Question 1118546: In a home theater system, the probability that the video components need repair within 1 year is 0.03, the probability that the electronic components need repair within 1 year is 0.007, and the probability that the audio components need repair within 1 year is 0.004. Assuming that the events are independent, find the following probabilities. (Round your answers to four decimal places.)
(a) At least one of these components will need repair within 1 year
(b) Exactly one of these component will need repair within 1 year
Answer by ikleyn(52864)   (Show Source):
You can put this solution on YOUR website! .
In a home theater system, the probability that the video components need repair within 1 year is 0.03,
he probability that the electronic components need repair within 1 year is 0.007,
and the probability that the audio components need repair within 1 year is 0.004.
Assuming that the events are independent, find the following probabilities. (Round your answers to four decimal places.)
(a) At least one of these components will need repair within 1 year.
According to the condition, the probability that
V-component will not need a repair within 1 year is (1-0.03),
E-component will not need a repair within 1 year is (1-0.007),
A-component will not need a repair within 1 year is (1-0.004).
Hence, the probability that NO ONE of the three components will not need a repair is the product (1-0.03)*(1-0.007)*(1-0.004).
Then the probability under the question has the complementary value
1 - (1-0.03)*(1-0.007)*(1-0.004) = 0.0406 = 4.06%.
(b) Exactly one of these components will need repair within 1 year.
(1-0.03)*(1-0.007)*0.004 + (1-0.03)*0.007*(1-0.004) + 0.03*(1-0.007)*(1-0.004) = 0.04030 = 4.030%.
Explanation.
The probability that only A-component will need a repair within 1 year is (1-0.03)*(1-0.007)*0.004.
The probability that only E-component will need a repair within 1 year is (1-0.03)*0.007*(1-0.004).
The probability that only V-component will need a repair within 1 year is 0.03*(1-0.007)*(1-0.004).
The probability that exactly one of these three independent events will happen within 1 year is the sum
of these particular probabilities, which coincides with my answer.
|
|
|
