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# An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

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

Toolbox:
• 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}$
edited Jun 18, 2013