If $P(A) = 0.8,P(B) = 0.5$ and $P(B|A) = 0.4$ , find
(i) $P(A \cap B)$
(ii) $P(A|B)$
(iii) $P(A \cup B)$
If $P(A) = 0.8,P(B) = 0.5$ and $P(B|A) = 0.4$ , find
(i) $P(A \cap B)$
(ii) $P(A|B)$
(iii) $P(A \cup B)$
Official Solution
.: Given, $P(B|A) = 0.4 \Rightarrow \cfrac{{P(A \cap B)}}{{P(A)}} = 0.4$
$\Rightarrow P(A \cap B) = P(A) \times 0.4 = 0.8 \times 0.4 = 0.32$
(ii) $P(A|B) = \cfrac{{P(A \cap B)}}{{P(B)}} = \cfrac{{0.32}}{{0.5}} = \cfrac{{16}}{{25}} = 0.64$
(iii) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= 0.8 + 0.5 - 0.32 = 1.3 - 0.32 = 0.98$
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