X-Git-Url: https://code.communitydata.science/stats_class_2020.git/blobdiff_plain/0dac693334fa3d4174e59b0eec1f3d6746cae3fd..refs/heads/master:/psets/pset2-worked_solution.rmd?ds=sidebyside

diff --git a/psets/pset2-worked_solution.rmd b/psets/pset2-worked_solution.rmd
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--- a/psets/pset2-worked_solution.rmd
+++ b/psets/pset2-worked_solution.rmd
@@ -245,9 +245,9 @@ The plot reveals a very strong positive relationship between average daily dista
 
 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. Here's a way to explain why: 
 
-Consider that 25 people can be combined into pairs ${25 \choose 2}$ ways (which you can read "as 25 choose 2"), which is equal to $\frac{25 \times 24}{2} = 300$ (and that little calculation is where those [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) I mentioned in my hint come in handy). 
+Consider that 25 people can be combined into pairs ${25 \choose 2}$ ways (which you can read as "25 choose 2"), which is equal to $\frac{25 \times 24}{2} = 300$ (and that little calculation is where those [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) I mentioned in my hint come in handy). 
 
-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 $P = (\frac{364}{365})$ of producing an *unmatched* set of birthdays.
+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 $P = (\frac{364}{365})$ of producing an *unmatched* set of birthdays.
 
 Putting everything together, I can employ the multiplication rule from *OpenIntro* Ch. 3 and get the following:
 $$P(any~match) = 1 - P(no~matches)$$