For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are below.P(A) = 0.74 P(B | A) = 0.85 P(B | A') = 0.82P(C | A ∩ B) = 0.76P(C | A ∩ B') = 0.55P(C | A' ∩ B) = 0.75P(C | A' ∩ B') = 0.25(a) Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch. (b) Compute P(A ∩ B ∩ C). (Round your answer to four decimal places.)(c) Compute P(B ∩ C). (Round your answer to four decimal places.)(d) Compute P(C). (Round your answer to four decimal places.)(e) Compute P(A | B ∩ C), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased. (Round your answer to four decimal places.) For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extendedwarranty. Relevant probabilities are below.P(A) = 0.74 P(BIA) = 0.85P(B | A') = 0.82P(CAN B) = 0.76P(CAN B') = 0.55P(C | A'n B) = 0.75P(C | A' B') = 0.25(a) Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch.A, 0.74BB', 0.15BB', 0.18C', 0.24C', 0.45C', 0.25(b) Compute P(AN BNC). (Round your answer to four decimal places.)(c) Compute P(BNC). (Round your answer to four decimal places.)(d) Compute P(C). (Round your answer to four decimal places.)(e) Compute P(A | Bn C), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased. (Round your answer to four decimal places.)

Given, P(A) = 0.74 P(B | A) = 0.85 P(B|A') = 0.82 P(C | A ∩ B) = 0.76 P(C | A ∩ B') = 0.55 P(C | A' ∩ B) = 0.75 P(C | A' ∩ B') = 0.25 a) Tree diagram : b) We have to find P(A ∩ B ∩ C) P(A ∩ B ∩ C) = P(A)×P(B|A)×P(C|A∩B) P(A ∩ B ∩ C) = 0.74×0.85×0.76 P(A ∩ B ∩ C) = 0.4780b) We have to find P(B ∩ C) P(B∩C)=P(A∩B∩B) + P(A'∩B∩C) P(B∩C)=(0.74×0.85×0.76)+(0.26×0.82×0.75) P(B∩C)=0.6379c) We have to find P(C) P(C) = P(A∩B∩C)+P(A∩B'∩C)+P(A'∩B∩C)+P(A'∩B'∩C) P(C) = (0.74×0.85×0.76)+(0.74×0.15×0.55)+(0.26×0.82×0.75)+(0.26×0.18×0.25) P(C) = 0.7107 d) We have to find P(A | B ∩ C) P(A | B ∩ C) = P(A∩B∩C)P(B∩C) From part a, P(A ∩ B ∩ C) = 0.4780 From part b, P(B∩C)=0.6379 So, P(A | B ∩ C) = 0.47800.6379 P(A | B ∩ C) = 0.7493

Answered: For customers purchasing a refrigerator…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are below.

P(A) = 0.74	P(B \| A) = 0.85	P(B \| A') = 0.82
P(C \| A ∩ B) = 0.76	P(C \| A ∩ B') = 0.55
P(C \| A' ∩ B) = 0.75	P(C \| A' ∩ B') = 0.25

(a) Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch.

(b) Compute P(A ∩ B ∩ C). (Round your answer to four decimal places.)

(c) Compute P(B ∩ C). (Round your answer to four decimal places.)

(d) Compute P(C). (Round your answer to four decimal places.)

(e) Compute P(A | B ∩ C), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased. (Round your answer to four decimal places.)