A) \[P(B/A)=P(B)-P(A)\]
B) \[P({{A}^{c}}\cup {{B}^{c}})=P({{A}^{c}})+P({{B}^{c}})\]
C) \[P{{(A\cup B)}^{c}}=P({{A}^{c}})\,P({{B}^{c}})\]
D) \[P(A/B)=P(A)\]
Correct Answer: C
Solution :
Since \[P(A\cap B)=P(A)\,P(B)\] It means \[A\] and \[B\] are independent events so \[{{A}^{c}}\] and \[{{B}^{c}}\] will also be independent. Hence \[P{{(A\cup B)}^{c}}=P({{A}^{c}}\cap {{B}^{c}})=P({{A}^{c}})\,P({{B}^{c}})\] (Demorgan?s law) As \[A\] is independent of \[B,\] hence \[P(A/B)=P(A)\], \[\{\because \,\,\,P(A\cap B)=P(B)P(A/B)\}\].
You need to login to perform this action.
You will be redirected in
3 sec
