#12
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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 ] 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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