A) 0 or 1
B) \[\frac{1}{2}\] or \[\frac{1}{3}\]
C) \[\frac{1}{2}\] or \[\frac{1}{4}\]
D) \[\frac{1}{3}\] or \[\frac{1}{4}\]
Correct Answer: B
Solution :
\[P(A\cap B)=\frac{1}{6}\] and \[P({{A}^{c}}\cap {{B}^{c}})=\frac{1}{3}\] Now \[P{{(A\cup B)}^{c}}=P({{A}^{c}}\cap {{B}^{c}})=\frac{1}{3}\] Þ \[1-P(A\cup B)=\frac{1}{3}\]\[\Rightarrow \,P(A\cup B)=\frac{2}{3}\] But \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow P(A)\,+P(B)=\frac{5}{6}\] ?..(i) \[\because \] A and B are independent events \ \[P(A\cap B)\,=\,P(A)\,P(B)\] Þ \[P(A)\,P(B)=\frac{1}{6}\] \[{{[P(A)\,-\,P(B)\,]}^{2}}={{[P(A)+P(B)]}^{2}}-4P(A)\,P(B)\] \[=\frac{25}{36}-\frac{4}{6}=\frac{1}{36}\] \[\Rightarrow \] \[P(A)-P(B)=\pm \,\frac{1}{6}\] ......(ii) Solving (i) and (ii), we get \[P(A)=\frac{1}{2}\] or \[\frac{1}{3}.\]
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