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    <name>
      <text>(not in quiz) 37.2.3.8 Number of students who like maths</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p><p> In a large school, {@percent@}% of the pupils like mathematics. A visitor to the school asks each of {@n_1@} pupils, chosen at random, whether they like mathematics.</p><ol class="HELM_parts">

<li><p>Calculate the probabilities of obtaining an answer yes from \(2\) of the pupils [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p></li>

<li><p>Find the probability that the visitor obtains the answer yes from at least \(2\) pupils: [[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p></li>

<li><p>Find the probability that the visitor obtains the answer yes from at least \(2\) pupils when the number of pupils questioned is increased to {@n_2@}: [[input:ans3]] [[validation:ans3]] [[feedback:prt3]]</p></li>

</ol>

<p><b>Note</b>: please round your answer up to \(4\) decimal places.</p>]]></text>
    </questiontext>
    <generalfeedback format="html">
      <text><![CDATA[<p><b>SOLUTION:</b></p><p>If \(X\) is the number of pupils who like maths, then \(X\) follows a binomial distribution with \(p=\dfrac{{@percent@}}{100}\)</p><ol class="HELM_parts">
  <li><p>In this case, \(n={@n_1@}\) so we have that:</p><p>\(\text{P}(X=2) =&nbsp;\&nbsp; ^{@n_1@}\!C_2 \left( \dfrac{{@percent@}}{100} \right)^2\left(1-\dfrac{{@percent@}}{100}\right)^{@n_1 - 2@} = {@ta_1@} \)</p></li>
  <li><p>Again, in this case&nbsp;\(n={@n_1@}\) so we have that:</p><p> \(\begin{align} \text{P}(X\geq 2) &amp;= 1-&nbsp;\text{P}(X = 0) -&nbsp;\text{P}(X = 1) \\ &amp;= 1 - \&nbsp; ^{@n_1@}\!C_0 \left( \dfrac{{@percent@}}{100} \right)^0\left(1-\dfrac{{@percent@}}{100}\right)^{@n_1@}&nbsp;\&nbsp; ^{@n_1@}\!C_1 \left( \dfrac{{@percent@}}{100} \right) \left(1-\dfrac{{@percent@}}{100}\right)^{@n_1 - 1@} \\&nbsp;&nbsp;&amp;= {@ta_2@} \end{align} \).</p></li>
  <li>Now, in this case, \(n={@n_2@}\) so we have that:<p></p><p>\(\begin{align} \text{P}(X\geq 2) &amp;= 1-&nbsp;\text{P}(X = 0) -&nbsp;\text{P}(X = 1) \\ &amp;= 1 -&nbsp;\&nbsp; ^{@n_2@}\!C_0 \left( \dfrac{{@percent@}}{100} \right)^0\left(1-\dfrac{{@percent@}}{100}\right)^{@n_2@} - \&nbsp; ^{@n_2@}\!C_1 \left( \dfrac{{@percent@}}{100} \right) \left(1-\dfrac{{@percent@}}{100}\right)^{@n_2 - 1@} \\&nbsp;&nbsp;&amp;= {@ta_3@} \end{align} \).</p></li></ol>]]></text>
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    <stackversion>
      <text>2020070100</text>
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    <questionvariables>
      <text>percent: 60 + 10*rand(3)
n_1: 4 + rand(2)
n_2: 7 + rand(2)
ta_1: decimalplaces(0.5*n_1*(n_1 -1)*((percent/100)^2)*(((100-percent)/100)^(n_1 -2)),4)
ta_2: decimalplaces(1 - ((100-percent)/100)^(n_1) - n_1*(percent/100)*(((100-percent)/100)^(n_1 -1)),4)
ta_3: decimalplaces(1 - ((100-percent)/100)^(n_2) - n_2*(percent/100)*(((100-percent)/100)^(n_2 -1)),4)</text>
    </questionvariables>
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      <text/>
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    <questionnote format="moodle_auto_format">
      <text>\(\text{P}(X=2) = {@ta_1@}\), \( \text{P}(X\geq 2) \) when n={@n_1@} = {@ta_2@}, \( \text{P}(X\geq 2) \) when n={@n_2@} = {@ta_3@}.</text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
    </prtcorrect>
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      <text><![CDATA[<span style="font-size: 1.5em; color:orange;"><i class="fa fa-adjust"></i></span> Your answer is partially correct.]]></text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:red;"><i class="fa fa-times"></i></span> Incorrect answer.]]></text>
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