Question

If $A$ and $B$ are mutually exclusive events, such that $ P(A)=0.25,\,\,\,\,\,\,P(B)=0.4, $ then $ P({{A}^{c}}\,\cap {{B}^{c}}) $ is equal to

• A. $ 0.45 $
• B. $ 0.55 $
• C. $ 0.9 $
• D. $ 0.35 $

Answer 1

Answer: (D)

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 $