Subjective Type

If $$ n(\xi)=40, n\left(A^{\prime}\right)=15, n(B)=12 $$ and $$ n\left((A \cap B)^{\prime}\right)=32, $$ find : (i) n(A) (ii) $$ n\left(B^{\prime}\right) $$ (iii) $$ n(A \cap B) $$ (iv) $$ n(A \cup B) $$ (v) $$ n(A-B) $$ (vi) $$ n(B-A) $$

Solution

From the question it is given that,
$$\begin{array}{l}n(\xi)=40 \\n\left(A^{\prime}\right)=15 \\n(B)=12\end{array}$$
$$n\left((A \cap B)^{\prime}\right)=32$$
(i) $$ \mathrm{n}(\mathrm{A}) $$We know that, $$ n(A)=n(\xi)-n\left(A^{\prime}\right) $$
$$\begin{array}{l}n(A)=40-15 \\n(A)=25\end{array}$$
(ii) $$ n\left(B^{\prime}\right) $$
We know that, $$ n\left(B^{\prime}\right)=n(\xi)-n(B) $$
$$\begin{array}{l}n\left(B^{\prime}\right)=40-12 \\n\left(B^{\prime}\right)=28\end{array}$$
(iii) $$ n(A \cap B)=n(\xi)-n\left((A \cap B)^{\prime}\right) $$
$$ =40-32 $$
$$ =8 $$
$$ \begin{aligned}(\text { iv }) \text { n }(A \cup B) &=n(A)+n(B)-n(A \cap B) \\ &=25+12-8 \\ &=37-8 \\ &=29 \end{aligned} $$
$$ \begin{aligned}(\mathrm{v}) \mathrm{n}(\mathrm{A}-\mathrm{B})=& n(\mathrm{A})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ &=25-8 \\ &=17 \end{aligned} $$
$$ \begin{aligned}(\mathrm{vi}) \mathrm{n}(\mathrm{B}-\mathrm{A}) &=\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ &=12-8 \\ &=4 \end{aligned} $$

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