Question

Three highways connect city A with city B. Two highways connect city B with city C.

During a rush hour, each highway is blocked by a traffic accident with probability 0.2,
independently of other highways.
(a) Compute the probability that there is at least one open route from A to C.
(b) How will a new highway, also blocked with probability 0.2 independently of other
highways, change the probability in (a) if it is built
(α) between A and B?
(β) between B and C?
(γ) between A and C?

Answer 1

(a) The probability that there is no open route from A to B is (0.2)^3 = 0.008.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:

`0.992*0.96=0.9523`

(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become:
`0.9984*0.96=0.95846`

β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:

`0.992*0.992=0.9841`

Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:

`P(A\cup B)=P(A)+P(B)-P(A\cap B)`

`0.9523+0.8-(0.9523*0.8)=0.99`
Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.