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• Notes

If $P(A)=0.3, P(B)=0.2,$ and $P(A \cap B)=0.1$

$P(A)=0.3$
$P(B)=0.2$
$P(A \cap B)=0.1$

$P\left(A^{\prime}\right)=1-P(A)=1-0.3=0.7$

b) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$=0.3+0.2-0.1=0.4$

c) $P\left(A^{\prime} \cap B\right)$

$P(B)=P(A \cap B)+P(A^{\prime} \cap B)$

$0.2=0.1+P(A^{\prime} \cap B)$

$\Rightarrow P(A^{\prime} \cap B) = 0.1$

d) $P\left(A \cap B^{\prime}\right)$

$P(A)=P(A \cap B)+P(A \cap B^{\prime})$

$0.3=0.1+P(A \cap B^{\prime})$

$P(A \cap B^{\prime})=0.2$

e) $P\left[(A \cup B)^{\prime}\right]$

$=1-P(A \cup B)=0.6$

f) $P\left(A^{\prime} \cup B\right)$

$=P\left(A^{\prime}\right)+P(B)-P\left(A^{\prime} \cap B\right)$

$=0.7+0.2-0.1=0.8$