# A and B are two events such that

Question:

A and B are two events such that

P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.

Find

(i) P(A ∩ B)

(ii) P(A′ ∩ B′)

(iii) P(A ∩ B′)

(iv) P(B ∩ A′)

Solution:

It is given that $P(A)=0.54, P(B)=0.69, P(A \cap B)=0.35$

(i) We know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.54+0.69-0.35=0.88$

(ii) $A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}[$ by De Morgan's law]

$\therefore P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)=1-0.88=0.12$

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

$=0.54-0.35$

$=0.19$

(iv) We know that $n\left(\mathrm{~B} \cap \mathrm{A}^{\prime}\right)=n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B})$

$\Rightarrow \frac{n\left(\mathrm{~B} \cap \mathrm{A}^{\prime}\right)}{n(\mathrm{~S})}=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}-\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}$

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

$\therefore \mathrm{P}\left(\mathrm{B} \cap \mathrm{A}^{\prime}\right)=0.69-0.35=0.34$