E1,E2 are events of a sample space such that P(E1)=14 , P(E2E1) =12 , P(E1E2) =14 Then P(E1E2¯) =

P(E1∩E2)=P(E1)P(E2)=14.12=18P(E2)=P(E1∩E2)P(E1)=1814=12P(E1/E2¯)=P(E1∩E2¯)P(E2¯)=P(E1)−P(E1∩E2)1−P(E2)=14

E1,E2 are events of a sample space such that P(E1)=14 , P(E2E1) =12 , P(E1E2) =14 Then P(E1E2¯) =

P(E1∩E2)=P(E1)P(E2)=14.12=18P(E2)=P(E1∩E2)P(E1)=1814=12P(E1/E2¯)=P(E1∩E2¯)P(E2¯)=P(E1)−P(E1∩E2)1−P(E2)=14

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$E_{1}, E_{2} a r e e v e n t s o f a s a m p l e s p a c e s u c h t h a t P (E_{1}) = \frac{1}{4}, P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}, P (\frac{E_{1}}{E_{2}}) = \frac{1}{4}$
$T h e n P (\frac{E_{1}}{\bar{E_{2}}}) =$

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$\frac{3}{4}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} P (E_{1} \cap E_{2}) = P (E_{1}) P (E_{2}) = \frac{1}{4} . \frac{1}{2} = \frac{1}{8} \\ P (E_{2}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2} \\ P (E_{1} / \bar{E_{2}}) = \frac{P (E_{1} \cap \bar{E_{2}})}{P (\bar{E_{2}})} = \frac{P (E_{1}) - P (E_{1} \cap E_{2})}{1 - P (E_{2})} = \frac{1}{4} \end{array}$