Respuesta :
Answer:
(a) Probability that at least one component needs service during the warranty period = 0.4044.
(b) Probability that exactly one of the components needs service during the warranty period = 0.3419.
Step-by-step explanation:
Given A1 be the event that the receiver functions properly throughout the warranty period.
A2 be the event that the speakers function properly throughout the warranty period.
A3 be the event that the CD player functions properly throughout the warranty period.
Also P(A1) = 0.91, P(A2) = 0.85, and P(A3) = 0.77.
Now P(A1)' means that the receiver need service during the warranty period which is 1 - P(A1) = 1 - 0.91 = 0.09.
Similarly,P(A2)' =1 - P(A2) =1 - 0.85 =0.15 and P(A3)' =1 - P(A3)=1 - 0.77 = 0.23
Note: The ' sign on the P(A1) represent compliment of A1 or not A1.
(a) The probability that at least one component needs service during the warranty period = 1 - none of the component needs service during the warranty period
And none of the component needs service during the warranty period means that all the three components functions properly during the warranty period .
So, Probability that at least one component needs service during the warranty period = 1 - P(A1) x P(A2) x P(A3) = 1 - (0.91 x 0.85 x 0.77) = 0.4044.
(b) Now to find the Probability that exactly one of the components needs service during the warranty period, there would be three cases for this:
- Receiver needs service and other two does not need during the warranty period.
- Speaker needs service and other two does not need during the warranty period.
- CD player needs service and other two does not need during the warranty period.
And we have to add these three cases to calculate above probability.
Probability that exactly one of the components needs service during the warranty period = P(A1)' x P(A2) x P(A3) + P(A1) x P(A2)' x P(A3) + P(A1) x P(A2) x P(A3)'
= 0.09 x 0.85 x 0.77 + 0.91 x 0.15 x 0.77 + 0.91 x 0.85 x 0.23
= 0.3419.
Answer:
(a) P (At least one component needs service) = 0.404
(b) P (Either component A₁ or A₂ or A₃) = 0.997
Step-by-step explanation:
Given:
[tex]A_{1}=[/tex] Event that the receiver functions properly throughout the warranty period.
[tex]A_{2}=[/tex] Event that the speakers function properly throughout the warranty period.
[tex]A_{3}=[/tex] Event that the CD player functions properly throughout the warranty period.
[tex]P(A_{1})=0.91,\ P(A_{2})=0.85\ and\ P(A_{3})=0.77[/tex]
(a)
Compute the probability that at least one component needs service during the warranty period:
P (At least one component needs service) = 1 - P (None of the one component needs service)
= 1 - {P ([tex]A_{1}[/tex]) × P ([tex]A_{2}[/tex]) × P ([tex]A_{3}[/tex])}
[tex]=1 - (0.91\times0.85\times0.77)\\=1-0.595595\\=0.404405\\\approx0.404[/tex]
Thus, the probability that at least one component needs service during the warranty period is 0.404.
(b)
Compute the probability that exactly one of the components needs service during the warranty period, i.e. P (Either A₁ or A₂ or A₃):
[tex]P(A_{1}\cup A_{2}\cup A_{3})=P(A_{1})+P(A_{2})+P(A_{3})-P(A_{1}\cap A_{2})-P(A_{2}\cap A_{3})-P(A_{3}\cap A_{1})+P(A_{1}\cap A_{2}\cap A_{3})\\=P(A_{1})+P(A_{2})+P(A_{3})-[P(A_{1})\tmes P(A_{2})]-[P(A_{2})\tmes P(A_{3})]-[P(A_{3})\tmes P(A_{1})] +[P(A_{1})\tmes P(A_{2})\times P(A_{3})]\\=0.91+0.85+0.77-(0.91\times0.85)-(0.85\times0.77)-(0.77\times0.91)+(0.91\times0.85\times0.77)\\=0.996895\\\approx0.997[/tex]
Thus, the probability that exactly one of the components needs service during the warranty period is 0.997
