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Lecture 10 – Introduction to
Probability
Topics
• Events, sample space, random variables
• Examples
• Probability distribution function
• Conditional probabilities
• Exponential distribution
• Poisson distribution
The Monte Hall Problem
There is $1,000,000 behind one door &
$0.00 behind the other two.
#1
#2
#3
You “reserve” a door (say #1) but it remains closed. Then,
Monte opens one of the other two doors (say #2).
The door Monte opens will be empty!
(Monte wants to keep the money.)
What’s the
best strategy?
1. Stay with the original door we chose?
2. Switch to the other unopened door?
We are interested in probability as it relates to
helping us make good or optimal decisions.
Another Example
Coins & Drawers
One drawer has 2 gold coins
One drawer has 1 gold & 1 silver coin
One drawer has 2 silver coins
You select a drawer at random and then randomly
select one coin
It turns out that your coin is gold
What is the probability that the other coin is gold ?
Probability Overview
The sample space, S
= set of all possible outcomes of an experiment
Events are subsets of the sample space.
An event occurs if any of its elements
occur when the experiment is run.
Two events A & B are mutually exclusive
if their intersection is null.
Elements of S are called realizations
outcomes, sample points, or scenarios
The choice of the sample space depends on the question
that you are trying to answer.
Roll Two Dice
Two possible sample spaces are:
S1 = { (1,1), (1,2), (1,3), …, (6,4), (6,5), (6,6) }
S2 = { 2, 3, 4, . . . , 11, 12 } (sum of the two values)
Examples of Events:
A1 = “the sum of the face values is 3”
Under S1 : A1 = { (1,2), (2,1) } ; Under S2 : A1 = { 3 }
A2 = “one die has value 3 & the other has value 1”
Under S1 : A2 = { (1,3), (3,1) } ; Under S2 : not an event
Probability
Intuitive: The proportion of time that an event occurs
when the same experiment is repeated
Mathematical: A probability measure P is defined on the
set of all events satisfying the following:
(1) P(A)  A  S
(2) P(S) = P(A1  A2  · · · ) = 1

(3) Mutually exclusive events Ai imply that
P(A1  A2  · · · ) = P(A1) + P(A2 ) + · · ·
_
_
(4) P(A) = 1 - P(A) where A = complement of A
(5) P(AB) = P(A) + P(B) - P(AB)
(6) If A & B are independent, then P(AB) = P(A)P(B)
(6) The conditional probability of event A given B is
P(AB)
P(A|B) =
P(B)
(7) If A & B are independent events then
P(A | B) = P(AB) =
P(B)
P(A)P(B)
= P(A).
P(B)
In this case B gives “no information” as to whether
A will occur.
Probability Calculations
Example: roll a balanced dice
Sample space : S = { (H), (T) }
Event H: head appears (H)
Event T: tail appears (T)
S = H  T and H  T = 
Intuitively: P(H) = P(T) = ½. How to prove it mathematically?
Proof:
According to (2)
According to (3)
Equal chance
1 = P(S) = P ( H  T )
= P (H) + P (T)
P (H) = P (T) = 1/2

Basic Conditional Probabilities
Example: You roll 2 dice and were told that the sum is
7. What is probability that the first die is 2?
Define
Event B: the sum of the two dice is 7
Event A: the first dice is 2
The conditional probability of event A given B is
P(AB)
P(A|B) =
P(B)
B = { (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) } and A  B = { (2,5) }
P(A | B) =
P (A  B )
P(B)
=
1/36
6/36
=
1
6
Computing Conditional Probabilities
You roll 2 dice and double your money if the sum is  8
and lose if the sum is  7
However, the rolls occur in sequence and
you do not get to observe the first roll.
You are simply told either “the value is  4” or “the value is 3”.
After being told this you get to place your bet.
A = event that you win = { (2,6), (3,5), (3,6), (4,4), (4,5), (4,6),
(5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6) }
B = you are told “ 4” after the first roll
= { (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3),
(5,4), (5,5), (5,6), (6,1),(6,2), (6,3), (6,4), (6,5), (6,6) }
_
The goal is to calculate: win = P(A|B) and lose = P(A|B).
P(A I B)
=
P(AB)
P(B)
A  B = { (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6),
(6,2), (6,3), (6,4), (6,5), (6,6) }
Each realization (i,j ) is equally likely
|AB|
P(A|B) =
36
|B|
=
12/36
18/36
= 2/3.
36
In a similar manner, we can show that
_
P(A | B)= 6/18 = 1/3
_
However, P(A|B) = 1 – P(A|B).
More Conditional Probability Calculations
P(A|B) = P(A  B) and P(B|A) = P(B  A)
P(B)
P(A)
P(A  B) = P(A|B)P(B) and P(B  A) = P(B|A)P(A)
This leads to Bayes’ Theorem
P(A|B) =
P(B|A)P(A)
P(B)
Example: Coins & Drawers
Drawer 1 has 2 gold coins
1
2
3
Drawer 2 has 1 gold & 1 silver coin
Drawer 3 has 2 silver coins
D1 = event that we pick drawer 1
G1 = event that the first coin we select is gold
P(D1|G1) = probability that both coins are gold
given that the first is gold
Coin & Drawer Computations
P(Di) = 1/3, i = 1, 2, 3
P(G1) = (1)(1/3) + (1/2)(1/3) + (0)(1/3)
P(D1|G1) = P(G1 | D1) P(D1)
P(G1)
=
(1)(1/3)
(1)(1/3) + (1/2)(1/3) + (0)(1/3)
= 2/3
Example: The Monte Hall Problem
#1
#2
#3
There is $1,000,000 behind one of the doors &
$0.00 behind the others.
Say you pick door #1 and then Monte opens one
of the other two doors.
If you don’t switch, P(win) = 1/3.
Optimal Strategy: Switch to the door that
remains closed; P(win) = 2/3.
Events:
D1 = you end up choosing door 1
D2 = you end up choosing door 2
D3 = you end up choosing door 3
L = prize is behind door 1
M = prize is behind door 2
R = prize is behind door 3
Event you win: W = (D1L) (D2 M)  (D3  R)
These are mutually exclusive so
P(W) = P(D1L) + P(D2 M) + P(D3  R)
= P(D1|L) P(L) + P(D2|M) P(M) + P(D3|R) P(R)
= (0) (1/3) + (1) (1/3) + (1) (1/3) = 2/3
Random Variables
R.V. is a real-valued function defined on the sample space S
Example: toss two dice,
Sample space: (1,1), . . . , (6,6)
Quantity of interest: sum of the two dice  2, 3, . . . ,12
RV: function that maps sample space
to the quantity of interest
Define : X  the RV, which is defined as the sum of 2 fair dice
P{X = 2} = P{ (1,1) } = 1/36
P(X = 3} = P{ (1,2), (2,1)} = 2/36
P{X = 4} = P{ (1,3), (2,2), (3,1) } = 3/36
...
P{X = 7} = P{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) } = 6/36
…
P{X = 12} = P{ (6,6) } = 1/36
Classification of Random Variables
Discrete random variables:
take finite or a countable num of possible values
Probability mass function (pmf):
Probability of an outcome, p(a) = P{X = a}
Examples: Bernoulli , Binomial, Geometric and Poisson
Continuous random variables:
takes uncountable number of possible values
Probability density function (pdf):
f(a) = P{X  a } =

a
-
f(x)dx
Examples: Uniform, Exponential, Gamma, Normal
Expected Value
Continuous RV, X:
E[X ] =

- x f(x)dx
Discrete random varible, X:
E[X ] = xS x p(x )
The Exponential Distribution
This is the most frequently used distribution for
modeling interarrival and service times in queueing
systems. In many applications, it is regarded as
doing a good job of balancing realism and
mathematical tractability.
• Time between customer arrivals at an ATM
• Time until a machine in a workcenter breaks
• Time between calls to a reservation system
Exponential Random Variable, T
Let  be parameter of the exponential distribution.
f(t ) =
{
e –t, t 0
0,
f
t<0
t
F(t ) = Pr{T  t ) =  e
– u
0
probability density
function (pdf)
du = 1 – e
– t
(t 0)
Expected value of T :
E[T ] =


0 t f(t )dt = 0 te –t dt
= 1/
Variance of T :
2
Var[T ] = E[T – 1/] =

0 (t – 1/)2e –tdt
= 1/
2
Memoryless Property of an
Exponential Random Variable
Pr{T  a + b | T  b } = Pr{T  a } a, b  0
This says that if we’ve already waited b units of time and the event
has not occurred (e.g., the next customer hasn’t arrived) then the
probability distribution governing the remaining time until its
occurrence is the same as it would be if the system were restarted.
That is, the process “forgets” that the event has not yet
occurred. This is often a good assumption when the time
of the next arrival is not “influenced” by the last arrival.
Proof:
P(A I B)
=
P(AB)
P(B)
P(T > a + b | T > b) =
P(T > a + b and T > b)
=
P(T > b)
–(a+b)
– ae – b
e
e
=
=
= e
e – b
e – b
= P(T > a)
P(T > a + b)
P(T > b)
– a
Using the Exponential Distribution
Calls arrive at an emergency hotline switchboard at a
rate of 3 per hour. It is now 8 AM.
(a) What is the probability that the first call arrives
after 8:30 AM?
(b) What is the probability that the first call arrives
between 8:15 and 8:45 AM?
(c) Given that no calls have arrived by 8:30 AM what
is the probability that one or more calls arrive by
9 AM?
Solutions
Let T = interarrival time random variable.
Assume T is exponentially distributed with  = 3, so
F(t) = P(T  t) = 1 – e
(a) P(T > ½) = e
–3t
–3(1/2)
3/4
and f(t) = 3e
–3t
= 0.223
(b) P(¼ < T < ¾) = 1/4 3e
-3u
du = F(¾) – F(¼)
= [ 1 - e-9/4] - [ 1 - e-3/4] = e-3/4 - e-9/4 = 0.367
(c) P(T  1 | T ½) = 1 - P(T > 1 | T ½)
= 1 - P(T  ½)  memoryless property
=1-e
–3(1/2) =
0.777
Relationship between Exponential Distribution
and the Poisson “Counting” Distribution
The exponential distribution is a continuous (time)
distribution that, in our setting, models the waiting time
until the next event (e,g., next customer arrival).
The Poisson distribution is a discrete (counting)
distribution which governs the total number of events
(e.g., arrivals) in the time interval (0, t ).
Fact: If inter-event times are independent exponential
RVs with rate  then the number of events that
occur in the interval (0, t ) is governed by a Poisson
distribution with rate . And vice-versa.
Poisson Distribution
Xt = # of events (e.g., arrivals) in the interval (0, t ) is
governed by the following probability mass function
n
Pr{ Xt = n } = (t) e
–t/n!
, n = 0, 1, . . .
E[ Xt ] = t
&
% length of interval
arrival rate
Note: Pr{ Xt = 0 } = Pr{ T > t } = e
– t
Example (revisited)
Calls arrive at an emergency hotline switchboard once
every 20 minutes on average. It is now 8:00 AM.
Use the Poisson distribution to answer the following
questions.
(a) What is the probability that the first call
arrives after 8:30 AM?
Solution: Xt ~ Poisson(3t ); X½ ~ Poisson(3/2),
where = # of arrivals in first ½ hour.
0
P(X½ = 0) = (3/2) e
–3/2
=e
–3/2
= 0.223
(b) What is the probability that the first call
arrives between 8:15 AM and 8:45 AM.
Solution: Let Y1 = #of arrivals in first 15 minutes
Y2 = # of arrivals between 8:15 and 8:45 AM
We must determine: Pr{ Y1 = 0 and Y2  1 }
However, Y1 ~ Poison(3/4), Y2 ~ Poisson(3/2)
and the events {Y1 = 0} and {Y2  1} are independent.
P(Y1 = 0 and Y2  1) = P(Y1 = 0)P(Y2  1) = P(Y1 = 0)[1 - P(Y2 = 0)]
0
(3/4)
e
=
0!
= 0.367
–3/4
0
(3/2)
e
[1
0!
–3/2
]=e
–3/4
-e
–9/4
(c) Given that no calls have arrived by 8:30 AM what’s the
probability that one or more calls arrive by 9:00 AM?
Solution: Pr{ Y2  1 | Y1 = 0 }, where
Y1 = #of arrivals between 8:00 and 8:30 AM
and
Y2 = # of arrivals between 8:30 and 9:00 AM
Y1 ~ Poisson(3/2), Y2 ~ Poisson(3/2)
Y1 and Y2 are independent because they are defined on
non-overlapping time intervals.
P(Y2  1 | Y1 = 0 ) = P(Y2  1) = 1 - P(Y2 = 0)
0
(3/2)
=1e
l
0.
–3/2
= 1-e
–3/2
= 0.777
Multiple Arrival Streams
If customers of type 1 arrive according to a Poisson
process with rate 1 and customers of type 2 arrive
according to an independent Poisson process with
rate 2 then customers arrive, regardless of type,
according to a Poisson process with rate  = 1 + 2.
Generalization: Let Ti ~ exp(i), i = 1,…,n
and define T = min{ T1,…,Tn }.
Then T ~ exp(), where  = 1 + 2 + · · · + n
Restated: The minimum of several independent
exponential rv’s is an exponential random variable
with a rate that is the sum of the individual rates.
What You Should Know About
Probability
• How to identify events and the
corresponding sample space.
• How to work with conditional probabilities.
• How to work with probability functions
(e.g., normal, exponential, uniform,
discrete).
• How to work with the Poisson distribution.