Transcript 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 6sided 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
–
–

Intuitively, we can understand why an event that is
guaranteed to occur should have a probability of 1 (or
100%)
–

Flipping Heads on a coin
Rolling a number divisible by 3 on a die
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
•
•
–
Events we are interested in can be symbolized as sets in
this sample space
•
–
Denoted by S
S = {1,2,3,4,5,6}
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
A S
Sets

The complement of A (written AC) is exactly
the opposite of A in S
–
AC occurs only if A does not
A
AC
S
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)
A B
S
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
AB
S
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
B
AB = 
S
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
A = {1,3,5}

AC = {2,4,6}
If A = {1,2,3,4} and B = {4,5,6}, write out
AC, BC, AB, and AB
AC = {5,6}
BC = {1,2,3}
AB = {4}
AB = 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(AB) = 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
Simple Sample Spaces

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
3
1
P ( 4) 

 .083
36 12

This is because the outcomes are equally likely
Simple Sample Spaces

What is the probability the dice total 7?
1/6 or about 16.7%

What is the probability that you roll doubles
(the dice show the same number)?
1/6 or about 16.7%
Simple Sample Spaces

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
1 2 3 4 5 6

Since we are equally likely to draw any ticket, the
probability of rolling any particular number is 1/6
Simple Sample Spaces
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?
1/52, or about 1.92%
•
If I turn the first card over and it’s the ace of diamonds (A♦), now
what is the probability of winning $1?
1/51, or about 1.96%
•
If the first card is Q♠, what is the probability of winning $1?
0
“And” Statements

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.
–

P( A | B) 
P( AB)
P( B)
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)
“And” Statements

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
“And” Statements

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?
(4/52)(4/51)(3/50)  0.0036%
–
What is the probability that all three cards are
diamonds?
(13/52)(12/51)(11/50)  1.29%
“And” Statements

Three dice are rolled. What is the probability that
they come up…
–
all aces (1/6)3  0.46%
–
no aces
–
at least one ace
–
not all aces 1 - (1/6)3  99.54%
(5/6)3  57.87%
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(AB) = P(A) + P(B)
 If A and B could both occur, then the
probability that at least one of the two
happens is P(AB) = 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?
13/52 + 13/52 = 26/52 = 50%
–
What is the chance of getting a face card or a seven?
12/52 + 4/52 = 16/52  30.77%
–
What is the chance of getting a heart or a seven?
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
AB
AB (or A  B)
General
Rule
P(AB) = P(A) + P(B) - P(AB)
P(AB) = P(A|B)P(B) =
P(B|A)P(A)
Special
Case, Rule
Mutually Exclusive,
P(AB) = 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
The Binomial Formula

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
The Binomial Formula

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
The Binomial Formula

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
–


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
–

This works because each of the simple outcomes is
equally likely
HHT, HTH, and THH
Then P(X = 2) = 3(1/8) = 3/8
The Binomial Formula

In general, the number of ways to shuffle k
heads around n flips is given by the
binomial coefficient
n
n!
  
 k  k!(n  k )!
–
Where x! = x(x-1)(x-2)…1
•
0! = 1by definition
The Binomial Formula

Special cases with the binomial coefficient
 n  n! n!
  
 1
 0  0!n! n!
 n
n!
n[(n  1)!]
  

n
(n  1)!
 1  1!(n  1)!
 n  n! n!
  
 1
 n  n!0! n!
 n 
n!
n[(n  1)!]

 

n
(n  1)!
 n  1 (n  1)!1!
The Binomial Formula

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
 n k
P( X  k )    p (1  p) nk
k 
–
This is called the binomial formula
The Binomial Formula

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
The Binomial Formula

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?
15
 (2 / 6)9 (4 / 6)6  2.23%
9
–
What is the probability that I roll a 3 or 4 exactly 2 times?
15
 (2 / 6) 2 (4 / 6)13  6%
2
–
What is the probability that I roll a 3 or 4 no more than 1 time?
15
15
0
15
 (2 / 6) (4 / 6)   (2 / 6)1 (4 / 6)14  1.94%
0
1
The Binomial Formula

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?
10
10
10
10
10
10
 (0.1)0 (0.9)15   (0.1) 2 (0.9)8   (0.1) 4 (0.9)6   (0.1)6 (0.9) 4   (0.1)8 (0.9) 2   (0.1)10 (0.9)0  55.37%
0
2
4
6
8
10

What is the probability of getting an odd number
of heads?
100 %  55.37%  44.63%
The Binomial Formula

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
The Binomial Formula

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?
 3
 (1 / 2) 2 (1 / 2)1  37.5%
 2
The Binomial Formula

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?
 2
 3
 4
 5
 6
(1 / 6) (1 / 2) 2 (1 / 2)0  (1 / 6) (1 / 2) 2 (1 / 2)1  (1 / 6) (1 / 2) 2 (1 / 2) 2  (1 / 6) (1 / 2) 2 (1 / 2)3  (1 / 6) (1 / 2) 2 (1 / 2) 4  25.78%
 2
 2
 2
 2
 2
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