(c) In a communication system the signal sent from point X to point Y arrives by two paths in parallel. Over each path the signal passes through two repeaters (in series). Each repeater in one path has a probability of failing (becoming an open circuit) of 0.005. This probability is 0.008 for each repeater on the other path. All repeaters fail independently of each other. Find the probability that the signal will
(c) In a communication system the signal sent from point X to point Y arrives by two paths in parallel. Over each path the signal passes through two repeaters (in series). Each repeater in one path has a probability of failing (becoming an open circuit) of 0.005. This probability is 0.008 for each repeater on the other path. All repeaters fail independently of each other. Find the probability that the signal will
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![(c) In a communication system the signal sent from point X to point Y arrives by
two paths in parallel. Over each path the signal passes through two repeaters (in
series). Each repeater in one path has a probability of failing (becoming an open
circuit) of 0.005. This probability is 0.008 for each repeater on the other path. All
repeaters fail independently of each other. Find the probability that the signal will
not arrive at point Y.
Suppose the signal sent from point X to point Y arrives by two identical
paths in parallel, in which each path has three repeaters; the first path has
repeater R₁, linked to two other parallel repeaters R₂,and R3. The second path has
identical arrangement as the first path with repeaters R4, R5 and R6. The
probabilities of the repeaters failing (independently) are; p₁ = P(R₁) = 0.005,
P2 = P(R₂) = P(R3) = P(R4) = 0.01, p3 = P(R5) = P(R) = 0.05. Find the
probability that the signal will not arrive at point Y.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe53828e0-fbca-4901-a6df-51fa058b3532%2F2e068109-89ac-4236-8ec5-b38ee42c2692%2Fvs62l3r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(c) In a communication system the signal sent from point X to point Y arrives by
two paths in parallel. Over each path the signal passes through two repeaters (in
series). Each repeater in one path has a probability of failing (becoming an open
circuit) of 0.005. This probability is 0.008 for each repeater on the other path. All
repeaters fail independently of each other. Find the probability that the signal will
not arrive at point Y.
Suppose the signal sent from point X to point Y arrives by two identical
paths in parallel, in which each path has three repeaters; the first path has
repeater R₁, linked to two other parallel repeaters R₂,and R3. The second path has
identical arrangement as the first path with repeaters R4, R5 and R6. The
probabilities of the repeaters failing (independently) are; p₁ = P(R₁) = 0.005,
P2 = P(R₂) = P(R3) = P(R4) = 0.01, p3 = P(R5) = P(R) = 0.05. Find the
probability that the signal will not arrive at point Y.
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