#### Transcript C2_Math3033

MATH 3033 based on Dekking et al. A Modern Introduction to Probability and Statistics. 2007 Slides by Tim Birbeck Instructor Longin Jan Latecki C2: Outcomes, events, and probability 2.1 – Sample Spaces Sample Space: A sample space is a set whose elements describe the outcomes of the experiment in which we are interested. Example: If we ask arbitrary people on the street what month they were born, the following is an obvious sample space: {Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec}. 2.1 – Sample Spaces Permutation: the order in which n different objects can be placed. Example: If we have three envelopes and number them 1, 2, and 3, the following sample space consists of every different permutation we can make using all three envelopes: {123, 132, 213, 231, 312, 321} 2.2 – Events Event: A subset of the sample space Example: In the birthday experiment, if we ask for the outcomes that only involve the months with 31 days, we would have the following event: L {Jan, Mar, May, Jul, Aug, Oct, Dec}. 2.2 – Events We can use set operators to combine events: Name Union Intersection Compliment Definition C A B C A B C Ac C {x : x A x B} C {x : x A x B} C {x : x A} Example: In the birthday experiment, if we intersect the event R, where the month has the letter ‘r’ in it, and the event L, where the month has 31 days, we get the following: L R {Jan, Mar, Oct, Dec} 2.2 – Events Disjoint / Mutually Exclusive: Two events that have no outcomes in common. A∩B = ∅ Example: In the birthday experiment, the event L, all the birthdays with 31 days, and the event {Feb} are mutually exclusive. We say the event A implies event B if the outcomes of A also lie in B. A ⊂ B 2.2 – Events DeMorgan’s laws: For any two events A and B we have: (A ∪ B)c = Ac ∩ Bc and (A ∩ B)c = Ac ∪ B 2.3 – Probability Probability function: A probability function P on a finite sample space Ω assigns to each event A in Ω a number P(A) in [0,1] such that: (i) P(Ω) = 1, and (ii) P(A ∪ B) = P(A) + P(B) if A and B are disjoint. The number P(A) is called the probability that A occurs. Example: In an experiment where we flip a perfectly weighted coin and record whether the coin lands on heads or tails, we could define the probability function P such that: P({H}) = P({T}) =1/2 2.3 – Probability Formally, we should write P({T}) and not P(T) because a probability function works on events and not outcomes. However, in practice, we often drop the curly braces for a singleton set. If we consider an experiment that only has two outcomes, such as success or failure, one outcome has a probability p to occur where 0 < p < 1, and the other outcome has a probability of 1 - p to occur. 2.3 – Probability To assign probability to an event, we can use the additivity property. Example: Ω = {123, 132, 213, 231, 312, 321} P(213) = P(231) = 1/6 T = {213, 231} P(T) = P(213) + P(231) = 1/6 + 1/6 = 1/3 2.3 – Probability If two sets are not disjoint, we must use following rule to determine probability: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Example: Ω = {123, 132, 213, 231, 312, 321 P(213) = P(231) = P(123) =1/6 S = {123, 213} T = {213, 231} S ∩ T = {213} P(S ∪T) = [P(123) + P(213)] + [P(213) + P(231)] - P(213) = [1/3] + [1/3] – 1/6 = 1/2 2.4 – Products of sample spaces Usually, one experiment is not sufficient, so an experiment is performed several times. To get a sample space of multiple experiments, we use the cross product. Example: Ω = Ω1 × Ω2 = {(ω1, ω2) : ω1 ∈ Ω1, ω2 ∈ Ω2} where Ω1 is the sample space of the first experiment and Ω2 is the sample space of the second experiment. Example: If we flip a coin twice the sample space would be: Ω = {H, T}× {H, T} = {(H,H), (H, T), (T,H), (T,T)}. 2.4 – Products of sample spaces A certain experiment may have 2 outcomes: success or failure. If we perform this experiment n times and let 0 represent failure and 1 represent success, we have the following sample space: Ω = Ω1 × Ω2 × … × Ωn Where Ω1 = Ω2 = … = Ωn = {1, 0} 2.4 – Products of sample spaces Example: If a certain Dr. Latecki goes on 5 dates and the date is either “successful” or “unsuccessful,” we can model the event where Dr. Latecki is only successful on 1 of his 5 dates as: A = {(0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)} Assuming Dr. Letecki’s chance of being “successful” is p and each date’s chance of success is independent of the previous date: P(A) = 5 (1 − p)4 p1 Because: (1 – p) is the probability of being unsuccessful and this must happen 4 times. p is probability of being successful and this must happen 1 time. There are 5 outcomes in the event A. 2.5 – An infinite sample space A probability function on an infinite (or finite) sample space Ω assigns to each event A in Ω a number P(A) in [0, 1] such that (i) P(Ω) = 1, and (ii) P(A1 ∪ A2 ∪ A3 ∪ ・ ・ ・) = P(A1) + P(A2) + P(A3) + ・・・ If A1,A2,A3, . . . are disjoint events. 2.5 – An infinite sample space Example: If we flip a coin until it lands on heads, the outcome of the experiment could be the number of times the coin needed to be flipped until heads came up. The sample space for this experiment would be: Ω = {1, 2, 3, . . . } If the chance of landing on heads is p, the chance of landing on tails is 1-p. Therefore: P(1) = p. P(2) = (1 – p)1p. P(3) = (1-p)2p P(n) = (1-p)n-1p