A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{3}{4}\]
D) \[\frac{3}{8}\]
Correct Answer: A
Solution :
Hence, A and B are mutually exclusive events.. \[P(A\cap B)=0\] Also, \[S=A\cup B\] \[\therefore \] \[P(\bar{A}\cup B)=1\] \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow \] \[1-\frac{1}{3}P(B)+P(B)-0\] \[\Rightarrow \] \[1=\frac{4}{3}P(B)\Rightarrow P(B)=\frac{3}{4}\] \[\therefore \] \[P(A)=\frac{1}{3}P(B)=\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}\]
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