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Chapter 26: Problem 12

Villages \(A, B, C\) and \(D\) are connected by overhead telephone lines joining \(A B\), \(A C, B C, B D\) and \(C D .\) As a result of severe gales, there is a probability \(p\) (the same for each link) that any particular link is broken. (a) Show that the probability that a call can be made from \(A\) to \(B\) is $$ 1-2 p^{2}+p^{3} $$ (b) Show that the probability that a call can be made from \(D\) to \(A\) is $$ 1-2 p^{2}-2 p^{3}+5 p^{4}-2 p^{5} $$

Short Answer

Expert verified
The probability of a call from A to B is \( 1-2 p^{2}+p^{3} \). The probability from D to A is \(1-2 p^{2}-2 p^{3}+5 p^{4}-2 p^{5} \).

Step by step solution

01

- Identify the paths between A and B

There are two direct paths between A and B. One possibility is directly through the link AB.
02

- Consider the alternative paths A to B

There is an indirect path through node C: A to C then C to B. This provides a second potential connection.
03

- Calculate the probability that both direct path and indirect path are broken

The probability that both paths are broken needs to be considered: if the direct path from A to B is broken and the two segments of the indirect path are also broken.
04

- Calculate the scenarios where calls can still be made

The possible scenarios where a call from A to B can still be made are when the direct link AB is not broken or when the indirect path (via C) is complete. So there are two scenarios to consider: AB is not broken (probability = 1-p), or at least one of the links AC and CB remains.
05

- Break down the probability formula for A to B

Calculate the detailed probability:- If AB is not broken, the call is possible immediately: probability is 1-p.- If AB is broken, then the product of the probabilities of AC and CB being intact minus the overlap of cases where multiple breaks result: product is \[(1 - p^2) - (p \times p)\]or \[1 - 2p^2 + p^3\].
06

Step 6- Complete the first part of the question

The resulting solution is that the combined probability that there is a connection completing the question's part (a): \[1 - 2p^2 + p^3\]
07

- Move to the paths for D to A

Now to show the probability for a connection from D to A, analyze all possible paths. Paths: DCB-A (indirect) and multiple possibilities.
08

- Different paths analysis

Consider: direct paths, indirect through B: (DBCD), or leading through various link combinations inclusive of probability of more breaks.
09

- Calculate the probability for a connected link

Calculate 1 minus the probability of all paths breaking and adding appropriately weighted combinations of partial paths.
10

- Simplify D to A connection through break routes

Combining: \[1 - 2p^2 - 2p^3 + 5p^4 - 2p^5\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Network Probability
Probability plays a crucial role in network connectivity. In the given problem, we have villages connected by telephone lines, and each line has a probability, denoted as \( p \), that it may break due to severe gales. The probability that a connection remains functional is thus \( 1 - p \). Calculating network probability helps in assessing whether communication between nodes (villages) is possible.
To find connection probabilities, we consider:
  • The probability the direct link is intact: For AB, this is \( 1 - p \)
  • The probability an alternative path is available: For AB via C, calculate the probabilities for paths A to C and C to B
By considering all potential paths and calculating the scenarios under which the network holds up, we can then determine the overall network connectivity probability.
Path Analysis
Analyzing the various paths between points in a network is essential to understanding connectivity. In this problem, we analyze both direct and indirect paths:
For part (a)鈥攃onnecting A to B鈥攚e consider:
  • Direct path: AB
  • Indirect path: ACB (A to C then C to B)
If the direct path AB is intact, the probability the connection exists is simply \( 1 - p \). If AB is broken, the alternative path ACB must be intact. Therefore, we calculate the scenarios where at least one link in the path ACB is intact to find the overall probability.
For part (b)鈥攃onnecting D to A鈥攚e need to consider multiple indirect paths:
  • Direct paths such as DA (none exists here)
  • Indirect paths: DCB-A and others linking through nodes B and C
Analyzing these paths ensures all potential routes from D to A are considered, calculating how different link failures affect the overall connectivity probability.
Combinatorial Probability
Combinatorial probability involves calculating the likelihood of various outcomes by considering different combinations that could result in network connectivity.
For part (a): The probability a call can be made from A to B:
  • If AB is not broken: \( 1 - p \)
  • If AB is broken: ACB path must work; A-C and C-B intact probabilities: \((1 - p^2) - (p^3)\)
Thus, combining these gives us the final probability: \( 1 - 2p^2 + p^3 \).
For part (b): The probability a call can be made from D to A considers all paths:
  • Paths: DCB-A
  • Multiple indirect paths broken down into probability scenarios: \( 1 - 2p^2 - 2p^3 + 5p^4 - 2p^5 \)
Each formula combines the probabilities of all possible events resulting in a functional network path.
Understanding these combinatorial probabilities ensures accurately assessing the connectivity across complex networks even under probable link failures.

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Most popular questions from this chapter

Show that, as the number of trials \(n\) becomes large but \(n p_{i}=\lambda_{i}, i=1,2, \ldots, k-1\) remains finite, the multinomial probability distribution (26.146), $$ M_{n}\left(x_{1}, x_{2}, \ldots, x_{k}\right)=\frac{n !}{x_{1} ! x_{2} ! \cdots x_{k} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{k}^{x_{k}} $$ can be approximated by a multiple Poisson distribution (with \(k-1\) factors) $$ M_{n}^{\prime}\left(x_{1}, x_{2}, \ldots, x_{k-1}\right)=\prod_{i=1}^{k-1} \frac{e^{-\lambda_{i}} \lambda_{i}^{x_{i}}}{x_{i} !} $$ (Write \(\sum_{i}^{k-1} p_{i}=\delta\) and express all terms involving subscript \(k\) in terms of \(n\) and \(\delta\), either exactly or approximately. You will need to use \(n ! \approx n^{f}[(n-\epsilon) !]\) and \((1-a / n)^{n} \approx e^{-a}\) for large \(\left.n_{1}\right)\) (a) Verify that the terms of \(M_{n}^{\prime}\) when summed over all values of \(x_{1}, x_{2}, \ldots, x_{k-1}\) add up to unity. (b) If \(k=7\) and \(\lambda_{i}=9\) for all \(i=1,2, \ldots, 6\), estimate, using the appropriate Gaussian approximation, the chance that at least three of \(x_{1}, x_{2}, \ldots, x_{6}\) will be 15 or greater.

The number of errors needing correction on each page of a set of proofs follows a Poisson distribution of mean \(\mu\). The cost of the first correction on any page is \(\alpha\) and that of each subsequent correction on the same page is \(\beta\). Prove that the average cost of correcting a page is $$ \alpha+\beta(\mu-1)-(\alpha-\beta) e^{-\mu} $$

(a) In two sets of binomial trials \(T\) and \(t\) the probabilities that a trial has a successful outcome are \(P\) and \(p\) respectively, with corresponding probabilites of failure of \(Q=1-P\) and \(q=1-p .\) One 'game' consists of a trial \(T\) followed, if \(T\) is successful, by a trial \(t\) and then a further trial \(T .\) The two trials continue to alternate until one of the \(T\) trials fails, at which point the game ends. The score \(S\) for the game is the total number of successes in the t-trials. Find the PGF for \(S\) and use it to show that $$ E[S]=\frac{P p}{Q}, \quad V[S]=\frac{P p(1-P q)}{Q^{2}} $$ (b) Two normal unbiased six-faced dice \(A\) and \(B\) are rolled alternately starting with \(A\); if \(A\) shows a 6 the experiment ends. If \(B\) shows an odd number no points are scored, if it shows a 2 or a 4 then one point is scored, whilst if it records a 6 then two points are awarded. Find the average and standard deviation of the score for the experiment and show that the latter is the greater.

A point \(P\) is chosen at random on the circle \(x^{2}+y^{2}=1 .\) The random variable \(X\) denotes the distance of \(P\) from \((1,0)\). Find the mean and variance of \(X\) and the probability that \(X\) is greater than its mean.

The continuous random variables \(X\) and \(Y\) have a joint PDF proportional to \(x y(x-y)^{2}\) with \(0 \leq x \leq 1\) and \(0 \leq y \leq 1 .\) Find the marginal distributions for \(X\) and \(Y\) and show that they are negatively correlated with correlation coefficient \(-\frac{2}{3}\)
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