Hint:- As we know that $P\left( {B|A} \right) = \dfrac{{P(A \cap B)}}{{P(A)}}$
It is given that $P(A) = 0.8,P(B) = 0.5$ and $P\left( {B|A} \right) = 0.4$ As we know that $P\left( {B|A} \right) = \dfrac{{P(A \cap B)}}{{P(A)}}$ For (i) $P(A \cap B)$ $P\left( {B|A} \right) = 0.4$ $P\left( {B|A} \right) = \dfrac{{P(A \cap B)}}{{P(A)}} = 0.4$ and in question it is given that $P(A) = 0.8$ $ \Rightarrow \dfrac{{P(A \cap B)}}{{0.8}} = 0.4$ $\therefore P(A \cap B) = 0.4 \times 0.8 = 0.32$ For (ii) $P(A|B)$ $P\left( {A|B} \right) = \dfrac{{P(A \cap B)}}{{P(B)}}$ and in question it is given that $P(B) = 0.5$and from above solution we get $P(A \cap B) = 0.32$ $ \Rightarrow P\left( {A|B} \right) = \dfrac{{0.32}}{{0.5}} = 0.64$ For (iii) $P(A \cup B)$ As we know $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $ \Rightarrow P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $ \Rightarrow P(A \cup B) = 0.8 + 0.5 - 0.64$ all the values are given above. $\therefore P(A \cup B) = 0.66$
Note:- This is a simple formula based question . You have to always keep in mind the basic formula by applying this hint you can easily achieve to the answer.
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