options("scipen" = 9999, "digits" = 16)Again, no notes. Mostly because this chapter is super short. Just for good measure, I’ll do the mathematician thing and say. Recall Bayes’ Theorem:
\[P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}.\]
Q1
Kansas City, despite its name, sits on the border of two US states: Missouri and Kansas. The Kansas City metropolitan area consists of 15 counties, 9 in Missouri, and 6 in Kansas. The entire state of Kansas has 105 counties and Missouri has 114. Use Bayes’ theorem to calculate the probability that a relative who just moved to a country in the KC metropolitan area also lives in a county in Kansas. Make sure to show \(P(\text{Kansas})\) (assuming your relative lives in either Kansas or Missouri), \(P(\text{Kansas City Metropolitan Area}),\) and \(P(\text{Kansas City Metropolitan Area} \mid \text{Kansas})\).
Assuming that our relative is equally likely to have moved to any of the counties in the KC metropolitan area is an interesting choice, but is necessary for doing this problem. So we’ll assume that. Now, assuming that our relative has moved to either Kansas or Missouri, the overall probability that they are in Kansas is \[P(\text{Kansas}) = \frac{105}{105 + 114} = \frac{105}{219}.\] The probability that our relative lives in Kansas City, assuming they live in either Kansas or Missouri (and again, that all counties are equally likely) is \[P(\text{KC}) = \frac{15}{219}.\] The probability that our relative lives in Kansas City, if we already knew that they lived in Kansas, is \[P(\text{KC} \mid \text{Kansas}) = \frac{6/219}{105/219}=\frac{6}{105}.\] Then, by Bayes’ Theorem, we get that the probability that our relative lives in Kansas, given that they live in Kansas City, is \[P(\text{Kansas} \mid \text{KC}) = \frac{P(\text{KC} \mid \text{Kansas}) \ P(\text{Kansas})}{P(\text{KC})} = \frac{\frac{6}{105} \frac{105}{219}}{ \frac{15}{219}} = \frac{6/219}{15/219} = \frac{6}{15} = 40\%.\]
Q2
A deck of cards has 52 cards with suits that are either red or black. There are four aces in a deck of cards: two red and two black. You remove a red ace from the deck and shuffle the cards. Your friend pulls a black card. What is the probability that it is an ace?
I like this problem more because it’s like orbs in an urn. We don’t have to make any other assumptions, this is just good normal probabilities. Anyways.
The probability that our friend draws an ace from the remaining 51 cards is \[P(\text{ace}) = \frac{3}{51}.\] (I am deliberately chosing not to reduce this fraction, because I don’t have to and because the solution is easier to understand if the fractions aren’t reduced.) The probability that our friend draws a black card from the remaining 51 cards (26 of which are black and 25 of which are red) is \[P(\text{black}) = \frac{26}{51}.\] Now, if our friend draws an ace, the probability that it is black (given that we know it is an ace already) is \[P(\text{black} \mid \text{ace}) = \frac{P(\text{black and ace})}{ P(\text{ace})} = \frac{2 / 51}{3 / 51} = \frac{2}{3}.\] Now we can use Bayes’ theorem to get that \[P(\text{ace} \mid \text{black}) = \frac{P(\text{black} \mid \text{ace}) P(\text{black})}{P(\text{ace})} = \frac{\frac{2}{3} \ \frac{3}{51}}{ \frac{26}{51}} = \frac{2 / 51}{26 / 51} = \frac{2}{26} = \frac{1}{13}.\] Since all the black cards are still in the deck, once we know that our friend has drawn a black card, the probability that is an ace is the same as if they had drawn from the full deck (if we hadn’t removed a card). How neat!