A traffic network is shown in the figure below. The reliability (probability of

Question

A traffic network is shown in the figure below. The reliability (probability of free-flowing without congestion) of link B is 0.76, link D is 0.82, and link E is 0.86: these are all statistically independent of each other and of other links in the network. The overall reliability of Link A is 0.65. However, if A is congested, then the reliability of C is 0.4, whereas if A is not congested, C has a reliability of 0.72. Please answer the following questions using the terminology "A = A is free-flowing: B = B is free-flowing" and so on. How do you write the event "A vehicle can travel from point 1 to point 2 without encountering congestion." in terms of A, B, C, D, and E? You may use "U" (capital u) to indicate unions, and "^" (caret, above the 6 on most keyboards) or "n" to indicate intersections. Continuing with the traffic question, find the probability of vehicles being able to travel freely from point 1 to point 2 under the assumption that routes A^B and C^D^E are mutually exclusive. (3 decimal places, 0.XXX) Continuing with the traffic question, if vehicles can flow freely, what is the probability that A is congested? (3 decimal places, 0.XXX)

Explanation / Answer

7:

Vehicle can take either root AB ot root CDE. So it can be written as

(A^B) U (C^D^E)

8:

Since A and B are independent so

P(A^B) = P(A)P(B) = 0.65*0.76 = 0.494

By the complement rule,

P(not A) = 1 -P(A) = 1 - 0.65 = 0.35

From the given information,

P(C|not A) = 0.4, P(C| A) = 0.72

By the law of total probability,

P(C) = P(C|A)P(A) + P(C|not A)P(not A) = 0.72 * 0.65 + 0.4 * 0.35 = 0.468 + 0.14 = 0.608

P(C^D^E) = P(C)P(D)P(E) = 0.608 * 0.82 * 0.86 = 0.4287616

So the probability of vehicles being able to travel freely from point 1 to point 2 so

P(traffic flow freely) =P[ (A^B) U (C^D^E) ] = P(A^B) + P(C^D^E) = 0.494 + 0.4287616= 0.9227616

Answer: 0.923

9)

The probability that A is congested (that is traffic move from path C) given that traffic flow freely is

P(A is congested | traffic move freely) = 0.4287616 / 0.9227616 = 0.4647

Answer: 0.465

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