STAT 418HW6 SolutionsSections 4.3- 4.630, 35, 38, 41, 42, 46, 49, TE 13, 1430.E(X) =∞Xn=12n(1/2)n= ∞a. No, E(X) represents the average in the long run. In a single game, there is a 1/2 + 1/4+ 1/8 + 1/16 + 1/32 = .96875 probability that X ≤ $32.b. Yes, playing for an arbitrarily large number of games without having to pay upfrontallows you to take advantage of E(X) = ∞.35. Let X be the amount in dollars that you win.a.P (X = 1.10) =252102= 4/9P (X = −1) = 1 − P (X = 1.1) = 1 − 4/9 = 5/9E(X) = 1.1 ∗ 4/9 − 5/9 = −.067b.V ar(X) = E(X2) − (EX)2= 1.12∗ 4/9 + 5/9 − .0672= 1.08938.a.E[(2 + X)2] = V ar(2 + X) + (E(2 + X))2= V ar(X) + (2 + EX)2= 14b.V ar(4 + 3X) = 32V ar(X) = 4541.P (X ≥ 7) =10Xn=710n12n1 −1210−n=176102442. Find p such that53p3(1 − p)2+54p4(1 − p) +55p5≥32p2(1 − p) +33p3⇔ 10p3(1 − 2p + p2) + 5p4− 5p5+ p5≥ 3p2− 3p3+ p3⇔ 10p − 20p2+ 10p3+ 5p2− 4p3≥ 3 − 2p⇔ 6p3− 15p2+ 12p − 3 ≥ 0.1Notice that 1 is a zero and therefore (p − 1) is a factor. The above is therefore equivalent to(p − 1)(6p2− 9p + 3) ≥ 0⇔ (p − 1)(p − 1)(6p − 3) ≥ 0⇔ 6(p − 1)2(p − .5) ≥ 0.Because 6(p − 1)2≥ 0, the condition on p is that p ≥ .5.46. Following Example 6e, the probability that the jury renders a correct decision isP (Correct decision) = P (Guilty)P (Correct decision|Guilty) + P (Not Guilty)P (Correct decision|Not Guilty)= .6512Xi=912i(1 − .2)i.212−i+ .3512Xi=412i(1 − .1)i.112−i,because you need at least 9 people to vote a guilty person guilty, and therefore at least 4people to vote an Not Guilty person Not Guilty.The percentage of people convicted isP (Convicted) = P (Guilty)P(Convicted|Guilty) + P (Not Guilty)P (Convicted|Not Guilty)= .6512Xi=912i(1 − .2)i.212−i+ .3512Xi=912i.1i(1 − .1)12−i,49.a.107(0.4)7(0.6)312+107(0.7)7(0.3)312= 0.1546b. Let A = {7 of 10 tosses are Heads}, Ci= {Coin i flipped}, i = 1, 2, and H1= {Firstflip Heads}. As shown in the solutions for problem 39 in Chapter 3,P (A|H1) = P (A|C1∩ H1)P (C1|H1) + P (A|C2∩ H1)P (C2|H1).P (C1|H1) =P (C1)P (H1|C1)P (C1)P (H1|C1) + P (C2)P (H1|C2)=.5 ∗ .4.5 ∗ .4 + .5 ∗ .7= .3636P (C2|H1) =P (C2)P (H1|C2)P (C1)P (H1|C1) + P (C2)P (H1|C2)=.5 ∗ .7.5 ∗ .4 + .5 ∗ .7= .6364P (A|H1) =96(0.4)6(0.6)3(.3636) +96(0.7)6(0.3)3(.6364) = 0.1968.TE 13. Because log is an increasing function, we may first take log(P (X = k)) and thenmaximize.log P (X = k) = lognk+ k log p + (n − k) log(1 − p)∂∂plog P (X = k) =kp−n − k1 − p= 0 ⇒knis the value of p that maximizes P (X = k).2TE 14. Let X = {number of children in a family}. Given: P (X = n) = αpn.a.P (X = 0) = 1 − P (X ≥ 1) = 1 −∞Xn=1αpn= 1 − α11 − p− 1= 1 −αp1 − pb. Let Y = {number of boys in a family}. Condition on X, the number of children: Fork > 0P (Y = k) =∞Xn=1P (Y = k|X = n)αpn=∞Xn=knk12k12n−kαpn= α∞Xn=knkp2n= α∞Xn=knkn − 1k − 1p2n=αk∞Xn=kn − 1k − 1np2n−1p2=αp2k∞Xn=kn − 1k − 1ddpp2n=αp2kddp∞Xn=kn − 1k − 1p2n=αp2kddpp2k1 −p2−k=α2p2k1 −p2−k+α2p2k+11 −p2−k−1=α2p2k1 −p2−k−1.P (Y = 0) = P (Y = 0|X = 0)P (X = 0) + P (Y = 0|X = n)P (X = n)= 1 −αp1 − p+∞Xn=1αpn12n= 1 −αp1 − p+ α∞Xn=1p2n= 1 −αp1 − p+αp/21 − p/2= 1 − αp/2(1 − p)(1 − p/2)= 1 −αp2 − 3p +