Re: mutual exclusivity
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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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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
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