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since when I first subtracted P(AnB) that was also including P(AnBnC). So next time when I am substracting P(AnC) and P(AnB), I am adding P(AnBnC) to avoind any duplicaiton

since when I first subtracted P(AnB) that was also including P(AnBnC). So next time when I am substracting P(AnC) and P(AnB), I am adding P(AnBnC) to avoind any duplicaiton

let me try.... AuBuC should include everying that is in A, B and C and common areas should be counted only once...

AnB is a part of A AnB is a part of B

so when we write A+B+C we are actully counting AnB twice .... so we have to substratc AnB once to get the final answer ... similar logic for AnC and BnC ....

AnBnC is a part of A, B, C, AnB, AnC and BnC.... so when we add A+B+C and substract - AnB-AnC-BnC we are adding AnBnC 3times and substraticg it 3 times .... so we have to add in once to get the final answer....

since when I first subtracted P(AnB) that was also including P(AnBnC). So next time when I am substracting P(AnC) and P(AnB), I am adding P(AnBnC) to avoind any duplicaiton

let me try.... AuBuC should include everying that is in A, B and C and common areas should be counted only once...

AnB is a part of A AnB is a part of B

so when we write A+B+C we are actully counting AnB twice .... so we have to substratc AnB once to get the final answer ... similar logic for AnC and BnC ....

AnBnC is a part of A, B, C, AnB, AnC and BnC.... so when we add A+B+C and substract - AnB-AnC-BnC we are adding AnBnC 3times and substraticg it 3 times .... so we have to add in once to get the final answer....

HTH

Make sense so far. Hope I dont go too deep into this. My problem is that I go very deep and eventually ned up in confusing myself. So far your explanation makes perfect sense.

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