Question

Now suppose that at breakfast time we either eat cereal, oatmeal, or rice. Suppose: • Having milk and having a side item are independent. • P[{have milk}] = 3/4. • P[{have side dish}] = 1/2. • Given we only have milk (no side dish): We eat cereal with prob 3/4 and oatmeal with prob 1/4. • Given we have milk and side dish: We eat cereal with prob 1/2, rice with prob 1/4, oatmeal with prob 1/4. • Given we only have side dish (no milk): We eat oatmeal with prob 1/2 and rice with prob 1/2. • Given we have no milk and no side dish: We eat oatmeal with prob 1. Let C,0, R be events of eating cereal, oatmeal, rice, respectively. Let M, D be events having milk and having a side dish, respectively. (D stands for “side [D]ish”). a) Compute the probability that we eat cereal. b) Compute the probability that we eat rice. Then, compute the probability we eat oatmeal. c) Compute the conditional probability we have a side dish, given we eat cereal. d) Compute the probability we have milk and eat cereal. Explain this result in relation to part (a). e) Compute the conditional probability we eat oatmeal, given we only have a side dish (no milk).

Answer 1

(a) P(C)= P(COMID) + PCMAD) +PCNMD) = P(CMND)P(M)P(DC)+P(CMOD)P(M)P(D)+P(CM'ND)P(M)P(D) (Since M and D are independent) = (3/4)*(3/4)*(1-1/2)+(1/2)*(3/4)*(1/2) +0*(1-3/4)*(1/2) = 0.46875 (6) P(R) = P(ROMAD) + P(ROMAD + PRO Mºn D) = P(RMND)P(M)P(DC)+P(R MOD)P(M)P(D)+P(RM'ND)P(M)P(D) = 0* (3/4) * (1 – 1/2) + (1/4) * (3/4) * (1/2) + (1/2) + (1 - 3/4) * (1/2) = 0.15625 P(O) = POMAD') +POMOD+POMAD) +P(OnMD) = POMOD) P(M)P(DC)+POMAD)P(M)P(D)+POM'ND)P(M)P(D) + POM° DP(M)P(D) = (1/4) * (3/4)*(1-1/2) + (1/4)*(3/4) * (1/2) + (1/2)*(1-3/4)*(1/2) + 1*(1 – 3/4) *(1 – 1/2) = 0.375 (c) P(DC) = ( PDI) PICND) PCDNM)P(D) P(M) + PICDOMPIDPC) PC) = ((1/2) * (3/4) * (1/2) +0*(1 – 3/4) * (1/2))/0.46875 = 0.4 P(C) (d) PM NC) = P(MODNC) + PM Dºnc) = (3/4) * (3/4) * (1 – 1/2) + (1/2) * (3/4) * (1/2) = 0.46875

(e) PODNMC) - PONDOM) - P( DMC) = 0.5