/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 115 Two sisters, Allison and Teri, h... [FREE SOLUTION] | 91Ó°ÊÓ

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Two sisters, Allison and Teri, have agreed to meet between 1 and 6 P.M. on a particular day. In fact, Allison is equally likely to arrive at exactly 1 P.M., 2 P.M., 3 P.M., 4 P.M., 5 P.M., or 6 P.M. Teri is also equally likely to arrive at each of these six times, and Allison's and Teri's arrival times are independent of one another. Thus there are 36 equally likely (Allison, Teri) arrival-time pairs, for example, \((2,3)\) or \((6,1)\). Suppose that the first person to arrive waits until the second person arrives; let \(w\) be the amount of time the first person has to wait. a. What is the probability distribution of \(w\) ? b. How much time do you expect to elapse between the two arrivals?

Short Answer

Expert verified
The probability distribution of \(w\) is \(6/36, 10/36, 8/36, 6/36, 4/36, 2/36\) for waiting times 0,1,2,3,4,5 hours respectively. The expected waiting time or elapsed time between the two arrivals is given by \(E(w) = 0*(6/36) + 1*(10/36) + 2*(8/36) + 3*(6/36) + 4*(4/36) + 5*(2/36)\)

Step by step solution

01

Identifying the Time Pairs and Possibilities

There are a total of 6 possible arrival time for each sister making a total of \(6*6=36\) arrival time pairs.
02

Calculate the Possible Waiting Times

The possible waiting times (in hours) can be 0, 1, 2, 3, 4, and 5. The waiting time is 0 if and only if both arrive at the same time. So, the frequency for 0 waiting time is 6 (considering times such as 1 and 1, 2 and 2, etc.) Since, for a waiting time of 1, 2, 3, 4, 5 hours, there are two possibilities: either Allison or Teri comes first. Therefore, the waiting time frequencies for 1,2,3,4,5 can be calculated as \(2(5,4,3,2,1)= 2(15) = 30\) respectively.
03

Construct the Probability Distribution table

The probability distribution of \(w\) will be calculated considering the frequency of waiting time over total number of outcomes. Given the waiting times as 0,1,2,3,4,5 hours and their respective frequencies 6,10,8,6,4,2, the corresponding probabilities would be \(6/36, 10/36, 8/36, 6/36, 4/36, 2/36\) respectively.
04

Determine Expected Time of Arrival

The expected time of arrival is calculated as the sum of product of each waiting time and its corresponding probability. Set \(E(w)\) as the expected time of arrival, hence \(E(w) = 0*(6/36) + 1*(10/36) + 2*(8/36) + 3*(6/36) + 4*(4/36) + 5*(2/36)\)

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Key Concepts

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

Independent Events
In probability theory, independent events play a crucial role. Two events are considered independent if the occurrence of one does not affect the occurrence of the other. In the context of our exercise, Allison's and Teri's arrival times are independent events. This means the time when Allison arrives has no bearing on the time when Teri arrives.
For example, if Allison arrives at 2 P.M., it does not change the probability of Teri arriving at any given time. Her probabilities remain equally distributed among 1 P.M., 2 P.M., 3 P.M., 4 P.M., 5 P.M., and 6 P.M. This independency forms the backbone of calculating the joint probabilities for the arrival-time pairs.
Understanding independence is essential because it allows us to calculate the probability of two or more events happening together simply by multiplying their individual probabilities. By knowing their independence, we can convenient construct the probability model for their meeting scenario.
Expectation Calculation
The concept of expectation is vital in understanding what is 'expected', on average, from a random experiment. In the given exercise, the expectation of the waiting time \(w\) describes how much time we predict would lapse on average when the first person who arrives waits for the second.
To calculate expectation, \(E(w)\), you must multiply each possible waiting time by its probability and sum these values. Using the step-by-step solution provided, \(E(w) = 0 \times (6/36) + 1 \times (10/36) + 2 \times (8/36) + 3 \times (6/36) + 4 \times (4/36) + 5 \times (2/36)\).
  • This weighted sum captures the average waiting time by factoring in both the duration of wait and the probability of that wait occurring.
  • Expectation is a form of weighted average - each outcome is weighed by its likelihood of occurrence.
Calculating expectation provides users with a succinct summary of a random variable behavior.
Random Variable
Random variables are tools used to quantify outcomes of probabilistic events. They assign numbers to each of these potential outcomes. In this exercise, \(w\) is the random variable representing the waiting time between Allison and Teri's arrivals.
Random variables can be discrete or continuous. In our case, \(w\) is discrete because it can take on only specific integer values: 0, 1, 2, 3, 4, and 5.
  • Discrete random variables have a probability distribution that lists all possible values the random variable can assume and their corresponding probabilities.
  • This is how we end up with a probability distribution table showing the probabilities for each waiting time.
Understanding random variables and how they relate to probability distributions is foundational for analyzing and predicting outcomes in probabilistic settings.

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