Question

In: Statistics and Probability

We are purchasing a new TV! Let A be the event that the TV was manufactured...

We are purchasing a new TV!

Let A be the event that the TV was manufactured in the U.S., B be the event that the TV has Wifi, and C the event that the customer purchased an extended warranty.

Relevant probabilities are:

P(A) = 0.75

P(B|A) = 0.9

P(B|A′) = 0.8

P(C|A ∩ B) = 0.8

P(C|A ∩ B′) = 0.6

P(C|A′ ∩ B) = 0.7

P(C|A′ ∩ B′) = 0.3

a. What is the probability that the TV was manufactured in the US, with Wifi, and the customer purchased an extended warranty?

b. What is the probability that the TV does NOT have Wifi or the customer did NOT purchase an extended warranty?

c. What is the probability that the customer purchased an extended warranty?

d. What is the probability that the TV does NOT have Wifi given that it was not manufactured in the US?

Expert Solution

a) the probability that the TV was manufactured in the US, with Wifi, and the customer purchased an extended warranty i.e P(A∩B∩C) = ?

By applying duplication rule twice, we will obtain as follows

P(A∩B∩C) = P(C|A∩B)P(A∩B) = P(C|A∩B)P(B|A)P(A) = 0.8×0.9×0.75 = 0.54

the probability that the TV was manufactured in the US, with Wifi, and the customer purchased an extended warranty i.e P(A∩B∩C) = 0.54

b) the probability that the TV does NOT have Wifi or the customer did NOT purchase an extended warranty i.e P( B∩C) = ?

By applying duplication rule,we will obtain as follows

P(B∩C) = P(B∩C∩A)+P(B∩C∩A′)

= P(C|A∩B)P(B|A)P(A) + P(C|A′∩B)P(B/A′)P(A′)

= (0.8×0.9×0.75) + (0.7×0.8×0.25)

= 0.54 + 0.14

= 0.68

the probability that the TV does NOT have Wifi or the customer did NOT purchase an extended warranty i.e P(B∩C) = 0.68

C) the probability that the customer purchased an extended warranty i.e P(C) = ?

P(C) = P(C|A∩B) P(B/A) P(A) + P(C|A∩B′) P(B′/A) P(A) + P(C|A′∩B) P(B/A′) P(A′) + P(C|A′∩B′) P(B′/A′)P(A′)

= (0.8×0.9×0.75) +(0.6×0.1×0.75) +(0.7×0.8×0.25) + (0.3×0.2×0.25)

= 0.54 + 0.045 + 0.14 + 0.015

= 0.74

the probability that the customer purchased an extended warranty i.e P(C) = 0.74

d) the probability that the TV does NOT have Wifi given that it was not manufactured in the US i.e P(A|B∩C) = ?

P(A|B∩C) = P(A∩B∩C)/P(B∩C) = 0.54/0.68 = 0.79

the probability that the TV does NOT have Wifi given that it was not manufactured in the US i.e P(A|B∩C) = 0.79

orchestra answered 2 years ago

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