mutual exclusivity

[jkkkk]

[ 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 ]

ahh ingenious, thanks.

edit:

Wait hold on a second, why does this not work?

A'B' = .2
A'B = .175
AB' = .325
AB= 1 - (A'B' + A'B + AB')

AB= .3

Funnily enough this is the answer I came up with earlier today though I arrived at it through a different method:

.65 * .5 = .325

.375 - .2 = .125 (gap of mutual inexclusivity)

.5 * .35 + .125 = .3

[mosdef]

[ 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.

[BruceZ]

[ 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.

[jkkkk]

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).

[BruceZ]

[ 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.

[jkkkk]

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]

[jtr]

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.

[Pokerscott]

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

[jtr]

Nice one.

[jkkkk]

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.