A) \[0.45\]
B) \[0.55\]
C) \[0.9\]
D) \[0.35\]
Correct Answer: D
Solution :
Given, A and B are mutually exclusive events. \[\Rightarrow \] \[P(A\cap B)=0\] By addition theorem of probability, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow \] \[P(A\cup B)=P(A)+P(B)\] \[\Rightarrow \] \[P(A\cup B)=0.25+0.4\] \[[\because \,\,P(A)=0.25,\,P(B)=0.4\,(given)]\] \[P(A\cup B)=0.65\] ??(i) Now, we have \[P({{A}^{c}}\cap {{B}^{c}})=P{{(A\cup B)}^{c}}\] \[\Rightarrow \] \[P({{A}^{c}}\cap {{B}^{c}})=1-P(A\cup B)\] \[=1-0.65=0.35\]
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