A) \[0.25\]
B) \[0.50\]
C) \[0.75\]
D) \[0.40\]
Correct Answer: A
Solution :
Given, \[P(A\cap {{B}^{c}})=0.25\] and \[P({{A}^{c}}\cap B)=0.5\] \[\therefore \] \[P\{{{(A\cup B)}^{c}}\}=1-P(A\cup B)\] \[=1-\{P(A\cap {{B}^{c}})+P({{A}^{c}}\cap B)+P(A\cap B)\}\] \[=1-(0.25+0.5+0)\] [ \[\because \] A and B are mutually exclusive events \[P(A\cap B)=0\]] \[=1-0.75=0.25\]You need to login to perform this action.
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