STAT 418HW4 SolutionsSections 3.4-3.537, 39, 43, 50, 52, 55, 64, 65, 78, TE 1, TE 637. Let H = {Heads on first flip}, F = {Fair coin}, and T = {Two-headed coin}. Notethat F and T form a partition.a. By Bayes rule,P (F |H) =P (H|F )P (F )P (H|F )P (F ) + P (H|T )P (T )=.5 · .5.5 · .5 + 1 · .5=13b. Let HH = {Heads, Heads in 2 flips of the coin}. By independence, P (HH|F ) =14.P (F |HH) =P (HH|F )P (F )P (HH|F )P (F ) + P (HH|B)P (B)=.25 · .5.25 · .5 + 1 · .5=15c. Because the two-headed coin can never flip tails, P (F |HHT ) = 1.39. Let A = {Accident prone}, A1= {Accident in first year}, and A2= {Accident insecond year}. Find P (A2|¯A1).P (A2|¯A1) =P (A2∩¯A1)P (¯A1).By the law of total probability,P (A2|¯A1) =P (A2∩¯A1∩ A)P (¯A1)+P (A2∩¯A1∩¯A)P (¯A1)=P (A2∩¯A1∩ A)/P (¯A1∩ A)P (¯A1)/P (¯A1∩ A)+P (A2∩¯A1∩¯A)/P (¯A1∩¯A)P (¯A1)/P (¯A1∩¯A)= P (A2|¯A1∩ A)P (A|¯A1) + P (A2|¯A1∩¯A)P (¯A|¯A1).Given:P (A2|¯A1∩A) = .4, P (A2|¯A1∩¯A) = .2, P (A1|A) = .4, P (¯A1|¯A) = 1−.2 = .8, P (A) = .3, P (A1) = .26.Therefore,P (A|¯A1) =P (¯A1|A)P (A)P (¯A1)=(1 − .4) · .31 − .26= .243P (¯A|¯A1) =P (¯A1|¯A)P (¯A)P (¯A1)=.8 · (1 − .3)1 − .26= .757P (A2|¯A1) = .4 · .243 + .2 · .757 = .2486.43. Let H = {Heads flipped}, F = {Fair coin}, B = {Biased coin}, and T = {Two-headedcoin}. Note that F , B, and T form a partition. Given:P (H|F ) =12, P (F ) =13, P (H|B) =34, P (B) =13, P (H|T ) = 1, P (T ) =131By Bayes rule,P (T |H) =P (H|T )P (T )P (H|T )P (T ) + P (H|B)P (B) + P(H|F)P(F)=4950. Let G = {Good risks}, A = {Average risks}, B = {Bad risks}, and C = {Accident}.Note that G, A, and B form a partition. Given:P (C|G) = .05, P (C|A) = .15, P (C|B) = .3, P (G) = .2, P (A) = .5, P (B) = .3.Find:P (C), P (G|¯C), P (A|¯C)By the law of total probability,P (C) = P (C|G)P(G) + P(C|A)P (A) + P (C|B)P(B) = .175.P (G|¯C) =P (¯C|G)P (G)1 − P (C)=(1 − .05) · .21 − .175= .23P (A|¯C) =P (¯C|A)P (A)1 − P (C)=(1 − .15) · .51 − .175= .5252. Let M, T , W, R, and F be the events that mail is received on that day. Let A ={Accepted} and J = {Rejected}. Note that A and J form a partition.a. By the law of total probability,P (M) = P (M|A)P (A) + P (M|J)P (J) = .15 · .6 + .05 · .4 = .11b.P (T |¯M) =P (T ∩¯M)P (¯M)=P (T )1 − P (M)=P (T |A)P (A) + P (T |J)P (J)1 − P (M)=.2 · .6 + .1 · .41 − .11= .180c. By Bayes rule,P (A|¯M ∩¯T ∩¯W ) =P (¯M ∩¯T ∩¯W |A)P (A)P (¯M ∩¯T ∩¯W |A)P (A) + P (¯M ∩¯T ∩¯W |J)P (J)=(1 − .15 − .2 − .25) · .6(1 − .15 − .2 − .25) · .6 + (1 − .05 − .1 − .1) · .4= .444.d. By Bayes rule,P (A|R) =P (R|A)P (A)P (R|A)P (A) + P (R|J)P (J)=.15 · .6.15 · .6 + .15 · .4= .6e. By Bayes rule,P (A|¯M ∩¯T ∩¯W ∩¯R ∩¯F ) =P (¯M ∩¯T ∩¯W ∩¯R ∩¯F |A)P (A)P (¯M ∩¯T ∩¯W ∩¯R ∩¯F |A)P (A) + P (¯M ∩¯T ∩¯W ∩¯R ∩¯F |J)P (J)=(1 − .15 − .2 − .25 − .15 − .1) · .6(1 − .15 − .2 − .25 − .15 − .1) · .6 + (1 − .05 − .1 − .1 − .15 − .2) · .4= .36255. Let x = the total number of sophomore girls. Let G = {Girl selected} and S ={Sophomore selected}. By the defintion of independence, sex and class, or equivalently, Gand S, are independent iff P (G|S) = P (G). Thereforex6 + x=6 + x16 + x16x + x2= 36 + 12x + x24x = 36x = 9.64. Let C = {Correct answer}, W = {Wife answers}, Wc= {Wife correct}, and Hc={Husband correct}. W and¯W form a partition of the entire sample space. Hc∩ Wc,¯Hc∩ Wc,Hc∩¯Wc, and¯Hc∩¯Wcalso form a partition of the sample space; when they are both corrector incorrect, they agree.Under situation (a), assuming choice is at random,P (C) = P (C|W )P (W ) + P (C|¯W )P (¯W ) = p ·12+ p ·12= p.Under situation (b),P (C) = P (C|Hc∩ Wc)P (Hc∩ Wc) + P (C|¯Hc∩ Wc)P (¯Hc∩ Wc) + P (C|Hc∩¯Wc)P (Hc∩¯Wc)+ P (C|¯Hc∩¯Wc)P (¯Hc∩¯Wc)= 1 · p · p +12· (1 − p) · p +12· p · (1 − p) + 0 · (1 − p) · (1 − p)= p2+ p − p2= p.Both strategies give the same probability of winning.65. Let A = {they agree}.a.P (C|A) =P (C ∩ A)P (A)=P (Hc∩ Wc)P (Hc∩ Wc) + P (¯Hc∩¯Wc)=p2p2+ (1 − p)2= .692b.P (C|¯A) =C ∩¯A)P (¯A)=P (C ∩ [(Hc∩¯Wc) ∪ (¯Hc∩ Wc)])1 − P (A)=P (C ∩ Hc∩¯Wc) + P (C ∩¯Hc∩ Wc)1 − P (A)=12· p · (1 − p) +12· (1 − p) · p1 − (2p2− 2p + 1)=p(1 − p)2p(1 − p)=12.378.a. Let F = {4 games are played}, A4= {A wins in 4 games}, and B4= {B wins in 4games}.P (F ) = P (A4) + P (B4)= p(1 − p)p2+ (1 − p)p3+ (1 − p)p(1 − p)2+ p(1 − p)3= 2p(1 − p)[p2+ (1 − p)2].b. Partition the sample space based on the outcome of the first two games. Let A = {Awins}, aa = {A wins first 2 games}, ab = {A wins first, B wins second}, ba = {B winsfirst, A wins second}, and bb = {B wins first 2 games}. By the law of total probability,P (A) = P (A|aa)P (aa) + P (A|ab)P (ab) + P (A|ba)P (ba) + P (A|bb)P (bb).If either ab or ba occur, the players are even and because the games are independent,this is equivalent to starting over. Therefore P (A) = P (A|ab) = P (A|ba). This yieldsP (A) = 1 · p2+ 2P (A)p(1 − p) + 0 · (1 − p)2P (A) − 2P (A)p(1 − p) = p2P (A) =p21 − 2p + 2p2.TE 1. Suppose P (A) > 0,P (A ∩ B|A ∪ B) =P (A ∩ B ∩ (A ∪ B))P (A ∪ B)=P ((A ∩ B) ∪ (A ∩ B)P (A ∪ B)=P (A ∩ B)P (A ∪ B).Because P (A ∪ B) ≥ P (A) and P (A) > 0,P (A ∩ B)P (A ∪ B)≤P (A ∩ B)P (A)=P (A ∩ B ∩ A)P (A)= P (A ∩ B|A)TE 6. By DeMorgan’s laws,P (∪ni=1Ei) = P (∩ni=1¯Ei) = 1 − P (∩ni=1¯Ei) = 1 −nYi=1(1 − P