How to solve this question? If n(U)=100,n(A)=30,n(B)=40 and n(Acap B)= 10. Find n(A^{c}cap B^{c}).

Answers (2)
N neha

As we know that ; 

\left ( A\cup B \right )^c=A^c \cap B^c

multiplying by n and analysing the same: 

n\left ( A\cup B \right )^c=n\left( A^c \cap B^c\right)

n\left ( A\cup B \right )^c=n(U)-n\left ( A\cup B \right )

n\left ( A\cup B \right )=n(A)+n(B)-n\left ( A\cap B \right )

n\left ( A\cup B \right )=30+40+10=80

n\left ( A\cup B \right )^c=100-80=20.

N neha

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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