or both, so the remaining stuff is the stuff neither in A nor in B. This is what we mean by (A ∪ B)' = A' ∩ B' or (A' ∩ B')' = A ∪ B.
  • #1
whitejac
169
0

Homework Statement


Let A and B be two events such that

P(A) = 0.4, P(B) = 0.7, P(A∪B) = 0.9

Find P((A^c) - B)

2. Homework Equations

I can't think of any relevant equations except maybe the Inclusion Exclusion property.
P(A∪B) = P(A) + P(B) - P(A∩B)
This leads us to another thing
P(A∩B^c) = P(A-B) = P(A) - P(A∩B)
And P(A^c) = 1 - P(A)

3. The Attempt at a Solution

The primary problem I'm having is exchanging the equation for one I can easily understand.
I know that P(A∩B) = 0.2 from the Inclusion Exclusion Property.
However, I guess I'm having trouble comprehending the principles behind this math. I can't simply plug things in and expect it to work with the venn diagrams right?
ex) I can't simply say A' = A^c = 1 - 0.4 = 0.6,
and then have it in the form P(A'-B) = P(A') - P(A'∩B),
Because then I have an equation with two unknowns...
Using the In-Ex Prop, I'd have P(A'-B) = P(A') - P(A') + P(B) - P(A'∪B)
Which basically leaves me in the same mess of too many variables.

The logive behind Unions and Intersections really confuses me mathematically. I can visualize the venn diagrams for the most part, but translating that into a math and a function leaves me lost.
 
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  • #2
whitejac said:

Homework Statement


Let A and B be two events such that

P(A) = 0.4, P(B) = 0.7, P(A∪B) = 0.9

Find P((A^c) - B)

2. Homework Equations

I can't think of any relevant equations except maybe the Inclusion Exclusion property.
P(A∪B) = P(A) + P(B) - P(A∩B)
This leads us to another thing
P(A∩B^c) = P(A-B) = P(A) - P(A∩B)
And P(A^c) = 1 - P(A)

3. The Attempt at a Solution

The primary problem I'm having is exchanging the equation for one I can easily understand.
I know that P(A∩B) = 0.2 from the Inclusion Exclusion Property.
However, I guess I'm having trouble comprehending the principles behind this math. I can't simply plug things in and expect it to work with the venn diagrams right?
ex) I can't simply say A' = A^c = 1 - 0.4 = 0.6,
and then have it in the form P(A'-B) = P(A') - P(A'∩B),
Because then I have an equation with two unknowns...
Using the In-Ex Prop, I'd have P(A'-B) = P(A') - P(A') + P(B) - P(A'∪B)
Which basically leaves me in the same mess of too many variables.

The logive behind Unions and Intersections really confuses me mathematically. I can visualize the venn diagrams for the most part, but translating that into a math and a function leaves me lost.

You can't say ##A' = A^c = 1 - 0.4 = 0.6## because ##A'## is an event, not a probability. However, you can say ##P(A') = P(A^c) = 1 - 0.4 = 0.6##. Then, your equation ##P(A' - B) = P(A') - P(A' \cap B)## is true. You just need to find ##P(A' \cap B)##. Look at a Venn diagram, to see what this "region" ##A' \cap B## looks like, then use the information given in the problem (plus some side calculations) to see how you would reckon its probability. It might be helpful to work out the individual probabilities of the four relevant regions: R1 = in A but not B, R2 = in B but not A, R3 = in both A and B, and R4 = outside both A and B. You have all the information you need to do these four computations, and you have already presented evidence that you know the basic way of approaching them.
 
  • #3
I've been redoing this problem since last night, here's what I came up with...

Using a Venn Diagram, I've begun with plotting P(A'). This would be everything outside of A, so B and the region outside of the overlapped circles. I've allocated the quantity 0.1 to this region because P(A∪B)' should be 1-0.9 which would be 0.1.

Then I subtracted the P(B) Portion, which would leave the 0.1 and give me P(A' - B).

Another way I came to this conclusion, and your reply helps cement my conviction, was by coming across this equality: P(A’ – B) = P(A’ ∩ B’) which would be all things that are neither in A or B.
 
  • #4
whitejac said:
I've been redoing this problem since last night, here's what I came up with...

Using a Venn Diagram, I've begun with plotting P(A'). This would be everything outside of A, so B and the region outside of the overlapped circles. I've allocated the quantity 0.1 to this region because P(A∪B)' should be 1-0.9 which would be 0.1.

Then I subtracted the P(B) Portion, which would leave the 0.1 and give me P(A' - B).

Another way I came to this conclusion, and your reply helps cement my conviction, was by coming across this equality: P(A’ – B) = P(A’ ∩ B’) which would be all things that are neither in A or B.

Exactly. You can also reason it out in words: A' = stuff not in A and so A' - B = stuff not in A but also outside B = the complement of A ∪ B. Note that (A ∪ B)' = A' ∩ B' (another "law" you should get to know by heart).
 

Related to Probability, Set Theory, Venn Diagrams

1. What is the difference between probability and statistics?

Probability is the branch of mathematics that deals with the likelihood of events occurring, while statistics is the study of data collection, organization, analysis, and interpretation. Probability is used to predict the likelihood of future events, while statistics is used to make sense of past data.

2. What is set theory and why is it important in probability?

Set theory is a branch of mathematics that deals with the study of sets, which are collections of objects. In probability, sets are used to represent possible outcomes of an experiment or event. Set theory is important in probability because it helps us to understand and calculate the likelihood of certain outcomes occurring.

3. How are Venn diagrams used in probability?

Venn diagrams are graphical representations of sets that are often used in probability to visually represent the relationships between different sets of data. They can be used to calculate probabilities by shading in the regions of the diagram that represent the desired outcomes.

4. What is the difference between independent and dependent events in probability?

In probability, independent events are those where the outcome of one event does not affect the outcome of another event. Dependent events, on the other hand, are those where the outcome of one event does affect the outcome of another event. This is important to consider when calculating probabilities.

5. How can probability be applied in real-life situations?

Probability can be applied in a variety of real-life situations, such as weather forecasting, stock market analysis, sports predictions, and medical research. It can also be used to make decisions in business, politics, and everyday life. Understanding probability can help us make more informed and rational decisions based on the likelihood of certain outcomes.

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