) Given the three events A, B, c which are partitioned of the sample space such that P (A )=0.3 ,2P (B^( c ) )=0.8 Find P(c).
2P (B^( c ) )=0.8→P (B^( c ) )=0.4P(B)=1-P (B^( c ) )=1-0.4=0.6Since A, B, c are the partitioned of the sample spaceP(A)+P(B)+P(C)=1→P(C)=0.1
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2P (B^( c ) )=0.8→P (B^( c ) )=0.4
need an explanation for this answer? contact us directly to get an explanation for this answerP(B)=1-P (B^( c ) )=1-0.4=0.6
Since A, B, c are the partitioned of the sample space
P(A)+P(B)+P(C)=1→P(C)=0.1