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If $P(A) = 0.8,P(B) = 0.5$ and$P\left( {B|A} \right) = 0.4$, find
(i)$P(A \cap B)$
(ii)$P(A|B)$
(iii)$P(A \cup B)$

Answer Verified Verified
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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