Patrick W. answered • 04/01/15
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High School Mathematics Teacher, Passionate Math Geek
Definitely practice talking through these problems and working with other people to get them to sink in. Much of finite math will click immediately, and much of what doesn't click will click immediately for other people. It's a lot about finding the right perspective and an explanation that makes sense to you. It sounds like I missed an opportunity to help you on your test, but I can talk you through these problems.
1.
Suppose that A and B are two events in a sample space S. You are told the following:
P(A') = 0.6
P(B) = 0.5
P(A∩B)= 0.1
Find P(A∪B)
I always had to draw a picture for these. This is super common because it's useful. Our sample space S will be a rectangle with an area of 1. The 1 is for 100%. There will be two intersecting circles inside S labeled A and B. I'm picturing this : http://www.dyclassroom.com/image/topic/aptitude/probability/probability.png
Look carefully at the diagram - there are four different areas, and we will be able to describe all of them. If we add up all four areas, we should end up with 1.
The first area I would label in our diagram is the intersection of A and B. It is given that this area is 0.1, when they say P(A∩B)= 0.1
Now, when they tell us that P(B)=0.5, they're saying the area of B is 0.5, but that includes the part which intersects with A, which we know is 0.1. The part of B that doesn't intersect with A must have an area of 0.4. We can express this by writing P(A'∩B)= 0.4
There are two areas left to describe. When they tell us that P(A') = 0.6, they're saying that the area of everything not A is 0.6. There are two areas outside of A: There's the part of B that doesn't intersect A, and there's the part outside of both circles. They must add up to 0.6, so the outside of both events must be 0.2.
What about the area of A that doesn't intersect with B? Well, P(A') = 0.6 means that 1-P(A)=0.6, so P(A) must equal 0.4. Since the intersection with B takes up 0.1 of this, the area of A not intersecting with B must have an area of 0.3.
Ok, do you have all four parts labeled? What are they asking?
When they say to find P(A∪B), they're saying what is the area of all of A and B together. We can either add up all three parts inside the circles ( 0.3+0.1+0.4 ) OR, we can see that the only part of S that is not in A OR B is the outside area of 0.2. Since S=1, 1-0.2=0.8
1-0.2=0.8
is like saying 1-P(A'∩B')=P(A∪B)
( 0.3+0.1+0.4 )=0.8
is like saying P(A∩B')+P(A∩B)+P(A'∩B)=P(A∪B)
2.
Suppose that A and B are two events in a sample space S.You are told the following:
P(A) = 0.3
P(B) = 0.7
A and B are independent events
Find Pr(A∩B).
When events are independent, it is a very specific special case. The word is used in mathematics the same way we use it in language - the events are independent as in one event does not affect the other. If you tell me that A is true it does not affect the likelihood of whether B is true, and vice versa.
That means that whether I'm looking at all of S or whether I'm looking just inside of A, the chance of B has to be the same. Well, starting from S, the chance of B is 70%. So I know that 70% of A must intersect with B, so that the chance of B within A is still 0.7. So what is 70% of 0.3?
0.7×0.3=0.21
This must be the area of the intersection between A and B. That means that the part of A that doesn't intersect B must be 0.3-0.21=0.09
Also, the part of B that doesn't intersect A must be 0.7-0.21=0.49
Label these three parts! It looks like the fourth outside part must be 1-0.09-0.21-0.49=0.21
Now look carefully at what we have. The parts of A add up to 0.3 and the parts of B add up to 0.7. This means that P(A)=0.3 and P(B)=0.7.
BUT, in this specific case, the part of A that intersects with B makes up 70% of A: 0.21/0.3=0.7
AND the part of B that intersects with A makes up 30% of B: 0.21/0.7=0.3
So it doesn't matter whether we're choosing B from all of S or from inside of A, the probability is the same.
3.
Suppose that A and B are two events in a sample space S. You are told the following:
A and B are disjoint events
P(A) = 0.4
P(B) = 0.3
Find P(A'∩B')
Disjoint simply means that there is no intersection. You can either draw a different diagram from the one we've been using, or you can simply label the area of the intersection 0.0.
The area of S is 1, so the fourth area in our diagram is 1-0.3-0.4. This is the same as saying the probability of not A AND not B.
5.
Suppose that A and B are two events in a sample space S. Further, you are told that:
P(A)=0.3
P(B)=0.2
P(A'∩B')=0.6
Find P(B|A)
Ok, same drill. Label our parts. You'll find that you don't usually want to start with A or B, since A and B both contain more than one area. The third line we're given, however, is talking about only one area. They're describing the part that is not A AND not B, so the outside area that is in neither event. Label this as 0.6
Ok, we've got 0.4 left to fill in the rest of the diagram, but we know we're going to have 0.2 and 0.3 ... Does this seem like a mistake? Won't that be too much area? Our events appear to be too large by 0.1 ...
It's not though - the intersection of A and B is in both events, so we don't need to count it twice (You'll see this again, sometimes triple-counting regions). We'll put that 0.1 discrepancy in the intersection, and it will cancel out nicely. If the intersection of A is 0.1, then the rest of A must be 0.3-0.1=0.2 and the rest of B must be 0.2-0.1=0.1.
If you've got all four regions labeled, we can talk about the meaning of P(B|A)
This is read as "the probability of B, given A"
It means that if we are given A, what is the chance of B within that?
Ignore everything that's not in A. That means we're only looking at two regions. How much total area do we have? We've got 0.3
Now, since we're in A, what are the chances of being in B? Well the region with B is 0.1, and 0.1/0.3=0.333
the answer: 0.3 Like the above problem I don't understand disjoint.
the answer: 0.3 Like the above problem I don't understand disjoint.
