Answer:
The probability is 0.00003986
Step-by-step explanation:
Given an event A :
[tex]P(A)=1-P(A^{c})[/tex]
Where [tex]A^{c}[/tex] is the event where A does not occur
Given the following events :
Path 1 : ''The signal arrives by path 1'' where the probability of each repeater to work is 1 - 0.002 = 0.998
Path 2 : ''The signal arrives by path 2'' where the probability of each repeater to work is 1 - 0.005 = 0.995
Given two events A and B :
P(A∪B) = P(A) + P(B) - P(A∩B)
And If A and B are independent events ⇒P(A∩B) = P(A).P(B)
P(Path 1) is the probability of both repeaters from path 1 working
[tex]P(Path 1)=0.998^{2}[/tex]
P(Path 2) is the probability of both repeaters from path 2 working
[tex]P(Path 2) =0.995^{2}[/tex]
P(Path 1 ∩ Path 2) = P(Path 1).P(Path 2) because all repeaters fail independently
P(Path 1 ∩ Path 2) = [tex](0.998^{2}).(0.995^{2})[/tex]
If we write the event A : ''The signal will not arrive at point b''
[tex]P(A) =1-P(A^{c})[/tex]
Where [tex]A^{c}[/tex] is the event where the signal arrives
[tex]P(A^{c})[/tex] = P [(Path 1) ∪ (Path 2)] = P(Path 1) + P(Path 2) - P( Path 1 ∩ Path 2)
[tex]P(A) =1-P(A^{c})\\P(A) = 1-[(0.998^{2})+(0.995^{2})-(0.998^{2})(0.995^{2})]\\P(A) = 0.00003986[/tex]
