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\\\text{If } n(U)=100, \ n(A)=30, \ n(B)=40 \ \text{and } \ n(A\cap B)= 10. \text{ Find } n(A^{c}\cap B^{c})

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The solution to this question can be solved by using the following formula:

n(U)=100       n(A)=30     n(B)=40       n(A\cap B)=10
                n(A^c\cap B^c)=?
  (A^c\cap B^c)=(A\cup B)^c      {De Morgan's Law}
                 
                       n(A\cup B)=n(A)+n(B)-n(A\cap B)
                                                40+30-10=60
                           n(A\cup B)^c=n(U)-n(A\cup B)
                                                  =100-60=40
                           n(A\cup B)^c=n(A^c\cap B^c)=40

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