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

SQ2.1 Which bet?

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 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.

+

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 I mentioned in my hint come in handy).

+

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: \[P(any~match) = 1 - P(no~matches)\]
\[P(no~matches) = (\frac{364}{365})^{300}\]
And I can let R do the arithmetic: