X-Git-Url: https://code.communitydata.science/stats_class_2020.git/blobdiff_plain/0dac693334fa3d4174e59b0eec1f3d6746cae3fd..efa6913590499d105277c2907cde2a96aa7bd51f:/psets/pset2-worked_solution.html diff --git a/psets/pset2-worked_solution.html b/psets/pset2-worked_solution.html index 7d426b0..8b9ac29 100644 --- a/psets/pset2-worked_solution.html +++ b/psets/pset2-worked_solution.html @@ -1814,8 +1814,8 @@ p + geom_point() </code></pre> <div id="sq2.1-which-bet" class="section level4"> <h4>SQ2.1 Which bet?</h4> <p>If you are willing to assume that birthdays in the class are independent (not a terrible assumption) and that birthdays are distributed randomly throughout the year (a terrible assumption, as it turns out!), you should take Bet #2. Hereâs a way to explain why:</p> -<p>Consider that 25 people can be combined into pairs <span class="math inline">\({25 \choose 2}\)</span> ways (which you can read âas 25 choose 2â), which is equal to <span class="math inline">\(\frac{25 \times 24}{2} = 300\)</span> (and that little calculation is where those <a href="https://en.wikipedia.org/wiki/Binomial_coefficient">binomial coefficients</a> I mentioned in my hint come in handy).</p> -<p>Generalizing the logic from part b of the textbook exercise last week problem, I have assumed that each of these possible pairings are independent and thus each one has a probability <span class="math inline">\(P = (\frac{364}{365})\)</span> of producing an <em>unmatched</em> set of birthdays.</p> +<p>Consider that 25 people can be combined into pairs <span class="math inline">\({25 \choose 2}\)</span> ways (which you can read as â25 choose 2â), which is equal to <span class="math inline">\(\frac{25 \times 24}{2} = 300\)</span> (and that little calculation is where those <a href="https://en.wikipedia.org/wiki/Binomial_coefficient">binomial coefficients</a> I mentioned in my hint come in handy).</p> +<p>Generalizing the logic from part b of the textbook exercise last week, I have assumed that each of these possible pairings are independent and thus each one has a probability <span class="math inline">\(P = (\frac{364}{365})\)</span> of producing an <em>unmatched</em> set of birthdays.</p> <p>Putting everything together, I can employ the multiplication rule from <em>OpenIntro</em> Ch. 3 and get the following: <span class="math display">\[P(any~match) = 1 - P(no~matches)\]</span><br /> <span class="math display">\[P(no~matches) = (\frac{364}{365})^{300}\]</span><br /> And I can let R do the arithmetic:</p>