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:
<em>A</em> = video components need repair within 1 year
<em>B</em> = electronic components need repair within 1 year
<em>C</em> = 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 <em>A</em>, <em>B</em> and <em>C</em> 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)
![=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](https://tex.z-dn.net/?f=%3D1-P%28A%5E%7Bc%7D%5Ccap%20B%5E%7Bc%7D%5Ccap%20C%5E%7Bc%7D%29%5C%5C%3D1-%5BP%28A%5E%7Bc%7D%29%5Ctimes%20P%28B%5E%7Bc%7D%29%5Ctimes%20P%28C%5E%7Bc%7D%29%5D%5C%5C%3D1-%5B%281-0.02%29%5Ctimes%20%281-0.007%29%5Ctimes%20%281-0.001%29%5D%5C%5C%3D1-0.97216686%5C%5C%3D0.02783314%5C%5C%5Capprox%200.0278)
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)
![=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](https://tex.z-dn.net/?f=%3DP%28A%5Ccap%20B%5E%7Bc%7D%5Ccap%20C%5E%7Bc%7D%29%2BP%28A%5E%7Bc%7D%5Ccap%20B%5Ccap%20C%5E%7Bc%7D%29%2BP%28A%5E%7Bc%7D%5Ccap%20B%5E%7Bc%7D%5Ccap%20C%29%5C%5C%3D%5B0.02%5Ctimes%20%281-0.007%29%5Ctimes%20%281-0.001%29%5D%2B%5B%281-0.02%29%5Ctimes%200.007%5Ctimes%20%281-0.001%29%5D%5C%5C%2B%5B%281-0.02%29%5Ctimes%20%281-0.007%29%5Ctimes%200.001%5D%5C%5C%3D0.02766642%5C%5C%5Capprox%200.0277)
Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.
