Lecture 10 – Introduction to Probability

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? • Stay with the original door we chose? • 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 silvercoin One drawer has 2 silvercoins 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 Thesample 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.

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 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)

P(A) ³ 0 " A  S • P(S) = P(A1È A2 È · · · ) = 1 • (3) Mutually exclusive events Aiimply that  P(A1È A2 È · · · ) = P(A1) + P(A2 ) + · · · _ _ (4) P(A) = 1 - P(A) where A = complement of A (5) P(AÈB) = 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 times that an event occurs when the same experiment is repeated. Mathematical: A probability measure P is defined on the set of all events and satisfies the following:

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

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

The conditional probability of event A given B is P (A B ) 1/36 1 P(AB) P(A | B) = = = P(A|B) = 6 P(B) 6/36 P(B) 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 B = { (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) } and A B = { (2,5) }

B = you are told “³ 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 You roll 2 dice and double your money if the sum is ³ 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 that “the value is ³ 4.” (event B) After being told this you get to place your bet. What is P(win)? Let event A = win. Need to define events A and B?) 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) }

P(A B) P(A B) = I P(B) |AB| 36 12/36 18/36 P(A|B) = = = 2/3. |B| 36 In a similar manner, we can show that _ P(A | B)= 6/18 = 1/3 _ However, P(A|B) = 1 – P(A|B). _ The goal is to calculate: win = P(A|B) and lose = P(A|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

1 Example: Coins & Drawers 2 3 Drawer 1 has 2 gold coins 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

(1)(1/3) = 2/3 = (1)(1/3) + (1/2)(1/3) + (0)(1/3) Coin & Drawer Computations P(Di) = 1/3, i = 1, 2, 3 P(G1) = (1)(1/3) + (1/2)(1/3) + (0)(1/3) P(G1 | D1) P(D1) P(G1) P(D1|G1) =

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.

These are mutually exclusive so P(W) = P(D1Ç L) + P(D2Ç M) + P(D3Ç 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 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 = (D1Ç L) È (D2Ç M) È (D3Ç R)

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= RV which is equal to 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

Continuous random variables: takes uncountable number of possible values Probability density function (pdf): f(a) = P{Xa} = ò- f(x)dx Examples: Uniform, Exponential, Gamma, Normal a 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 RV, X: E[X] = òx f(x)dx ¥ - Expected Value Discrete random variable, X: E[X] = xSx 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 fails • Time between calls to a reservation system

Expected value of T: E[T] = òtf(t)dt = òte –t dt = 1/ ¥ ¥ 0 0 Variance of T: Var[T] = E[T– 1/]2 = ò (t– 1/)2e –tdt = 1/2 ¥ 0 Exponential Random Variable, T Let  be parameter of the exponential distribution. e –t, t³ 0 0, t < 0 probability density function (pdf) f f(t ) = t F(t) = P{T£t) = ò e –u du = 1 – e –t (t³ 0) 0

Memoryless Property of an Exponential Random Variable P{Ta + b|Tb} = P{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.

P(AB) P(A B) = I P(B) P(T > a + b and T > b) P(T > a + b) P(T > a + b | T > b) = = P(T > b) P(T > b) Proof of Memoryless Property: P(T ≤ t) = 1 –P(T > t) = e –(a+b) =e –ae –b = e –a e –be –b = P(T > 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?

3/4 -3u = 3e du = F(¾) –F(¼) (b) P(¼ < T < ¾) ò 1/4 - - - - -3/4 -9/4 -9/4 -3/4 = e e = 0.367 = [ 1 e ] [ 1 e ] (c) P(T£ 1 | T³ ½) = 1 -P(T > 1 | T³ ½) ¬ memoryless property = 1 -P(T > ½) = 1 -e –3(1/2) = 0.777 Solutions for Exponential Distribution Problems 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) = 3e –3t (a) P(T > ½) = 1 –P(T£ ½) e –3(1/2) = 0.223

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.

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

Example (revisited) Calls arrive at anemergency hotline switchboard once every 20 minutes on average. It is now 8:00 AM. Use the Poisson distribution to answer the following questions. • What is the probability that the first call arrives after 8:30 AM? Solution: Xt ~ Poisson(3t); X½ ~ Poisson(3/2), where X½ = # of arrivals in first ½ hour. P(X½ = 0) = (3/2)0e –3/2/0! = e –3/2 = 0.223

P(Y1 = 0 and Y2³ 1) = P(Y1 = 0)P(Y2³ 1) = P(Y1 = 0)[1 -P(Y2= 0)] 0 0 (3/4) (3/2) e –3/4 e –3/2 ) = e –3/4 - e –9/4 - = (1 0! 0! = 0.367 (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: P{ Y1 = 0 and Y2³ 1 } However, Y1 ~ Poison(3/4), Y2 ~ Poisson(3/2) and the events {Y1 = 0} and {Y2³ 1} are independent.

P(Y2³ 1 | Y1 = 0 ) = P(Y2³ 1) = 1 -P(Y2 = 0) 0 (3/2) = 1 - e –3/2 = 1 - e –3/2 = 0.777 0. l (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: P{ Y2³ 1 | Y1 = 0 }, where Y1 = #of arrivals between 8:00 and 8:30 AM Y2 = # of arrivals between 8:30 and 9:00 AM and Y1 ~ Poisson(3/2), Y2 ~ Poisson(3/2) Y1 and Y2 are independent because they are defined on non-overlapping time intervals.

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.

Lecture 10 – Introduction to Probability