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Let E_{1} and E_{2} be two events such that the conditional probabilities \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}_{2}\right)=\frac{1}{2},P\left(E_{2} \mid E_{1}\right)=\frac{3}{4} \text { and } P\left(E_{1} \cap E_{2}\right)=\frac{1}{8}. Then:

Option: 1

P\left(E_{1} \cap E_{2}\right)=P\left(E_{1}\right) \cdot P\left(E_{2}\right)


Option: 2

\mathrm{P}\left(\mathrm{E}_{1}^{\prime} \cap \mathrm{E}_{2}^{\prime}\right)=\mathrm{P}\left(\mathrm{E}_{1}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{E}_{2}\right)


Option: 3

P\left(E_{1} \cap E_{2}^{\prime}\right)=P\left(E_{1}\right) \cdot P\left(E_{2}\right)


Option: 4

P\left(E_{1}^{\prime} \cap E_{2}\right)=P\left(E_{1}\right) \cdot P\left(E_{2}\right)


Answers (1)

\mathrm{\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{2}\right)}=\frac{1}{2}, \frac{P\left(E_{2} \cap E_{1}\right)}{P\left(E_{1}\right)}=\frac{3}{4}}
\mathrm{P\left(E_{1} \cap E_{2}\right)=\frac{1}{8}}

\mathrm{\Rightarrow P\left(E_{2}\right)=\frac{1}{4}, P\left(E_{1}\right)=\frac{1}{6}, P\left(E_{1} \cup E_{2}\right)=\frac{1}{6}+\frac{1}{4}-\frac{1}{8}=\frac{7}{24} }

Clearly, option (A) is wrong

\mathrm{{P}\left(E_{1}^{\prime} \cap E_{2}^{\prime}\right)=P\left(E_{1} \cup E_{2}\right)^{\prime}=1-\frac{7}{24}=\frac{17}{24} }

So option B is also wrong

P(E_1 \cap E_2\,')= P(E_1)- P(E_1 \cap E_2)=\frac{1}{24}= P(E_1).P(E_2)

Option C is correct

Posted by

Kshitij

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