A) \[2/3\]
B) \[1/6\]
C) \[5/6\]
D) \[1/3\]
Correct Answer: D
Solution :
Given, \[P(A)=\frac{1}{2},P(B)=\frac{1}{3}\] \[\because \] A and B are in depended events. \[\therefore \] \[P(A\cap B)=P(A).P(B)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}\] Now, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\] \[=\frac{3+2-1}{6}=\frac{4}{6}=\frac{2}{3}\] \[\therefore \] P (neither A nor B ) \[=P(\bar{A}\cap \bar{B})=P(\overline{A\cup B})\] \[=1-P(A\cup B)\] \[=1-\frac{2}{3}=\frac{1}{3}\]
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