Answer to #1

1. Answer to #1 L source-receptor distance i hydraulic gradient K hydraulic conductivity n effective soil porosity BD soil bulk density foc fraction organic carbon Koc organic partition coefficient Porosity and conductivity would likely be positively related Porosity and density would likely be inversely related Thus, conductivity and density should be inversely related too Fraction organic carbon and the partition coefficient would likely be positively related

2. Answer to #2 L = [80,120] m // source-receptor distance i = [0.0003,0.0008] m per m // hydraulic gradient K = [300,3000] m per yr // hydraulic conductivity n = [0.2,0.35] // effective soil porosity BD = [1500,1750] kg per m3 // soil bulk density foc = [0.0001,0.005] // fraction organic carbon Koc = [5 ,20] m3 per kg // organic partition coefficient Kd = foc * Koc R = 1 + BD * Kd / n V = K * i / (n * R) T = L/V T [ 20.952380952, 408800] yr func TT() return L/(K * i / ($1 * (1 + BD * foc * Koc / $1))) b = left(n) e = TT(b) w = width(n)/100 for k=1 to 100 do begin s = [ (k-1), k] * w + b e = env(e, TT(s)) end e [ 31.480562448, 235352] yr

3. Answer to #3 Iwater = [1.5, 2.5] liters per day // water intake Ifish = [0, 8] g per day // dietary ingestion of fish tissue B = [0.9, 2.1] liters per g // bioaccumulation factor W = [60, 90] kg // receptor biomass D = [0, 6] mg per kg per day // tolerable dose // forward equation // D = (Iwater * C) / W + (Ifish * B * C) / W // condense repeated parameters // D = C * (Iwater + Ifish * B) / W // solve for C by factoring the quotient (Iwater+Ifish*B)/W out of D C = factor((Iwater + Ifish * B) / W, D) C [ 0, 18.652849741] mg liters1 // put C back into original forward equation (and ensure correct units) d = (Iwater * C) / W + (Ifish * C * B) / W + 0 mg per kg per day d [ 0, 6] mg kg1 day1

4. Answer to #5 3.5105 [3.499105, 3.504105] [3.50105, 1.35104] [3105, 1.4104] [2.9105, 4.1105] [ 2.5105, 1.905 104] Vesely et al. (all independent) Mixed dependencies Frank to Fréchet All Fréchet Mixed dependence with intervals All Fréchet 105 104 103 Probability of tank rupturing under pumping

5. Plan of attack for #6 plane =random ship=random raft =random pet =random zoo=random pregnant =random malefemale=random repro=random female =random parthenog=random dogs =random cats =random pigs=random eggs =random rodents=random hides =random quick =random disperse=random (plane ||| ship ||| raft ||| pet ||| zoo) |&| (pregnant ||| (malefemale |&| repro) ||| (female |&| parthenog)) |&| (dogs |&| cats |&| pigs) |&| (eggs ||| rodents) |&| (hides ||| quick ||| disperse) • Use the events from the slide “Snakes on a plane” • Write as Boolean expression • Represent ANDs as conjunctions & • Represent ORs as disjunctions V (plane V ship V raft V pet V zoo) & (pregnant V (malefemale & repro) V (female & parthenog)) & (dogs & cats & pigs) & (eggs V rodents) & (hides V quick V disperse) • Specify the temporal context, e.g., a year • Estimate marginal event probabilities • Maybe as submodels, e.g., plane = flights * stowaway per flight • Decide which events are independent

6. Answer to #7 Assuming a = 0.29, b = 0.22, and r = 1.0, yields a * b + r * sqrt(a*(1-a)*b*(1-b)) Pearson AND 0.2517692528 [ max(0, a+b–1), min(a, b) ] Fréchet AND [ 0, 0.22] The probability assuming Pearson correlation r =1 is larger than the largest possible probability. This happens because the Pearson correlation for two events with these marginals cannot be as large as one (so we are making a false assumption which leads to the error). The extreme Pearson correlations for them are (-a*b)/sqrt(a*b*(1-a)*(1-b)) // when they don’t overlap -0.33941721344 (b-a*b)/sqrt(a*b*(1-a)*(1-b)) // when b’s inside a 0.83098697084 We can check this with this calculation r = 0.83098697084 a * b + r * sqrt(a*(1-a)*b*(1-b)) 0.22