If A and B are mutually exclusive events, prove thatP(A⁄B ̅ )=(P(A))/(1-P(B))
∵P(A⁄B ̅ )=(P(A∩B ̅))/(P(B ̅))=(P(A-B))/(1-P(B))=(P(A)-P(A∩B))/(1-P(B))∵A,B mutually exclusive events→∴P(A∩B)=0∴P(A⁄B ̅ )=(P(A))/(1-P(B))
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∵P(A⁄B ̅ )=(P(A∩B ̅))/(P(B ̅))=(P(A-B))/(1-P(B))=(P(A)-P(A∩B))/(1-P(B))
need an explanation for this answer? contact us directly to get an explanation for this answer∵A,B mutually exclusive events→∴P(A∩B)=0
∴P(A⁄B ̅ )=(P(A))/(1-P(B))