For customers purchasing a refrigerator at a certain appliancestore, let A be the event that the refrigerator wasmanufactured in
the U.S., B be the event that therefrigerator had an icemaker, and C be the event that thecustomer purchased an extended warranty. Relevant probabilities arebelow.
(a) Construct a tree diagram consisting of first-, second-, andthird-generation branches and place an event label and appropriateprobability next to each branch.
(b) Compute P(A intersectB intersectC).
(c) Compute P(B intersectC).
2
(d) Compute P(C).
3
(e) Compute P(A | B intersectC), theprobability of a U.S. purchase given that an icemaker and extendedwarranty are also purchased.
P(A) = 0.73 P(B | A) = 0.90 P(B | A' ) =0.76
P(C | A intersectB) =0.84 P(C | A intersectB' ) =0.65
P(C | A' intersectB) =0.71 P(C | A' intersectB' ) =0.32
(a) Construct a tree diagram consisting of first-, second-, andthird-generation branches and place an event label and appropriateprobability next to each branch.
(b) Compute P(A intersectB intersectC).
(c) Compute P(B intersectC).
2
(d) Compute P(C).
3
(e) Compute P(A | B intersectC), theprobability of a U.S. purchase given that an icemaker and extendedwarranty are also purchased.
P(A) = 0.73 P(B | A) = 0.90 P(B | A' ) =0.76
P(C | A intersectB) =0.84 P(C | A intersectB' ) =0.65
P(C | A' intersectB) =0.71 P(C | A' intersectB' ) =0.32
1 answer:
Answer:
Part a : The tree diagram is attached with the solution.
Part b: The probability of P(A∩B∩C) is 0.5519.
Part c: The probability of P(B∩C) is 0.6976.
Part d: The probability of P(C) is 0.7658.
Part e: The probability of P(A∣B∩C) is 0.7658.
Explanation:
The solution is as follows.
Data Given is as
P(A) = 0.73
P(B | A) = 0.90
P(B | A' ) =0.76
P(C | A∩B) =0.84
P(C | A∩B' ) =0.65
P(C | A' ∩B) =0.71
P(C | A' ∩B' ) =0.32
Part a:
The remaining values of Probabilities for the tree formation are calculated as follows
P(A')=1-P(A)=1-0.73=0.27
P(B'|A)=1-P(B|A)=1-0.90=0.10
P(B' | A' )=1-P(B | A' ) =1-0.76=0.24
P(C' | A∩B) =1-P(C | A∩B)=1-0.84=0.16
P(C' | A∩B' ) =1-P(C | A∩B' ) =1-0.65=0.35
P(C' | A' ∩B) =1-P(C | A' ∩B) =1-0.71=0.29
P(C' | A' ∩B' ) =1-P(C | A' ∩B' )=1-0.32=0.68
so the probability tree is given as attached in the diagram.
Part b:
P(A∩B∩C)=P(A)P(B∣A)P(C∣A∩B)
From the above data in part a
P(A)=0.73, P(B|A)=0.90 and P(C∣A∩B)=0.84
=0.73×0.90×0.84
=0.55188
≈0.5519
The probability of P(A∩B∩C) is 0.5519.
Part c
P(B∩C)=P(A∩B∩C)+P(A'∩B∩C)
= [P(A)P(B∣A)P(C∣A∩B)+{1−P(A)}P(B∣A′)P(C∣A′∩B)]
From the above data
P(A)=0.73,P(B∣A)=0.9,P(C∣A∩B)=0.84,P(B∣A′)=0.76 and P(C∣A′∩B)=0.71
P(B∩C)= [P(A)P(B∣A)P(C∣A∩B)+{1−P(A)}P(B∣A′)P(C∣A′∩B)]
=(0.73×0.9×0.84)+(1−0.73)×0.76×0.71
=0.55188+0.145692
=0.697572
≈0.6976
The probability of P(B∩C) is 0.6976.
Part d
P(C)=P(A∩B∩C)+P(A′∩B∩C)+P(A∩B ′∩C)+P(A′∩B ′∩C)
=P(A)P(B∣A)P(C∣A∩B)+[1−P(A)]P(B∣A′)P(C∣A′∩B)+P(A)[1−P(B∣A)]P(C∣A∩B′)+
[1−P(A)][1−P(B∣A′)]P(C∣A′∩B′)
Here from above data
P(A)=0.73,P(B∣A)=0.9,P(C∣A∩B)=0.84,P(B∣A′)=0.76,P(C∣A∩B′)=0.65,P(C∣A′∩B′)=0.32 and P(C∣A′∩B)=0.71
P(C)=P(A)P(B∣A)P(C∣A∩B)+[1−P(A)]P(B∣A′)P(C∣A′∩B)+P(A)[1−P(B∣A)]P(C∣A∩B′)+
[1−P(A)][1−P(B∣A′)]P(C∣A′∩B′)
={0.7x0.9x0.84}+{(1-0.73)x0.76×0.71}+{0.73x(1-0.9)x0.32}+{(1-0.73)x(1-
0.76)x0.32}
=0.55188+0.145692+0.04745+0.020736
=0.765758
≈0.7658
The probability of P(C) is 0.7658.
Part e
From part (b),P(A∩B∩C)=0.55188
From part (c), P(B∩C)=0.697572
<em>By conditional probability rule, </em>
P(A∣B∩C)= P(A∩B∩C)/P(B∩C)
= 0.55188/0.697572
=0.791144
≈0.7911
The probability of P(A∣B∩C) is 0.7658.
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