MATH SOLVE

3 months ago

Q:
# In a communication system the signal sent from point a to point b arrives by two paths in parallel. Over each path the signal passes through two repeaters connected in series. Each repeater in one path has a probability of failing (becoming an open circuit) of 0.002. For the other path, the probability of a repeater failing is 0.005. All repeaters fail independently of each other. Find the probability that the signal will not arrive at point b.

Accepted Solution

A:

Answer:The probability is 0.00003986Step-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 occurGiven the following events : Path 1 : ''The signal arrives by path 1'' where the probability of each repeater to work is 1 - 0.002 = 0.998Path 2 : ''The signal arrives by path 2'' where the probability of each repeater to work is 1 - 0.005 = 0.995Given 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 independentlyP(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]