A) \[\frac{1}{2}\]
B) \[\frac{2}{5}\]
C) \[\frac{2}{7}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
[b] Event A and B are mutually exclusive. Hence \[P(A\cap B)=\phi =0\] \[\therefore \,\,\,\,P(A\cup B)=P(A)+P(B)\] ? (1) \[P(A)=0.2\] [Given] \[P(B)=P(\bar{A}\cap B)+P(A\cap B)\] \[P(B)=P(\bar{A}\cap B)\] \[[\because P(A\cap B)=0]\] \[=0.3\] \[P(A\cup B)=0.2+0.3=0.5\] \[P(A|A\cup B)=\frac{P(A)}{P(A\cup B)}=\frac{0.2}{0.5}=\frac{2}{5}\] \[P(A|(A\cup B))=\frac{2}{5}\]You need to login to perform this action.
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