fundamentals of probability

Fundamentals of Probability

Math 1680

Overview

Introduction Sets Properties of Probability Simple Sample Spaces “And” Statements “Or” Statements The Binomial Formula Summary

Introduction

Historically, probability was developed by gamblers– Wanted to increase their winnings (or at least

decrease their losses) In this course, we will focus on the gaming

applications of probability– Bear in mind that probability is also used in

finance, medicine, and genetics

Introduction

Because of the emphasis on games, it will be in everyone’s best interest to acquire a standard deck of cards and a couple of 6-sided dice to play with– Playing with these will sharpen your intuition

with the material to come – Very helpful to have when answering

homework problems

Introduction

A probability is a numerical value assigned to denote the likelihood that an event will occur

– Flipping Heads on a coin– Rolling a number divisible by 3 on a die

Intuitively, we can understand why an event that is guaranteed to occur should have a probability of 1 (or 100%)

– Conversely, an event that cannot occur should have probability 0 (or 0%)

To see how probabilities are assigned to nontrivial events, we need to develop a little machinery with sets

Sets

Consider rolling a single die– We can list out all of the possible outcomes in a sample

space• Denoted by S• S = {1,2,3,4,5,6}

– Events we are interested in can be symbolized as sets in this sample space

• The sample space above is itself an example of a set

– The numbers inside S are called the elements of S• Represent the outcomes of the experiment

Sets

A is a subset of S if every element of A is also an element of S

A S

SA

Sets

The complement of A (written AC) is exactly the opposite of A in S– AC occurs only if A does not

A SAC

Sets

The union of A with B is the set of outcomes contained in A or B put together– occurs if A occurs or B occurs (or both)

SBA

Sets

The intersection of A with B is the set of outcomes contained in both A and B– AB occurs if both A and B occur

S

AB

Sets

If AB is empty then we say A and B are disjoint – In terms of events, this means that A and B are mutually

exclusive• They can’t both happen

A SB

AB =

Sets

Consider rolling a fair 6-sided die.– Sample space S = {1,2,3,4,5,6}

If A is the event “roll an odd number,” write out A and AC

If A = {1,2,3,4} and B = {4,5,6}, write out AC, BC, AB, and AB

A = {1,3,5} AC = {2,4,6}

AC = {5,6} BC = {1,2,3} AB = {4} AB = S

Properties of Probability

A probability takes an event (in the form of a set) and assigns a number value between 0 an 1

Axioms of probability– P(S) = 1– If A and B are mutually exclusive, then

P(AB) = P(A) + P(B)

Properties of Probability

Complement rule– P(AC) = 1 – P(A)

If calculating P(A) looks unfriendly, take a moment and see if calculating P(AC) is any easier

– If so, you can use the complement rule to get the answer you want without doing it the hard way!

Simple Sample Spaces

Possible outcomes when rolling two dice

Each singular possibility is equally likely– This is a simple sample

space


In this case, the probability of rolling a total number is equal to the total number of ways to get that number, divided by the number of possible outcomes

– The probability that the dice total 4 is

This is because the outcomes are equally likely

083.12

1

36

3)4( P


What is the probability the dice total 7?

What is the probability that you roll doubles (the dice show the same number)?

1/6 or about 16.7%

1/6 or about 16.7%


We will often picture chance processes as a box model– Takes each possible outcome and makes a “ticket” out of it– Equally likely to draw any one ticket

In the dice context, rolling a single die could be modeled with the box

Since we are equally likely to draw any ticket, the probability of rolling any particular number is 1/6

1 2 3 4 5 6


If there is more than one of a type of ticket in the box, use superscripts to denote how many of that ticket type there are

The “sum of two dice” box model is

2 1 3 2 4 3 5 4 6 5 7 6 8 5 9 4 10 3 11 2 12 1

“And” Statements

Quite often, the probability of one outcome occurring is dependent (or conditional) on the outcome of a prior event

– If I deal two cards off of a well-shuffled standard deck, I’ll give you a dollar if the second card is the queen of spades (Q♠)

• If the first card hasn’t been turned over, what is the probability of winning $1?

• If I turn the first card over and it’s the ace of diamonds (A♦), now what is the probability of winning $1?

• If the first card is Q♠, what is the probability of winning $1?

1/52, or about 1.92%

1/51, or about 1.96%

0


In general, we say the conditional probability of A occurring given that B occurs is the probability that both A and B occur, divided by the probability that B occurs in the first place.

–

We can rewrite this equation to give ourselves a general rule for finding the probability that A and B occur

– Could be at once or in sequence, depending on the context– P(AB) = P(A|B)P(B)

)(

)()|(

BP

ABPBAP


In some cases, the occurrence of one event has no effect on the outcome another

– Flipping a coin or rolling a die, for example– Events like these are independent– Mathematically, A and B are independent if and only if

P(A|B) = P(A)• Substituting this into the rule for intersections in the previous

paragraph gives us a special case rule for “and” statements– If A and B are independent, then P(AB) = P(A)P(B)

• If A and B are independent, then so are their complements


If I deal 3 cards off of a well-shuffled standard deck…– What is the probability that the first card is an

ace, the second card is a jack, and the third card is another ace?

– What is the probability that all three cards are diamonds?

(4/52)(4/51)(3/50) 0.0036%

(13/52)(12/51)(11/50) 1.29%


Three dice are rolled. What is the probability that they come up…

– all aces

– no aces

– at least one ace

– not all aces

(1/6)3 0.46%

(5/6)3 57.87%

1 - (1/6)3 99.54%

1 - (5/6)3 42.13%

“Or” Statements

If A and B are mutually exclusive events, then the probability that at least one of the two (equivalent to one or the other or both) will happen is P(AB) = P(A) + P(B)

If A and B could both occur, then the probability that at least one of the two happens is P(AB) = P(A) + P(B) – P(AB)– Why is this?

“Or” Statements

If two dice are rolled, the probability that the red die is a 6 can be shown by counting the number of outcomes where the red die shows 6– Similar for green

“Or” Statements

Due to the overlap at double-6, the probability of rolling at least one 6 with two dice is 11/36

– P(1st or 2nd is 6) = P(1st is 6) + P(2nd is 6) – P(both 6’s)

= 1/6 + 1/6 – 1/36 = 11/36 When finding the probability at least one of two

events occurs, add the separate probabilities up and subtract off the probability of the intersection

– Since exclusive events have no intersection, this is why you can just add up the separate probabilities

“Or” Statements

A card is dealt off the top of a well-shuffled standard deck

– What is the chance of getting a heart or a spade?

– What is the chance of getting a face card or a seven?

– What is the chance of getting a heart or a seven?

13/52 + 13/52 = 26/52 = 50%

12/52 + 4/52 = 16/52 30.77%

13/52 + 4/52 – 1/52 = 16/52 30.77%

“Or” Statements

Memorize these properties, and use them to your advantage

Keyword Or And

Operation Add Multiply

Set Notation

AB AB (or A B)

General Rule

P(AB) = P(A) + P(B) - P(AB) P(AB) = P(A|B)P(B) = P(B|A)P(A)

Special Case, Rule

Mutually Exclusive,

P(AB) = P(A) + P(B)

Independent,

P(AB) = P(A) P(B)

The Binomial Formula

If I flip a fair coin 1 time, the possible outcomes are heads (H) and tails (T)– If I am interested in counting heads, the

possible outcomes are 1 (for heads) and 0 (for tails)

– If X = number of heads…• P(X = 0) = 1/2• P(X = 1) = 1/2


If I flip a fair coin 2 times, the possible outcomes are HH, HT, TH, and TT– If I am interested in counting heads, the

possible outcomes are 0, 1, and 2– If X = number of heads…

• P(X = 0) = 1/4• P(X = 1) = 2/4• P(X = 2) = 1/4


If I flip a fair coin 3 times, the possible outcomes are HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT

– If I am interested in counting heads, the possible outcomes are 0, 1, 2, and 3

– If X = number of heads…• P(X = 0) = 1/8• P(X = 1) = 3/8• P(X = 2) = 3/8• P(X = 3) = 1/8


In the 3-coin case, one way of arriving at P(X = 2) is to find P(HHT) and multiply it by the number of ways to shuffle the H’s and T’s around and still have 2 heads

– This works because each of the simple outcomes is equally likely

P(HHT) = P(H)P(H)P(T) = P(H)2P(T) = (1/2)2(1/2) = 1/8

There are 3 ways to shuffle 2 heads around 3 flips– HHT, HTH, and THH

Then P(X = 2) = 3(1/8) = 3/8


In general, the number of ways to shuffle k heads around n flips is given by the binomial coefficient

– Where x! = x(x-1)(x-2)…1• 0! = 1by definition

)!(!

!

knk

n

k

n


Special cases with the binomial coefficient

1!

!

!0!

!

n

n

n

n

n

n

1!

!

!!0

!

0

n

n

n

nnn

n

nn

n

nn

)!1(

])!1[(

)!1(!1

!

1

nn

nn

n

n

n

n

)!1(

])!1[(

!1)!1(

!

1


Use the binomial coefficient to find P(X = k)– If p = P(H)

• P(k H’s and n-k T’s) = P(k H’s)P(n-k T’s) = P(H)kP(T)n-k

= pk(1-p)n-k

– Multiply by the number of ways to shuffle the k heads around to get

– This is called the binomial formula

knk ppk

nkXP

)1()(


The binomial formula applies only under the following conditions– You play a sequence of independent games

• Coin flips, dice rolls, etc.

– You play n times– You are interesting in counting wins

• In particular, you want exactly k wins out of the n games


I roll a die 15 times and count the number of times I roll a 3 or a 4

– What is the probability that I roll a 3 or 4 exactly 9 times?

– What is the probability that I roll a 3 or 4 exactly 2 times?

– What is the probability that I roll a 3 or 4 no more than 1 time?

%23.2)6/4()6/2(9

15 69

%6)6/4()6/2(2

15 132

%94.1)6/4()6/2(1

15)6/4()6/2(

0

15 141150


I flip a coin 10 times– The coin is weighted so that the probability of getting

heads is 1/10 What is the probability of getting an even number

of heads?

What is the probability of getting an odd number of heads?

%37.55)9.0()1.0(10

10)9.0()1.0(

8

10)9.0()1.0(

6

10)9.0()1.0(

4

10)9.0()1.0(

2

10)9.0()1.0(

0

10 01028466482150

%63.44%37.55%100


We play a game where I roll a fair 6-sided die– Whatever number I roll, you flip a fair coin that

many times and count the number of heads Are the die rolls and number of heads from

the coin independent?

No


We play a game where I roll a fair 6-sided die– Whatever number I roll, you flip a fair coin that

many times and count the number of heads Suppose I roll a 3

– What is the probability of getting 2 heads?

%5.37)2/1()2/1(2

3 12


We play a game where I roll a fair 6-sided die– Whatever number I roll, you flip a fair coin that many

times and count the number of heads If I haven’t rolled yet, what is the probability of

getting 2 heads?

%78.25)2/1()2/1(2

6)6/1()2/1()2/1(

2

5)6/1()2/1()2/1(

2

4)6/1()2/1()2/1(

2

3)6/1()2/1()2/1(

2

2)6/1( 4232221202

Summary

A probability is a numerical value assigned to an event to quantify the likelihood of that event’s occurrence

– Probabilities are always between 0 and 1– A probability of 1 denotes a “sure thing”

Probabilities are usually determined by a combination of counting methods and use of formulae governing logical statements

– Or– And– Not

Summary

If a game’s outcomes are broken down into equally likely simple events, the probability a general event occurs is equal to the number of simple events satisfying the conditions, divided by the total possible number of simple events

Summary

If the probability of one event is determined in part by knowledge of another event, we say those events are conditional– Otherwise, the events are independent

If two events cannot occur simultaneously, we say the events are mutually exclusive

Summary

Most problems you will encounter have the following format

– An event is given that we are interested in finding the probability for

• This event will be some combination of unions, intersections, and complements of simpler events that we can calculate

– By using the rules and properties from this section, we can rewrite the probability we were initially interested in in terms of the simpler probabilities

• In a sense, you create a formula for each problem from the building blocks learned in this section

Summary

If you play a sequence of n independent games and count the number of wins, the binomial formula gives the probability of winning exactly k out of n games

fundamentals of probability

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