1. Week 11 - Wednesday

2.  What did we talk about last time?  Exam 2 post-mortem  Combinations

4.  This is a puzzle we should have done with sequences  Consider the following sequence, which should be read from left to right, starting at the top row 1 1 2 1 1 2 1 1 1 1 1 2 2 1  What are the next two rows in the sequence?

6.  How many ways are there to choose 5 people out of a group of 12?  What if two people don't get along? How many 5 person teams can you make from a group of 12 if those two people cannot both be on the team?

7.  How many five-card poker hands contain two pairs?  If a five-card hand is dealt at random from an ordinary deck of cards, what is the probability that the hand contains two pairs?

8.  What if you want to take r things out of a set of n things, but you are allowed to have repetitions?  Think of it as putting r things in n categories  Example: n = 5, r = 4  We could represent this as x||xx|x|  That's an r x's and n – 1 |'s 12345 xxxx

9.  So, we can think of taking an r-combination with repetitions as choosing r items in a string that is r + n – 1 long and marking those as x's  Consequently, the number of r-combinations with repetitions is

10.  Let's say you grab a handful of 10 Starbursts  Original Starbursts come in  Cherry  Lemon  Strawberry  Orange  How many different handfuls are possible?  How many possible handfuls will contain at least 3 cherry?

11.  This is a quick reminder of all the different ways you can count things: Order MattersOrder Doesn't Matter Repetition Allowed nknk Repetition Not Allowed P(n,k)P(n,k)

12. Student Lecture

14.  Hopefully, you are all familiar with Pascal's Triangle, the beginning of which is:  If we number rows and columns starting at 0, note that the value of row n, column r is exactly 1 11 121 1331 14641 1510 51 1615201561

15.  Pascal's Triangle works because of Pascal's Formula:  We can easily show its truth:

16.  a + b is called a binomial  Using combinations (or Pascal's Triangle) it is easy to compute (a + b) n  We could prove this by induction, but you probably don't care

17.  Compute (1 – x) 6 using the binomial theorem

19.  Let A and B be events in the sample space S  0 ≤ P(A) ≤ 1  P(  ) = 0 and P(S) = 1  If A  B = , then P(A  B) = P(A) + P(B)  It is clear then that P(A c ) = 1 – P(A)  More generally, P(A  B) = P(A) + P(B) – P(A  B)  All of these axioms can be derived from set theory and the definition of probability

20.  What is the probability that a card drawn randomly from an Anglo-American 52 card deck is a face card (jack, queen, or king) or is red (hearts or diamonds)?  Hint:  Compute the probability that it is a face card  Compute the probability that it is red  Compute the probability that it is both

21.  Expected value is one of the most important concepts in probability, especially if you want to gamble  The expected value is simply the sum of all events, weighted by their probabilities  If you have n outcomes with real number values a 1, a 2, a 3, … a n, each of which has probability p 1, p 2, p 3, … p n, then the expected value is:

22.  A normal American roulette wheel has 38 numbers: 1 through 36, 0, and 00  18 numbers are red, 18 numbers are black, and 0 and 00 are green  The best strategy you can have is always betting on black (or red)  If you bet $1 on black and win, you get $1, but you lose your dollar if it lands red or green  What is the expected value of a bet?

23.  Given that some event A has happened, the probability that some event B will happen is called conditional probability  This probability is:

24.  Given two, fair, 6-sided dice, what is the probability that the sum of the numbers they show when rolled is 8, given that both of the numbers are even?

25.  Let sample space S be a union of mutually disjoint events B 1, B 2, B 3, … B n  Let A be an event in S  Let A and B 1 through B n have non-zero probabilities  For B k where 1 ≤ k ≤ n

26.  Bayes' theorem is often used to evaluate tests that can have false positives and false negatives  Consider a test for a disease that 1 in 5000 people have  The false positive rate is 3%  The false negative rate is 1%  What's the probability that a person who tests positive for the disease has the disease?  Let A be the event that the person tests positively for the disease  Let B 1 be the event that the person actually has the disease  Let B 2 be the event that the person does not have the disease  Apply Bayes' theorem

27.  If events A and B are events in a sample space S, then these events are independent if and only if P(A  B) = P(A)∙P(B)  This should be clear from conditional probability  If A and B are independent, then P(B|A) = P(B)

30.  Finish probability  Graph basics

31.  Work on Homework 8  Due Friday  Start reading Chapter 10