$\begin{array}{1 1} \frac{5}{9} \\ \frac{4}{9} \\ \frac{1}{3} \\ \frac{2}{3} \end{array} $

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- In any problem that involves determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
- Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

Given, the total number of questions = 300 + 200 + 500 + 400 = 1400.

Let E be the event that it is an easy question. Then, the number of easy questions we can select = 300 + 500 = 800.

\(\Rightarrow \;P(E)=\large \frac{800}{1400}=\frac{4}{7}\)

Let D be the event that it is an difficult question. Then, the number of difficult questions we can select = 200 + 400 = 600, i.e, \(200\;in\;T/F\;and\;400\;in \) multiple choice.

\(\Rightarrow \;P(D)=\large \frac{600}{1400}\) = $\large\frac{3}{7}$

Let T be the event that it is an T/F question. Then, the number of T/F questions we can select = 300 + 200 = 500, i.e, \(300\;\)easy and\( \;200\;\)difficult.

\(\Rightarrow \;P(T)=\large \frac{500}{1400}=\frac{5}{14}\)

=\(500\)

Let M be the event that it is an multiple choice question. Then, the number of difficult questions we can select =500 + 400 = 900, i.e, 500 easy and 400 difficult.

\(\Rightarrow \;P(M)=\large \frac{900}{1400}=\frac{9}{14}\)

\(\Rightarrow P(E\cap\;M)=\)P(easy and also multiple choice) = 500.$

\(\Rightarrow \;P(E\cap\;M)=\large \frac{500}{1400} = \frac{5}{14}\)

Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

\(\Rightarrow \;P(E/M)=\Large \frac{\frac{5}{14}}{\frac{9}{14}}\)\(=\large \frac{5}{9}\)

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