Two Plus Two Older Archives - mutual exclusivity

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- - mutual exclusivity (http://archives2.twoplustwo.com/showthread.php?t=384974)

mutual exclusivity

p(A)= .5 p(B) = .35

p(A' and B') = .2

Find the probability that both A and B occur.

Re: mutual exclusivity

0.05?

Is this a trick question?

Re: mutual exclusivity

How did you get to 0.05?

Re: mutual exclusivity

Perhaps I should be more clear.

p(A') is where A does NOT happen so .5 for A' and .65 for B'

Re: mutual exclusivity

[ QUOTE ]
How did you get to 0.05?

[/ QUOTE ]

Of course he is right:

Code:<hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | | | .35
-------------------------------------------
B = 0 | | .2 |
--------------------------------------------
| .5 | | 1.00
</pre><hr />
Now you go ahead and fill in the missing values:
Code:<hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | (3) .05 | | .35
-------------------------------------------
B = 0 | (2) .45 | .20 |(1) .65
--------------------------------------------
| .50 | | 1.00
</pre><hr />

Re: mutual exclusivity

[ QUOTE ]
[ QUOTE ]
How did you get to 0.05?

[/ QUOTE ]

Of course he is right:

Code:<hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | | | .35
-------------------------------------------
B = 0 | | .2 |
--------------------------------------------
| .5 | | 1.00
</pre><hr />
Now you go ahead and fill in the missing values:
Code:<hr /><pre>
| A = 1 | A = 0 |
-------------------------------------------
B = 1 | (3) .05 | | .35
-------------------------------------------
B = 0 | (2) .45 | .20 |(1) .65
--------------------------------------------
| .50 | | 1.00
</pre><hr />

[/ QUOTE ]

Ok now now explain your calculations, I'm quite lost on this one.

Re: mutual exclusivity

I've managed to figure out that you've .35 * .5 - .2

But I don't understand why you've done it this way, if we take .5 and .65 and * them together we get .325 so there is a .125 gap which means they are not mutually exclusive, why do we just - the .2?

Re: mutual exclusivity

Let's try this a different way.

Prob ( A or B ) = 1 - Prob ( A' and B') = 0.8

Prob ( A or B ) = Prob (A) + Prob (B) - Prob (A and B)

Prob (A and B) = 0.5 + 0.35 - 0.8 = 0.05

Basically, for A or B to happen, add up the probabilities of A and B, but the subtract the "A and B" case so that you don't double count.

Re: mutual exclusivity

.35 + .5 does not equal the probability of A or B happening, they are not two sides of a coin, they are independent events each with differing probabilities.

Re: mutual exclusivity

There are four possibilities. AB, A'B, AB' and A'B'. You have four equations:

AB+AB' = 0.35
AB+A'B = 0.70
A'B' = 0.2
AB+A'B+AB'+A'B' = 1

If we add the first three and subtract the fourth, we get:

AB = 0.35 + 0.70 + 0.20 - 1 = 0.25.

To check, that means:

AB = 0.25
AB' = 0.10
A'B = 0.45
A'B' = 0.20

Those add up to 1 and meet the stated conditions.

Re: mutual exclusivity

[ QUOTE ]
.35 + .5 does not equal the probability of A or B happening

[/ QUOTE ]

If you read my post carefully you will see that this is not what I wrote.

Re: mutual exclusivity

[ QUOTE ]
There are four possibilities. AB, A'B, AB' and A'B'. You have four equations:

AB+AB' = 0.35
AB+A'B = 0.70
A'B' = 0.2
AB+A'B+AB'+A'B' = 1

If we add the first three and subtract the fourth, we get:

AB = 0.35 + 0.70 + 0.20 - 1 = 0.25.

To check, that means:

AB = 0.25
AB' = 0.10
A'B = 0.45
A'B' = 0.20

Those add up to 1 and meet the stated conditions.

[/ QUOTE ]

I think you read his problem as P(A) = .5*P(B) = .35, which would make P(A) = 0.35 and P(B) = 0.7. If you look at the spacing in the original text box (which you get when you quote it) it says P(A) = 0.5 and P(B) = 0.35.

In either case, we can easily solve this without setting up simultaneous equations. We know that P(A) + P(B) computes the probability of the union of A and B plus the intersection of A and B since the intersection is double counted. In other words:

P(A or B) = P(A) + P(B) - P(A and B)

P(A) + P(B) = P(A or B) + P(A and B)

so

P(A) + P(B) + P(A' and B') = P(A or B) + P(A and B) + P(A' and B')

P(A) + P(B) + P(A' and B') = 1 + P(A and B)

Since P(A' and B') = 0.2, if P(A) = 0.5 and P(B) = 0.35 then this equals 1.05, and P(A and B) = 0.05. However, if P(A) = 0.35 and P(B) = 0.7, then this equals 1.25, and P(A and B) = 0.25 as Aaron got.

Re: mutual exclusivity

p(A) is indeed .5 and p(B) is indeed .35

I can't get my head around the 0.05 answer, it just doesn't add up for me, I'm probably wrong but heres why I think its not 0.05:

In a perfect world probability AB = .5 * .35 = .175

if we take the chance of them both not happening in a perfect world it is .5 * .65 = .325

We are told the world isn't perfect and that neither of them happening is .2. It is for some reason less likely that they both don't happen.

So if the probability of them both not happening goes down then doesn't it follow logic that the chance that both of them happen goes up? I don't see why the probability of both of them happening goes from .175 to .05.

The additional .125 of mutual unexclusivity needs to be added somewhere, I don't see how it works against P(AB).

Re: mutual exclusivity

[ QUOTE ]
p(A) is indeed .5 and p(B) is indeed .35

I can't get my head around the 0.05 answer, it just doesn't add up for me, I'm probably wrong but heres why I think its not 0.05:

[/ QUOTE ]

I guarantee you're wrong; didn't you see my proof? What step don't you agree with? The three numbers you are given add up perfectly to

P(A) + P(B) + P(A' and B') = 1 + P(A and B).

Think about it. If you count everything in set A, plus everything in set B, plus everything in neither, then you have counted everything with the stuff in both A and B counted twice. I spelled out the proof of this, which just comes from this fundamental relation which you have to know:

P(A or B) = P(A) + P(B) - P(A and B).

This is always true. If A and B were mutually exclusive, then P(A and B) = 0, by definition. Mutual exclusivity means that both A and B cannot both happen.

[ QUOTE ]
In a perfect world probability AB = .5 * .35 = .175

if we take the chance of them both not happening in a perfect world it is .5 * .65 = .325

[/ QUOTE ]

This is only true when A and B are independent. If they were mutually exclusive, then P(A and B) = 0.

[ QUOTE ]
So if the probability of them both not happening goes down then doesn't it follow logic that the chance that both of them happen goes up?

[/ QUOTE ]

Of course not. It just means that the chance that one of them happens goes up. If the chance of A and B remain the same while the chance of them both not happening goes down, then the chance of them both happening must also go down.

If you can't see it, draw a Venn diagram, with 2 overlapping circles inside a box. As you move the two circles apart, the area outside the circles gets smaller. This is A' and B'. Also, the area of overlap gets smaller. This is A and B. The area inside the circles goes up. This is A or B.

[ QUOTE ]
I don't see why the probability of both of them happening goes from .175 to .05. The additional .125 of mutual unexclusivity needs to be added somewhere, I don't see how it works against P(AB).

[/ QUOTE ]

Again, you are confusing independence and mutual exclusivity. The "extra .125" is for the case where A and B are independent, not mutually exclusive. Compared to mutual exclusivity, P(AB) increased from 0 to 0.05.

Re: mutual exclusivity

I think I have this worked out now, I somehow thought that if p of failing two different tests is for some reason lower then your going to see more people passing both, of course this is not true because pA and pB remain constant and the spread becomes larger.

oh well, i finally got my head around it [img]/images/graemlins/smile.gif[/img]

Re: mutual exclusivity

Hi, JKKK.

I didn't check this thread for a while and it seems BruceZ has answered your question very capably as always.

Just to let you know: I got my original answer of 0.05 using a very simple graphical method. You know what a Venn diagram is right?

1) If we draw a framing rectangle to denote the whole universe of possibilities, all the mutually exclusive outcomes (i.e., the area of the rectangle) must be 1.

2) We know that P(A) = 0.5 and P(B) = 0.35. We also know that A and B are not mutually exclusive. So we can draw two circles, one representing A with an area of 0.5, and one representing B with an area of 0.35. Plus they overlap a bit. (Note that it's not necessary to draw anything to scale, of course, just that we note the area the circles are nominally supposed to have.)

3) P (A' and B') = 0.2, you said. This means that the area of the rectangle that's not within either circle is 0.2.

4) What you want to know is the area of the intersection of the two circles. This corresponds to the probability of A and B happening. Call this area X.

5) Finally we have four regions on our diagram that must add up to 1. We have the region exterior to the circles (0.2), circle A less its intersection with B (0.5 - X), circle B less its intersection with A (0.35 - X) and the intersection itself (X). Thus 0.2 + (0.5 - X) + (0.35 - X) + X = 1. Or, 1.05 - X = 1. Thus X = 0.05.

I guess it doesn't look that simple having spelled it all out now, but I promise you with a piece of paper it's really easy.

Cheers,
--JTR.

Re: mutual exclusivity

ok, my take on explaining this graphically and easily [img]/images/graemlins/smile.gif[/img]

1. Draw a square

2. Draw a vertical line down the middle of the square. Left half is A, right half is A'

3. Draw a horizontal line in the square (I drew it about 1/3 up from the bottom so it would look like about 35%). Below the horizontal line is B, above is B'.

4. Your square is now divided into 4 rectangular regions.

5. Finding your answer...
-The top right box is P(A' and B') and equals 20%.
-If the top right box is 20% that means the bottom right box is 30% (remember the right two boxes are A' and must equal 50%)
-If the bottom right box is 30% then the bottom left box must be 5% (remember the bottom two boxes are B and must add to 35%).
-But the bottom left box is P(A and B). Congrats you now know P(A and B) is 5%!

Pokerscott

Re: mutual exclusivity

The penny started to drop when fim used a rather colourful example of a population that have either, both or neither penises or vaginas.

I think the easiest analogy he used was to imagine two sticks that create angles, AB is opposite A'B' so what happens to AB when A'B' is shortened?

At first I didn't understand how this applied to the question of mutual exclusivity, but now I do, thanks guys.

Re: mutual exclusivity

[ QUOTE ]
ok, my take on explaining this graphically and easily [img]/images/graemlins/smile.gif[/img]

1. Draw a square

2. Draw a vertical line down the middle of the square. Left half is A, right half is A'

3. Draw a horizontal line in the square (I drew it about 1/3 up from the bottom so it would look like about 35%). Below the horizontal line is B, above is B'.

4. Your square is now divided into 4 rectangular regions.

5. Finding your answer...
-The top right box is P(A' and B') and equals 20%.
-If the top right box is 20% that means the bottom right box is 30% (remember the right two boxes are A' and must equal 50%)
-If the bottom right box is 30% then the bottom left box must be 5% (remember the bottom two boxes are B and must add to 35%).
-But the bottom left box is P(A and B). Congrats you now know P(A and B) is 5%!

Pokerscott

[/ QUOTE ]

There is an issue with this drawing which didn’t prevent you from getting the correct answer, but nevertheless it is important to discuss. Aside from the numbers, the drawing implies that A and B are independent, which they are not. The vertical line down the middle makes A occupy 1/2 of the total square, meaning that its probability is 1/2, and since B goes all the way across horizontally, it gets cut in half by the A/A'line, so that the fraction of B that overlaps A is also 1/2. That means that P(A) = P(A | B), or in words, the probability of being in A is the same as the probability of being in A given that you are in B. This means that A and B are independent by definition since it means that being in B has no bearing on the chance that we are in A.

It is equivalent to saying that P(A and B) = P(A)*P(B), which can also be taken as the definition of independence, and this relation can also be seen from your drawing since P(A)*P(B) is the width of A multiplied by the width of B, which is the area of P(A and B). You wrote the number 0.05 in this box because that was the amount left over once you put the numbers in the other boxes, but there is no way that you could make the area of that box 0.05 if, as you say, the vertical line goes down the middle, and the other line is 35% from the bottom. The area is 0.5*0.35. In fact, no matter where you draw the vertical line, and no matter where you draw a horizontal line, as long as the lines go all the way across, you can never make P(A and B) anything other than P(A)*P(B), and they will always represent independent events.

This aspect of your drawing doesn't impact your solution, and it's OK as long as we only interpret the drawing as a table of numbers. However, when you draw the line for B 35% from the bottom, that implies that the areas are part of the model, and that A and B are independent, which they are not. This could be corrected by using a circle instead of a square, and then drawing a diameter for A/A', and another chord for B/B' that does not get bisected by A/A'.

The main reason I mention this is because many people have difficulty grasping the concept of independence. You have made a drawing which illustrates this concept beautifully, though this was unintentional. Often people confuse the concepts of independence and mutual exclusivity. In a diagram such as this one, mutual exclusivity would be shown by drawing A and B so that they do not overlap, so that P(A intersect B) = P(A and B) = 0. Clearly this bears no relationship to independence, and in fact, events with non-zero probability can never be both independent and mutually exclusive. I don’t know why the title of this thread is “mutual exclusivity”, since that applies to nothing in this problem.

Re: mutual exclusivity

[ QUOTE ]

This aspect of your drawing doesn't impact your solution, and it's OK as long as we only interpret the drawing as a table of numbers. However, when you draw the line for B 35% from the bottom, that implies that the areas are part of the model, and that A and B are independent, which they are not.

[/ QUOTE ]

No argument here. I probably shouldn't have made hints to in the middle and 1/3 from the bottom. I was just using it as a convenient way to segment the problem so you could easily see what P(A' and B') implied and could use that information to figure out P(A and B).

It is pretty obvious that the area inside each square does not correspond to the probability in the square (the left and right bottom boxes are the same size in the drawing but one is 5% and one is 30%!).

Anyway, you are right, think of it just as a convenient way of have tabulating the four events.

Pokerscott

Re: mutual exclusivity

Its called mutual exclusivity because A and B are not mutually exclusive, did I miss something?

Re: mutual exclusivity

Infact now I can see now how this has nothing to do with mutual exclusivity, my statistics teacher has done a thorough job of confusing me here.