STAT 418HW2 SolutionsSections 3.1-2.42, 5, 6, 11, 12, 15, 18, 21, 23, 282. Let E = {First die lands on a 6} and Fi= {Sum of the dice = i}.P (E|F2) = P (E|F3) = P (E|F4) = P (E|F5) = P (E|F6) = 0.P (E|F7) =P (E ∩ F7)P (F7)=1/366/36=16P (E|F8) =1/365/36=15P (E|F9) =1/364/36=14P (E|F10) =1/363/36=13P (E|F11) =1/362/36=12P (E|F12) = 15. Let E = {First 2 balls are white} and F = {Last 2 balls are black}. By the Multipli-cation Rule,P (E ∩ F ) =615514913812.6. Let E = {First and third balls are white} and F = {Exactly 3 white balls are drawn}.Note that in both cases, there are 2 ways we can place the black ball if both E and F occurand there are 4 ways we can place the ball if F occurs.a. Without Replacement:P (E|F ) =P E ∩ F )P (F=21812711610494181271161049=12.b. With Replacement:P (E|F ) =P E ∩ F )P (F=218123412418123412=12.11. Let B = {Both cards are aces}, A = {At least one card is an ace}, and As= {Ace ofSpades is drawn}.a. Consider the reduced sample space where the Ace of Spades is drawn:P (B|As) =3511b.P (B|A) =P (A ∩ B)P (A)=P (A|B)P (B)1 − P (¯A)=(1)42/5221 −482/522=13312. Let Pj= {She passes the jthexam}.a. By the Multiplication Rule,P (P1∩ P2∩ P3) = P (P1)P (P2|P1)P (P3|P1P2) = (.9)(.8)(.7) = .504b.P (¯P2|P1∩ P2∩ P3) =P (P1∩ P2∩ P3∩¯P2)P (P1∩ P2∩ P3)=P (¯P2)P (P1∩ P2∩ P3)=P (P1∩¯P2) + P (¯P1∩¯P2)1 − (P1∩ P2∩ P3)=(.9)(.2) + 01 − .504= .3629.15. Let E = {Ectopic Pregnancy} and S = {Smoker}. Given: P (S) = .32, andP (E|S)P (E|¯S)= 2.Find: P (S|E). Because S and¯S form a partition of the entire sample space, by Bayes TheoremP (S|E) =P (E|S)P (S)P (E|S)P (S) + P (E|¯S)P (¯S)=P (E|S)P (E|¯S)P (S)P (E|S)P (E|¯S)P (S) +P (E|¯S)P (E|¯S)P (¯S)=2 ∗ .322 ∗ .32 + 1 − .32= .4848.18. Let I = {Independent}, L = {Liberal}, and C = {Conservative}. L, I, and C forma partition of the entire sample space.a. By Bayes Theorem,P (I|V ) =P (V |I)P (I)P (V |I)P (I) + P (V |L)P (L) + P (V |C)P (C)=.35 ∗ .46.35 ∗ .46 + .62 ∗ .3 + .58 ∗ .24= .3311.2b. By Bayes Theorem,P (L|V ) =P (V |L)P (L)P (V |I)P (I) + P (V |L)P (L) + P (V |C)P (C)=.62 ∗ .3.35 ∗ .46 + .62 ∗ .3 + .58 ∗ .24= .3826.c. By Bayes Theorem,P (C|V ) =P (V |C)P (C)P (V |I)P (I) + P (V |L)P (L) + P (V |C)P (C)=.58 ∗ .24.35 ∗ .46 + .62 ∗ .3 + .58 ∗ .24= .2863.d. By the law of total probability,P (V ) = P (V |I)P (I)+P (V |L)P (L)+P (V |C)P (C) = .35∗.46+.62∗.3+.58∗.24 = .486221.a. P( Husband earns less than $25,000)= P( Husband earns less than $25,000 and Wife earns less than $25,000) +P(Husband earns less than $25,000 and Wife earns more than $25,000)=212+36212+36+198+54= .496.b. Examine the reduced sample space:P( Wife earns more than $25,000 | Husband earns more than $25,000)=54198+54= .214.c. Examine the reduced sample space:P( Wife earns more than $25,000 | Husband earns less than $25,000)=36212+36= .145.23.a. Let W1= {White ball drawn from Urn I and put in Urn II} and W2= {White balldrawn from Urn II}. Note that W1and W1form a partition of S, the entire samplespace. By the law of total probability,P (W2) = P (W2|W1)P (W1) + P (W2|W1)P (W1) =2326+1346=49.b. Find: P (W1|W2). By Bayes Theorem,P (W1|W2) =P (W2|W1)P (W1)P (W2|W1)P (W1) + P (W2|W1)P (W1)=2/94/9=12.328.a. Let As= {21st card is the ace of spades} and A = {20th card is the first ace}.P (As|A) =P (As∩ A)P (A)=481931115219331321÷4819415219331=34 ∗ 32=3128.b. Let DC= {21st card is the 2 of clubs} and A = {20th card is the first ace}.P (DC|A) =P (DC∩ A)P (A)=471941115219331321÷4819415219331=47 ∗ 46 ∗ · · · ∗ 2948 ∗ 47 ∗ · · · ∗ 30 ∗ 32=2948 ∗