#1
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mutual exclusivity
p(A)= .5 p(B) = .35
p(A' and B') = .2 Find the probability that both A and B occur. |
#2
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Re: mutual exclusivity
0.05?
Is this a trick question? |
#3
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Re: mutual exclusivity
How did you get to 0.05?
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#4
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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' |
#5
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Re: mutual exclusivity
[ QUOTE ]
How did you get to 0.05? [/ QUOTE ] Of course he is right: <font class="small">Code:</font><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: <font class="small">Code:</font><hr /><pre> | A = 1 | A = 0 | ------------------------------------------- B = 1 | (3) .05 | | .35 ------------------------------------------- B = 0 | (2) .45 | .20 |(1) .65 -------------------------------------------- | .50 | | 1.00 </pre><hr /> |
#6
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Re: mutual exclusivity
[ QUOTE ]
[ QUOTE ] How did you get to 0.05? [/ QUOTE ] Of course he is right: <font class="small">Code:</font><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: <font class="small">Code:</font><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. |
#7
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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? |
#8
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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. |
#9
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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.
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#10
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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. |
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