Tag: complement of sets

We have the De Morgan's theorem,

$A-(B\cup C)=A\cap (B\cup C)^{C}=A\cap (B^{C}\cap C^{C})=(A\cap B^{C})\cap (A\cap C^{C})=(A-B)\cap (A-C)$

$A\cup B=B-A+A-B+A\cap B\\implies B-A=A-B=\phi \implies A=B$

We know that $\displaystyle \left ( A\cup B \right )'=$ $\displaystyle A'\cap B'$ so we need to find $\displaystyle \left ( A\cup B \right ) $  first.
$n\displaystyle \left ( A\cup B \right ) =$ $n(A) + n(B) - n$$\displaystyle (A\cap B)$ $= 200 + 300 - 100 = 400$
$n\displaystyle \left ( A\cup B \right )' =$ $n(U) - n$$\displaystyle \left ( A\cup B \right )'=$$  700 - 400 = 300$
$\displaystyle \Rightarrow $ $n\displaystyle (A'\cap B')$$= 300$.

Since $A\subset B$, $A\cap B$ will be all the elements of A. So, the number of elements will be 5.
Since $A\subset B$, $A\cup B$ will be all the elements of A & B. So, the number of elements will be 8.

Students passed in Hindi $ = 70\% =n(H)$

Students passed in English $ = 80\% =n(E)$

Students passed in Both $ = 65\% =n(H\cap E)$

No. of students passed $=n(H \cup E)= n(H)+n(E)-n(H \cap E) = 70+80-65 =85\%$

No. of Students failed $= 100\% - n(H \cup E) = 100\% - 85\% = 15\%$

Fail: $n(M) =50\%$

$n(P) =50\%$

$n(M \cap P) =20\%$

$n(M\cup P) =n(M)+n(P)-n(M\cap P)$

$n(M\cup P) =50+50-20 = 80\%$

Pass:

n(Pass in both subjects) $=100\%$ -n(Fail in both the subjects)

$=100\% -80\% = 20\%$

$(A\cup B)' = A'\cap B'$ is called De Morgan's law.

n(Watch new on TV) $=65\%=n(T)$

n(Reads newspaper) $=40\%=n(N)$

n(Watch news on TV & read newspapers ) $=25\%=n(T\cap N)$

$n(N \cup T)= n(T)+n(N)-n(N\cap T)$

$n(N \cup T)= 65+40-25 = 80\%$

$n(N\cap T)^{c}=100-80=20\%$