Let $A$ and $B$ are independent events with $P(A) = 0.3$ and $P(B) = 0.4$ . Find
(i) $P(A \cap B)$
(ii) $P(A \cup B)$
(iii) $P(A|B)$
(iv) $P(B|A)$
Let $A$ and $B$ are independent events with $P(A) = 0.3$ and $P(B) = 0.4$ . Find
(i) $P(A \cap B)$
(ii) $P(A \cup B)$
(iii) $P(A|B)$
(iv) $P(B|A)$
Official Solution
.: Since $A$ and $B$ are independent events
(i) $P(A \cap B) = P(A) \times p(B) = 0.3 \times 0.4 = 0.12$
(ii) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= P(A) + P(B) - P(A)P(B)$
$= 0.3 + 0.4 - 0.3 \times 0.4 = 0.7 - 0.12 = 0.58$
(iii) $P(A/B) = \cfrac{{P(A \cap B)}}{{P(B)}} = \cfrac{{P(A)P(B)}}{{P(B)}} = P(A) = 0.3$
(iv) $P(B/A) = \cfrac{{P(B \cap A)}}{{P(A)}} = \cfrac{{P(B)P(A)}}{{P(A)}} = P(B) = 0.4$
No comments yet — start the discussion.