Lecture 10 – Introduction to Probability Topics Events, sample space, random variables Examples Probability distribution function Conditional probabilities Exponential distribution Poisson distribution

There is $1,000,000 behind one door & $0.00 behind the other two. 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.) #1#2#3 The Monte Hall Problem

1. Stay with the original door we chose? 2. Switch to the other unopened door? What’s the best strategy? We are interested in probability as it relates to helping us make good or optimal decisions.

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 ? Another Example

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.

Two possible sample spaces are: S 1 = { (1,1), (1,2), (1,3), …, (6,4), (6,5), (6,6) } S 2 = { 2, 3, 4,..., 11, 12 } (sum of the two values) Examples of Events: A 1 = “the sum of the face values is 3” Under S 1 : A 1 = { (1,2), (2,1) } ; Under S 2 : A 1 = { 3 } A 2 = “one die has value 3 & the other has value 1” Roll Two Dice Under S 1 : A 2 = { (1,3), (3,1) } ; Under S 2 : not an event

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(A 1  A 2  · · · ) = 1 (3) Mutually exclusive events A i imply that P(A 1  A 2  · · · ) = P(A 1 ) + P(A 2 ) + · · · (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) _ _ Probability Intuitive:The proportion of time that an event occurs when the same experiment is repeated 

(7) If A & B are independent events then P(A | B) = P(A  B) P(B) = P(A)P(B) P(B) In this case B gives “no information” as to whether A will occur. = P(A). (6) The conditional probability of event A given B is P(A|B) = P(A  B) P(B)

Probability Calculations Sample space : S = { (H), (T) } Example: roll a balanced dice Event H: head appears (H) Event T: tail appears (T) S = H  T and H  T =  = P (H) + P (T) Equal chance P (H) = P (T) = 1/2 1 = P(S) = P ( H  T ) According to (2) According to (3) Proof: Intuitively: P(H) = P(T) = ½.How to prove it mathematically? 

Basic Conditional Probabilities The conditional probability of event A given B is P(A|B) = P(A  B) P(B) Example:You roll 2 dice and were told that the sum is 7. What is probability that the first die is 2? 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 = Define Event B: the sum of the two dice is 7 Event A: the first dice is 2

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) } Computing Conditional Probabilities

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 P(A|B) = |A  B| |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(B) and P(B|A) = P(B  A) 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) P(A|B) =

Example: Coins & Drawers Drawer 1 has 2 gold coins Drawer 2 has 1 gold & 1 silver coin Drawer 3 has 2 silver coins D 1 = event that we pick drawer 1 G 1 = event that the first coin we select is gold P(D 1 |G 1 ) = probability that both coins are gold given that the first is gold

Coin & Drawer Computations = (1)(1/3) (1)(1/3) + (1/2)(1/3) + (0)(1/3) = 2/3 P(D i ) = 1/3, i = 1, 2, 3 P(G 1 ) = (1)(1/3) + (1/2)(1/3) + (0)(1/3) P(D 1 |G 1 ) = P(G 1 | D 1 ) P(D 1 ) P(G 1 )

Example : The Monte Hall Problem There is $1,000,000 behind one of the doors & $0.00 behind the others. #1#2#3 If you don’t switch, P(win) = 1/3. Optimal Strategy:Switch to the door that remains closed; P(win) = 2/3. Say you pick door #1 and then Monte opens one of the other two doors.

Events:D 1 = you end up choosing door 1 D 2 = you end up choosing door 2 D 3 = 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 = (D 1  L)  (D 2  M)  (D 3  R) These are mutually exclusive so P(W) = P(D 1  L) + P(D 2  M) + P(D 3  R) = P(D 1 |L) P(L) + P(D 2 |M) P(M) + P(D 3 |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 Examples: Bernoulli, Binomial, Geometric and Poisson Probability mass function (pmf): Probability of an outcome, p(a) = P{ X = a} Continuous random variables: takes uncountable number of possible values Probability density function (pdf) : f ( a ) = P{ X  a } =  -  f(x) d x Gamma, Normal Examples: Uniform, Exponential, Gamma, Normal a

Expected Value Continuous RV, X : E [ X ] =  x f ( x )d x --  Discrete random varible, X : E [ X ] =  x  S x p ( x )

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. The Exponential Distribution Time between customer arrivals at an ATM Time until a machine in a workcenter breaks Time between calls to a reservation system

F ( t ) = Pr{ T  t ) =  0 t e – u d u = 1 – e – t ( t  0) f ( t ) = { e – t, t  0 0, t < 0 probability density function (pdf)  Exponential Random Variable, T Let be parameter of the exponential distribution. 0  Variance of T : Var[ T ] = E[ T – 1/ ] 2 =  ( t – 1/ ) 2 e – t d t = 1/ 2 Expected value of T : E [ T ] =  t f ( t )d t =  t e – t d t = 1/ 0  0 

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 ) P( T > a + b | T > b ) = P( T > b ) = P( T > a + b and T > b )P( T > a + b ) P( T > b ) = e – (a+b) = e – a e – b = e – a e – b e – b

Using the Exponential Distribution Calls arrive at an emergency hotline switchboard at a rate of 3 per hour. It is now 8 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? (b) What is the probability that the first call arrives between 8:15 and 8:45 AM? (a) What is the probability that the first call arrives after 8:30 AM?

Let T = interarrival time random variable. Assume T is exponentially distributed with = 3, so F(t) = P(T  t) = 1 – e – 3t and f(t) = 3 e – 3t (a) P( T > ½) = e – 3(1/2) = (b) P(¼ < T < ¾) = 3e -3u du = F(¾) –  F(¼)  1/4 3/4 = [ 1  e -9/4 ]  [ 1  e -3/4 ] = e -3/4  e -9/4 = (c) P( T  1 | T  ½) = 1  P( T > 1 | T  ½  = 1  P(T  ½)  memoryless property = 1  e – 3(1/2) = Solutions

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.

X t =# of events (e.g., arrivals) in the interval (0, t ) is governed by the following probability mass function Note: Pr{ X t = 0 } = Pr{ T > t } = e – t Pr{ X t = n } = ( t ) n e – t / n !, n = 0, 1,... E[ X t ] = t  arrival rate  length of interval Poisson Distribution

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: X t ~ Poisson(3 t ); X ½ ~ Poisson(3/2), where = # of arrivals in first ½ hour. P( X ½ = 0) = (3/2) 0 e –3/2 = e –3/2 = 0.223

Solution: Let Y 1 = #of arrivals in first 15 minutes Y 2 = # of arrivals between 8:15 and 8:45 AM We must determine: Pr{ Y 1 = 0 and Y 2  1 } However, Y 1 ~ Poison(3/4), Y 2 ~ Poisson(3/2) and the events { Y 1 = 0} and { Y 2  1} are independent. (b) What is the probability that the first call arrives between 8:15 AM and 8:45 AM. P( Y 1 = 0 and Y 2  1) = P( Y 1 = 0)P( Y 2  1) = P( Y 1 = 0) [ 1  P( Y 2  0) ] = (3/4) 0 0! [1  (3/2) 0 0! e –3/4 e –3/2 ] = e –3/4  e –9/4 = 0.367

Solution: Pr{ Y 2  1 | Y 1 = 0 }, where Y 1 = #of arrivals between 8:00 and 8:30 AM Y 2 = # of arrivals between 8:30 and 9:00 AM and Y 1 ~ Poisson(3/2), Y 2 ~ Poisson(3/2) Y 1 and Y 2 are independent because they are defined on non-overlapping time intervals. (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? = 1  (3/2) 0 0. l e –3/2 = 1  e –3/2 = P( Y 2  1 | Y 1 = 0 ) = P( Y 2  1) = 1  P( Y 2 = 0)

Generalization:Let T i ~ exp( i ), i = 1,…, n and define T = min{ T 1,…, T n }. Then T ~ exp( ), where = · · · + n 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 = 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.