Question

Suppose that 60% of all adults regularly consume coffee, 45% regularly consume carbonated soda, and 70% regularly consume at least one of these two products. (a) What is the probability that a randomly selected adult regularly consumes both coffee and soda? (b) What is the probability that a randomly selected adult doesn't regularly consume at least one of these two products? Need Help? Read It Watch It Talk to a Tutor

Answer 1

P. (coffee) = 0 60 psoda) - 0.45 Pl coffee U soda) = 0.70 (at least one of two) P (consume both) = P(coffee n soda p(coffeen soda)= P(coffee) + P(soda) - P(coffee U soda) - 0.60 + 0.45 -0.70 = 0.35 el doeen't consume at least one of two). - P1 - Pl conjume at least both) - 1- P ( coffee U Soda) = -0.70 = 0.30 P (stop at 1st signal) = P(A)= 0.35 pl stop at 2nd signal) = P(B) = 0.45. 0.50 plstop at at least one) = P(AUB) = a) P(ANB)= P(A) + P(B) - P(AUB) = 0.35 +0.45 -0.50 - 030 Plan B') = ? using law of total probability, PIA) - PLANB) + P(A18') PLANB) = P(A)- PLANB) = 0.35 - 0:30 = 0.05 el stop at exactly one signal). = P(AUB) - PLANB) = 0.50 - 0.30 2020

Total failed keyboards = 25. electrical defects = 13 mechanical defects = 12 # of ways to select 5 keyboarde out of 25 (order not a) considerad) # of ways = 265 25 51201 53136 # of ways to select 5 keyboarde out of 25 so that exactly have a electrical defect # of ways = 130, X12C3 2 from electrically defected keyboard en 3 from remaining keyboards 131 12 1 2100 x Z 319 = 17160

prob = # of ways Total # of ways # of way to select 5 out of 25 sot that at least 4 will have mechanical defect exactly # of ways - 1264 x 13c, + 120 x 30 = 7227 7227 prob = 53130 0.1360