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<quiz>
  <question type="stack">
    <name>
      <text>35.3.3.4 Exercise on Parallel circuits</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p></p><p>A circuit has four independent switches \(A, B, C\) and \(D\) wired in parallel as shown below.</p><p></p><p style="text-align: center;"><img src="@@PLUGINFILE@@/35.3.3%20Exercise.svg" class="img-responsive atto_image_button_text-bottom" alt="" width="191" height="84"></p><p style="text-align: center;"><b>Switches in Parallel</b></p><p></p><p>Current can only flow through the bank of switches if at least one is closed.&nbsp; The probabilities that \(A, B, C\) and \(D\)&nbsp; are closed are \(\displaystyle{{@aa@}, {@bb@}, {@cc@}}\) and \(\displaystyle{{@dd@}}\) respectively. Calculate the probability that current can flow through the bank of switches.</p><p>[[input:ans1]] [[validation:ans1]]</p><p>Hint: First consider the opposite case that the current does not
flow.</p>]]></text>
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    </questiontext>
    <generalfeedback format="html">
      <text><![CDATA[<p>Denoting the switches by \(A, B, C\)&nbsp; and \(D\) we have:</p><p>\[\begin{eqnarray*}\text{P}(A\cup B\cup C\cup D)&amp;=&amp;\text{P} (A)+\text{P}(B)+\text{P} (C)+\text{P} (D)-\text{P} (A\cap B)- \text{P} (B\cap C)\\&amp;-&amp; \text{P} (C\cap D- \text{P} (D\cap A)- \text{P} (A\cap C)-\text{P} (B\cap D)\\&amp;+&amp;\text{P} (A\cap B\cap C)+ \text{P} (B\cap C\cap D)+\text{P} (C\cap D\cap A)+\text{P} (D\cap A\cap B)\\&amp;-&amp; \text{P} (A\cap B\cap C\cap D)\end{eqnarray*}\]</p><p>Using the fact that the switches operate independently and substituting gives:</p><p>\[ \text{P} (A\cup B\cup C\cup D)=3-3.35+1.65-0.3024=0.9976\]</p><p>Hence, the probability that current can flow through the bank of switches is \(\displaystyle {@ta@}\).</p><p>Alternative solution:</p><p>Consider the opposite case where the current does <i>not</i> flow through, then all of the switches
need to be open. If \(A\), \(B\), \(C\) and \(D\) represent the event that the respective switch is closed, then the
event \(A’ \cap B’ \cap C’ \cap D’\) represents all four switches being closed. As each switch is
independent, we have:
\[
P(A’\cap B’\cap C’\cap D’) = P(A’)P(B’)P(C’)P(D’)
= {@naa@}\times{@nbb@}\times{@ncc@}\times{@ndd@}= \frac{3}{1250}. \]
Therefore the probability that the current flows is</p><p>\begin{eqnarray*}\text{P}(A\cup B\cup C\cup D)&amp;=&amp;1-\text{P}(A'\cap B'\cap C'\cap D')\\&amp;=&amp;1-&nbsp;\frac{3}{1250}=0.9976.\end{eqnarray*}</p>]]></text>
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    <penalty>0.1</penalty>
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    <idnumber/>
    <stackversion>
      <text>2021072200</text>
    </stackversion>
    <questionvariables>
      <text>listprobs:[9/10,8/10,7/10,6/10]
aa:rand(listprobs)
naa:1-aa
lista:delete(aa,listprobs)
bb:rand(lista)
nbb:1-bb
listb:delete(bb,lista)
cc:rand(listb)
ncc:1-cc
dd:delete(cc,listb)[1]
ndd:1-dd
ta: aa+bb+cc+dd-(aa*bb+aa*cc+aa*dd+bb*cc+bb*dd+cc*dd)+(aa*bb*cc+aa*bb*dd+aa*cc*dd+bb*cc*dd)-aa*bb*cc*dd</text>
    </questionvariables>
    <specificfeedback format="html">
      <text>[[feedback:prt1]]</text>
    </specificfeedback>
    <questionnote format="moodle_auto_format">
      <text>probabilities that \(A, B, C, D\)  are closed are \(\displaystyle{{@aa@}, {@bb@}, {@cc@}, {@dd@}}\) so \( P(flow) = {@ta@}\)</text>
    </questionnote>
    <questiondescription format="moodle_auto_format">
      <text/>
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    <questionsimplify>1</questionsimplify>
    <assumepositive>0</assumepositive>
    <assumereal>0</assumereal>
    <prtcorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
    </prtcorrect>
    <prtpartiallycorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:orange;"><i class="fa fa-adjust"></i></span> Your answer is partially correct.]]></text>
    </prtpartiallycorrect>
    <prtincorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:red;"><i class="fa fa-times"></i></span> Incorrect answer.]]></text>
    </prtincorrect>
    <decimals>.</decimals>
    <scientificnotation>*10</scientificnotation>
    <multiplicationsign>dot</multiplicationsign>
    <sqrtsign>1</sqrtsign>
    <complexno>i</complexno>
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      <type>algebraic</type>
      <tans>ta</tans>
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        <expectedscore>1.0000000</expectedscore>
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