CS322Week 10 - Monday

Last time

What did we talk about last time? More permutations Addition rule Inclusion and exclusion

Questions?

Logical warmup

A bundle of 120 wires has been laid underground between two telephone exchanges 10 miles apart

Unfortunately, it was discovered that the individual wires are not labeled

Visually, there is no way of knowing which wire is which, making connections at either end impossible

Your job is to label the wires at both ends Walking is your only transportation You have a battery and a light bulb to test continuity You have tape and a pen for labeling the wires

What is the shortest distance in miles you will need to walk to correctly identify and label each wire?

Exam 2 Post Mortem

CombinationsStudent Lecture

Combinations

Subsets of sets

How many different subsets of size r can you take out of a set of n items? Subset of size 3 out of a set of size 5? Subset of size 4 out of a set of size 5? Subset of size 5 out of a set of size 5? Subset of size 1 out of a set of size 5?

This is called an r-combination, written

r

n

Permutations and combinations

In r-permutations, the order matters In r-combinations, the order doesn't Thus, the number of r-combinations

is just the number of r-permutations divided by the possible orderings

)!(!!

!),(

rnrn

rrnP

r

n

Combinations example

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?

Poker examples

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?

r-combinations with repetitions

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

1 2 3 4 5

x xx x

r-combinations with repetitions

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

r

nr 1

Example

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?

Handy dandy guide to counting

This is a quick reminder of all the different ways you can count things:

Order Matters Order Doesn't Matter

Repetition Allowed nk

Repetition Not Allowed P(n,k)

k

nk 1

k

n

Binomial Matters

Pascal's Triangle

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

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

r

n

Pascal's Formula

Pascal's Triangle works because of Pascal's Formula:

We can easily show its truth:

r

n

r

n

r

n

1

1

)!1(!)!1(

)!1(!)1(!

)!1(!)1(!

)!1(!!

)!(!!

)!1()!1(!

1

rnrn

rnrnn

rnrrnn

rnrrn

rnrn

rnrn

r

n

r

n

Binomial Theorem

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

kknn

k

n bak

nba

0

)(

Binomial Example

Compute (1 – x)6 using the binomial theorem

More on Probability

Probability axioms

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(Ac) = 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

Union probability example

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

Expected value

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 a1, a2, a3, … an, each of which has probability p1, p2, p3, … pn, then the expected value is:

n

kkkpa

1

Expected value: Roulette

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

What is the expected value of a bet?

Conditional probability

Given that some event A has happened, the probability that some event B will happen is called conditional probability

This probability is:

)()(

)|(APBAP

ABP

Conditional probability example

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?

Bayes' Theorem

Let sample space S be a union of mutually disjoint events B1, B2, B3, … Bn

Let A be an event in S Let A and B1 through Bn have non-

zero probabilities For Bk where 1 ≤ k ≤ n

)()|(...)()|()()|()()|(

)|(2211 nn

kkk BPBAPBPBAPBPBAP

BPBAPABP

Applying Bayes' theorem

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 B1 be the event that the person actually has the disease

Let B2 be the event that the person does not have the disease

Apply Bayes' theorem

Independent events

If events A and B are events in a sample space S , then these events are independent if and only ifP(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)

)()()()(

)()()|(

BAPBPAPAPBAP

BPABP

Upcoming

Next time…

Keep reading Chapter 9 Finish probability

Reminders

Work on Homework 8 Due Friday