Question

Three highways connect city A with city B. Two highways connect city B with city 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:
$0.992\times0.96=0.9523$

(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:$0.9984\times0.96=0.95846$

β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:
$0.992\times0.992=0.9841$

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:
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$0.9523+0.8-(0.9523\times0.8)=0.99$
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.