Probability definitions, axioms and theorems CEE 3030 A LECTURE PREPARED BY Classical approach to probability Sample space S is known Let n = number of total outcomes in sample space Let h = number of outcomes in event A Let P(A) = probability of event A Gilberto E. Urroz January 2006 P A = h n Example 1 - Throwing a coin Sample space S = { H, T }, event A = head = {H} In this case n and h = 1, thus P A = h n = 1 2 =0.5 If interested in the number of tails, S = {0, 1}, then: P(0) = ½ and P(1) = ½ Example 2 - Throwing a single die Sample space S = { 1, 2, 3, 4, 5, 6 }, thus n = 6 Define events: A = getting an even number = { 2, 4, 6 }, h A = 3 B = getting an odd number = { 1, 3, 5 }, h B = 3 C = getting an multiple of 3 = { 3, 6 }, h C The probabilities are: P(A) = h A / n = 3/6 = ½ = 0.5 P(B) = h B / n = 3/6 = ½ = 0.5 P(C) = h C / n /6 = 1/3 = 0.333... Example 3 - Throwing two coins Sample space S = {(H,H), (H,T), (T,H), (T,T)}, thus n = 4 P[(H,H)] =... = P[(T,T)] = ¼ Example 4 - Throwing two dice: A = 1T = {(H,T),(T,H)}, h A B H = {(H,H)}, h B = 1 C T = {(T,T)}, h C = 1 D T or 2H = {(T,T),(H,H)}, h D P(A) = h A /n /4 = 0.50 P(B) = h B /n = 1/4 = 0.25 P(C) = h C /n = 1/4 = 0.25 P(D) = h D /n /4 = 0.50
Example 4 (continued)- Throwing two dice: h 2 = h 12 = 1, h 3 = h 11, h 4 = h 10 = 3, h 5 = h 9 = 4, h 6 = h 8 = 5, h 7 = 6. Also, n = 6 x 6 = 36. Example 4 (continued)- Throwing two dice: h 2 = h 12 = 1, h 3 = h 11, h 4 = h 10 = 3, h 5 = h 9 = 4, h 6 = h 8 = 5, h 7 = 6. Also, n = 6 x 6 = 36. Probabilities: P(2) = P(12) = 1/36 = 0.02777 P(3) = P(11) /36 = 1/18 = 0.05555 P(4) = P(10) = 3/36 = 1/12 = 0.08333 P(5) = P(9) = 4/36 = 1/9 = 0.1111 P(6) = P(8) = 5/36 = 0.13888 P(7) = 6/36 = 1/6 = 0.1666. Example 4 (continued) Throwing two dice: Probabilities: Example 5 Computer room: Select up to 3 Macintosh and up to 2 PCs Sample space S = {(1M,1P), (1M,2P), (2M,1P), (2M,2P), (3M,1P), (3M,2P)} P(2) = P(12) = 0.02777 P(3) = P(11) = 0.05555 P(4) = P(10) = 0.08333 P(5) = P(9) = 0.1111 P(6) = P(8) = 0.13888 P(7) = 0.1666. Results summarized next... X = discrete random variable Example 5 (continued) Computer room: S = {(1M,1P), (1M,2P), (2M,1P), (2M,2P), (3M,1P), (3M,2P)} n = 6 Events: M 1 = 1 Mac is selected = {(1M,1P),(1M,2P)} M 2 = 2 Macs are selected = {(2M,1P),(2M,2P)} M 3 = 3 Macs are selected = {(3M,1P),(3M,2P)} PC 1 = 1 PC is selected = {(1M,1P),(2M,1P),(3M,1P)} PC 2 = 1 PC is selected = {(1M,2P),(2M,2P),(3M,2P)} Counts and probabilities h(m 1 ) = h(m 2 ) = h(m 3 ), h(pc 1 ) = h(pc 2 ) = 3 Frequency approach to probability If not possible to enumerate all possible outcomes, e.g., occurrence of floods Frequency approach requires determining: h = number of times in which event A occurs n = total number of recorded outcomes available The probability is calculated as. P A = h n P(M 1 ) = P(M 2 ) = P(M 3 ) /6 = 1/3 P(PC 1 ) = P(PC 2 ) = 3/6 = 1/2
Examples of frequency approach Example 6 Annual river flooding in low-lying areas Event A = flood occurs on a given year Records show: n = 150 (150 years of records) H 0 (20 years flooding recorded) Examples of frequency approach Example 7 Left-turning cars at intersection Event A = car turns left during observation period Records show: n 50 (250 cars tallied) H 5 (25 cars turned left) P A = h n 0 150 =0.133... P A = h n 5 250 =0.10 Notes of frequency approach 1. The larger the number of records available => the better probability estimates 2. Frequency approximations can be used to verify classical approach values (e.g., throwing a die and recording the outcomes) Personal probability estimates Personal estimates of probability of an outcome Difficult to quantify, but sometimes only information available Example: dam operator gives estimates of the flooding probability in a given dam ( I've seen overtopping of the dam 6 out of the 20 years I've been operating it ), thus P(overtopping) = 6/20 = 0.30 Probability axioms and theorems 1 In these formulas S represents the sample space, Ø is the null event or empty set, and A, B, C, etc., are events of interest. 0 P(A) 1, P(S) = 1, P(Ø) = 0 P(A ) = 1 P(A) If A B = Ø (mutually exclusive events), Probability axioms and theorems 2 For any events A and B, P(A or B) = P(A B) = P(A) + P(B) P(A B) Since A = (A B) (A B ), and (A B) and (A B ) are mutually exclusive, P(A) = P(A B)+ P(A B ) P(A or B) = P(A B) = P(A) + P(B)
control testing 1 Experiments in rainfall simulator Perform n = 100 experiments Rainfall intensity in/hr Test duration = 1 hour Three mechanisms of failure identified: Sliding: slope saturated with water slides downhill Riling: small water paths (rills) populate the slope surface Sinkage: soil compacts and sinks Also, record case of no failure control testing 2 Observations indicate: A = hill slope fails by sliding (15 tests out of 100) P(A) = 15/100 = 0.15 B = hill slope fails by riling (20 tests out of 100) P(B) 0/100 = 0.20 C = hill slope fails by sinkage (5 tests out of 100) P(C) = 5/100 = 0.05 D = hill slope does not fail (60 tests out of 100) P(D) = 60/100 = 0.60 control testing 3 Only one failure mechanism (or no-failure) is identified in each test => events A, B, C, D are mutually exclusive P(D ) = P(failure) = 1 - P(D) = 1-0.60 = 0.40 P(failure) = P(A B C) = P(A) + P(B) + P(C) = 0.15 + 0.20 + 0.05 = 0.40 control testing 4 P(hill slope fails by sliding or riling) = P(A or B) = P(A B) = P(A) + P(B) = 0.15 + 0.20 = 0.35 P(hill slope fails by sliding or sinkage) = P(A or C) = P(A C) = P(A) + P(C) = 0.15 + 0.05 = 0.20 P(hill slope fails by riling or sinkage) = P(B or C) = P(B C) = P(B) + P(C) = 0.20 + 0.05 = 0.25 Probability example water quality sampling 1 Water specimens taken for 100 consecutive days Specimens used to determine concentrations of Cadmium (Cd) and Mercury (Hg). Events: A = significant concentration of Cd (5 out of 100) P(A) = 0.05 B = significant concentration of Hg(10 out of 100) P(B) = 0.10 A B = significant concentrations of Cd and Hg (3 out of 100) P(A B) = 0.03 Probability example water quality sampling 2 Since P(A B) 0, A B Ø, => events A and B are not mutually exclusive i.e., finding a significant level of Cd does not preclude finding a significant level of Hg. Probability of significant levels of Cd or of Hg: P(A or B) = P(A B) = P(A) + P(B) - P(A B) = = 0.05 + 0.10-0.03 = 0.12
Probability example water quality sampling 3 Other probabilities: P(no significant Cd) = P(A ) = 1 - P(A) = 1-0.05 = 0.95 P(no significant Hg) = P(B ) = 1 - P(B) = 1-0.10 = 0.90 Also, from P(A) = P(A B)+ P(A B ) P(A B ) = P(A) - P(A B) = 0.05-0.03 = 0.02 = P(significant Cd but not significant Hg) Probability example water quality sampling 3 Other probabilities: Since (A B ) = A (B ) = A B, [with (B ) = B] then P(A B) = P[(A B ) ] = 1 - P(A B ) = 1-0.02 = 0.98 = P(not significant Cd or significant Hg) Also, P(A B) = P(A ) + P(B) - P(A B), thus, P(A B) = P(A ) + P(B) - P(A B) = 0.95 + 0.10-0.98 = 0.07 = P(not significant Cd but significant Hg) Probability example water quality sampling 3 Other probabilities: Since (A B) = (A ) B = A B [Using (A ) = A], then P(A B ) = P[(A B) ] = 1 - P(A B) = 1-0.07 = 0.93 = P(significant Cd or not significant Hg)