During the winter season at the Fearsome Peaks Ski Resort,...

During the winter season at the Fearsome Peaks Ski Resort, each of the two roads from Area A to Area B has probability $p$ of being blocked by snow. The same can be said of the two roads that lead from Area B to Area C; that is, all roads, independently, have probability $p$ of being blocked by snow. (a) What is the probability that there is an open path from Area A to Area C? (b) Having calculated the probability in part (a) when $p=1 / 2$, the owners of FPSR decide to build a direct road from Area A to Area C, which, independently of the other roads, is blocked with probability $p$. What is the new probability there is an open road from Area A to Area C? (c) If $p=0.25$, calculate the probabilities of part (a) and part (b).

During the winter season at the Fearsome Peaks Ski Resort, each of the two roads from Area A to Area B has probability $p$ of being blocked by snow. The same can be said of the two roads that lead from Area B to Area C; that is, all roads, independently, have probability $p$ of being blocked by snow.
(a) What is the probability that there is an open path from Area A to Area C?
(b) Having calculated the probability in part (a) when $p=1 / 2$, the owners of FPSR decide to build a direct road from Area A to Area C, which, independently of the other roads, is blocked with probability $p$. What is the new probability there is an open road from Area A to Area C?
(c) If $p=0.25$, calculate the probabilities of part (a) and part (b).

Key Concepts

Independent Events

This concept involves events whose outcomes do not affect each other. In the problem, every road being blocked or open happens independently of the others. Calculations such as multiplying probabilities of open roads (or blocked roads) rely on the fact that these events do not influence one another’s outcomes.

Series and Parallel Systems

These terms describe the configuration of components in a network. In a series system, the overall functionality depends on every component working (for example, both a road from A to B and a road from B to C must be open to form a complete path). In parallel systems, the system works if at least one component functions (for instance, having two roads connecting the same areas means that if one is blocked, the other might still be open). Understanding these arrangements is key to analyzing the connectivity probabilities in the network.

Complement Principle

The complement principle is a method used to calculate the probability of an event by subtracting the probability of its complement from 1. This is particularly useful in network reliability problems, as calculating the likelihood that no path exists (all roads blocked) can be simpler than directly summing the probabilities of all possible open paths.

Network Reliability

This concept pertains to the probability that a given network remains operational despite the possibility of individual component failures. In the question, it involves determining the probability that there exists at least one open route from Area A to Area C, considering multiple potential paths and the redundancy offered by parallel roads. This analysis is central in fields such as transportation planning and communication networks.

Transcript

00:01 For question number 19, the corresponding distribution is a console distribution because an average rate is given.

00:09 That is not a probability for which you would use the binomial distribution.

00:13 So the average rate is once every 60 hours.

00:21 Once every 60 hours.

00:28 So it's equal to 1.

00:31 To point b the average rate is once every 60 hours so lambda is equal to 1 by 108 over 60 equals to 1 .a notes that we multiply the rate by 108 over 60 because we have 108 hours instead of 60 hours so the formula for solve probability is for p when x equal to k is equal to lambda power k by e power negative lambda over factorization k.

01:18 To evaluate at k equal to 0 and 1 and 2 and 3, when x equal to 0, we get 1 .840 by e.

01:39 4 negative 1 .8 over factorization 0 .5 ,000, which is approximately equal 0 .1 ,633.

01:51 But u and x equal to 1, equal to 1 .8 1, 1 .1, y, equal negative, 1 .8 over 1 ,000, over factorization 1, will approximate equal 0 .275.

02:09 For p .1 .x equal to 2, equal to 1 .8 2 by e.

02:17 Power 2 .1 million 5 .8 over factorization 2.

02:20 Well, a proximate equal 0 .2 ,678.

02:27 For view p x equal to stream, equal to 1 .8 by e .p.

02:37 Negative 1 .8 or v.

02:39 Factorization 3 or atuipsunate equal 0 .1 .6.

02:44 0 .1 .6707.

02:49 For p.

02:50 When x equal to 4 or ap equal to 1 .8 power 4 by e...