Respuesta :
Answer:
(a) The probability that at least one of these components will need repair within 1 year is 0.0278.
(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.
Step-by-step explanation:
Denote the events as follows:
A = video components need repair within 1 year
B = electronic components need repair within 1 year
C = audio components need repair within 1 year
The information provided is:
P (A) = 0.02
P (B) = 0.007
P (C) = 0.001
The events A, B and C are independent.
(a)
Compute the probability that at least one of these components will need repair within 1 year as follows:
P (At least 1 component needs repair)
= 1 - P (No component needs repair)
[tex]=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278[/tex]
Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.
(b)
Compute the probability that exactly one of these component will need repair within 1 year as follows:
P (Exactly 1 component needs repair)
= P (A or B or C)
[tex]=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277[/tex]
Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.
