If A,B independent events, prove that A^c,B^C are independent.
∵P(A^c∩B^C )=〖P(A∪B)〗^c=1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-P(A)-P(B)+P(A∩B)∵A,B independent∴P(A∩B)=P(A)P(B)∵P(A^c∩B^C )=P(A^c )-P(B)+P(A)(B)=P(A^c )-P(B)[1-P(A)]=P(A^c )-P(B)P(A^c)=P(A^c )[1-P(B)]=P(A^c )P(B^c )
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∵P(A^c∩B^C )=〖P(A∪B)〗^c=1-P(A∪B)
need an explanation for this answer? contact us directly to get an explanation for this answer=1-[P(A)+P(B)-P(A∩B)]=1-P(A)-P(B)+P(A∩B)
∵A,B independent
∴P(A∩B)=P(A)P(B)
∵P(A^c∩B^C )=P(A^c )-P(B)+P(A)(B)
=P(A^c )-P(B)[1-P(A)]
=P(A^c )-P(B)P(A^c)
=P(A^c )[1-P(B)]=P(A^c )P(B^c )