The solution does not make sense to me. My logic wouldbe: any of the four aces c
ID: 2954944 • Letter: T
Question
The solution does not make sense to me. My logic wouldbe: any of the four aces could be in pile 1 so P[E1]=4/52. any of the three aces could be in pile 2 soP[E2|E1]=3/39 any of the 2 aces left could be in pile 3 so P[E3|E1 andE2]=2/26 only 1 of the aces left will be in pile 4 so P[E4| E1 and E2and E3] = 1/13 I am getting a different answer ( .000035). What isthe error in my reasoning? ANSWER: the trick here is to consider both aces and non aces in theanswer. I just figured that out. PE4|E1 E2 E3] = Binomial[1,1] *Binomial[12,12] /Binomial[13,13] We chooose 1 ace from the 1 remaining ace, 12 non aces fromthe 12 non aces, and 13 cards from 13. Similarly P[E1]=Binomial[4,1]*Binomial[48,12] / Binomial[52,13] P[E2|E1] =Binomial[3,1]*Binomial[36,12] /Binomial[39,13] P[E3|E1E2]=Binomial[2,1]*Binomial[24,12] / Binomial[26,13] P[E4||E1 E2E3]=Binomial[1,1]*Binomial[12,12] / Binomial[13,13] The solution does not make sense to me. My logic wouldbe: any of the four aces could be in pile 1 so P[E1]=4/52. any of the three aces could be in pile 2 soP[E2|E1]=3/39 any of the 2 aces left could be in pile 3 so P[E3|E1 andE2]=2/26 only 1 of the aces left will be in pile 4 so P[E4| E1 and E2and E3] = 1/13 I am getting a different answer ( .000035). What isthe error in my reasoning? ANSWER: the trick here is to consider both aces and non aces in theanswer. I just figured that out. PE4|E1 E2 E3] = Binomial[1,1] *Binomial[12,12] /Binomial[13,13] We chooose 1 ace from the 1 remaining ace, 12 non aces fromthe 12 non aces, and 13 cards from 13. Similarly P[E1]=Binomial[4,1]*Binomial[48,12] / Binomial[52,13] P[E2|E1] =Binomial[3,1]*Binomial[36,12] /Binomial[39,13] P[E3|E1E2]=Binomial[2,1]*Binomial[24,12] / Binomial[26,13] P[E4||E1 E2E3]=Binomial[1,1]*Binomial[12,12] / Binomial[13,13] the trick here is to consider both aces and non aces in theanswer. I just figured that out. PE4|E1 E2 E3] = Binomial[1,1] *Binomial[12,12] /Binomial[13,13] We chooose 1 ace from the 1 remaining ace, 12 non aces fromthe 12 non aces, and 13 cards from 13. Similarly P[E1]=Binomial[4,1]*Binomial[48,12] / Binomial[52,13] P[E2|E1] =Binomial[3,1]*Binomial[36,12] /Binomial[39,13] P[E3|E1E2]=Binomial[2,1]*Binomial[24,12] / Binomial[26,13] P[E4||E1 E2E3]=Binomial[1,1]*Binomial[12,12] / Binomial[13,13]Explanation / Answer
What is the question? I don't have that book.