/*! 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 19 A library subscribes to two diff... [FREE SOLUTION] | 91Ó°ÊÓ

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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one \(P(\) Wed. \()=.3\), \(P(\) Thurs. \()=.4, \quad P\) (Fri. \()=.2\), and \(P(\) Sat. \()=.1\). Let \(Y=\) the number of days beyond Wednesday that it takes for both magazines to arrive (so possible \(Y\) values are 0,1 , 2, or 3 ). Compute the pmf of \(Y\). [Hint: There are 16 possible outcomes; \(Y(W, W)=0, Y(F, T h)=2\), and so on.]

Short Answer

Expert verified
The PMF of \(Y\) is: \(P(Y=0)=0.09\), \(P(Y=1)=0.4\), \(P(Y=2)=0.32\), \(P(Y=3)=0.19\).

Step by step solution

01

Identify Possible Outcomes

There are four possible arrival days (Wed, Thurs, Fri, Sat) for each magazine. Since they work independently, there are a total of \(4 \times 4 = 16\) possible outcomes for when both magazines can arrive. These outcomes include all combinations such as (Wed, Wed), (Wed, Thurs), (Thurs, Wed), etc.
02

Calculate Days Beyond Wednesday

Define \(Y\) as the number of days beyond Wednesday it takes for both magazines to arrive. For each outcome, calculate the \(Y\) value by taking the latest day from both magazine arrivals and subtract Wednesday (0 days). For example, if one arrives on Wed and the other on Thurs, use Thurs as the later day giving \(Y = 1\).
03

List the Possible Values of Y

The possible values of \(Y\) are \(0, 1, 2,\) and \(3\) corresponding to both magazines arriving on Wednesday progressing to both arriving by Saturday.
04

Determine Probability of Each Y

Calculate the probability for each \(Y\) value by summing the probabilities of job combinations matching \(Y\).- \(Y = 0\): Both arrive on Wednesday (\(P(\text{Wed})^2 = 0.3 \times 0.3 = 0.09\)).- \(Y = 1\): At least one arrives by Thursday while waiting only from Wednesday: - Combinations: (Wed, Thurs), (Thurs, Wed), (Thurs, Thurs) - Probability: \(2\times(0.3 \times 0.4) + (0.4 \times 0.4) = 0.24 + 0.16 = 0.4\)- \(Y = 2\): At least one arrives by Friday (but no later than Sat): - Combinations: (Wed, Fri), (Thurs, Fri), (Fri, Wed), (Fri, Thurs), (Fri, Fri) - Probability: \(2\times(0.3\times0.2) + 2\times(0.4\times0.2) + (0.2\times0.2) = 0.12 + 0.16 + 0.04 = 0.32\)- \(Y = 3\): At least one arrives by Saturday: - Combinations: (Wed, Sat), (Thurs, Sat), (Fri, Sat), (Sat, Wed), (Sat, Thurs), (Sat, Fri), (Sat, Sat) - Probability: \(3\times(0.3\times0.1) + (0.4\times0.1) + 2\times(0.2\times0.1) + (0.1\times0.1) = 0.09 + 0.04 + 0.04 + 0.01 = 0.19\)

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Key Concepts

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

Independent Events
In probability, independent events are those that do not influence each other's outcomes. When events are independent, the occurrence of one event does not affect the probability of the other event occurring. This concept is crucial when calculating probabilities for multiple events, like the arrival of two magazines in our exercise.
For the magazines, each can arrive on different days: Wednesday, Thursday, Friday, or Saturday. Since the arrival days are independent events, the arrival day of one magazine doesn't impact the other. This independence simplifies calculations, allowing us to find the combined probabilities by multiplying their individual probabilities.
For example, the probability of both magazines arriving on Wednesday is the product of their individual probabilities. Independently, each one has a 0.3 probability of arriving on Wednesday. Thus, the probability of both arriving on Wednesday together is: \[P(\text{Wed, Wed}) = 0.3 \times 0.3 = 0.09\] This principle is used repeatedly in the problem to find the total probabilities of all different outcomes.
Discrete Random Variables
A discrete random variable is a type of random variable that takes on a countable number of distinct values. These are values that can be listed out, typically integers like our variable \(Y\), which represents the number of days beyond Wednesday that it takes for both magazines to arrive.
Discrete random variables often appear in scenarios dealing with counts or specific categories. In our scenario, \(Y\) can take on values such as 0, 1, 2, or 3. These values cover all the possible days past Wednesday until Saturday that it might take for the two magazines to be received.
The nature of discrete random variables is essential in setting up probability distributions, where each possible value of the variable is assigned a probability. Knowing what values \(Y\) can take helps establish this distribution clearly, facilitating further probability calculations.
Probability Distribution
The concept of a probability distribution refers to a statistical function that describes all the possible values and likelihoods that a discrete random variable can take. In the context of our exercise, we calculate the probability distribution of \(Y\), the number of days beyond Wednesday for both magazines to arrive.
Each value of \(Y\) (0, 1, 2, or 3) corresponds to a specific probability calculated by examining all possible combinations of magazine arrivals that result in that particular \(Y\).
  • For \(Y = 0\), both magazines arrive on Wednesday, with a probability of 0.09.
  • For \(Y = 1\), at least one arrives by Thursday, calculated to have a probability of 0.4.
  • For \(Y = 2\), at least one arrives by Friday, with a resulting probability of 0.32.
  • For \(Y = 3\), at least one arrives by Saturday, which occurs with a probability of 0.19.
Ensuring each probability adds to 1 confirms the accuracy of the probability distribution:\[0.09 + 0.4 + 0.32 + 0.19 = 1\]This distribution fully characterizes the random variable \(Y\), providing insights into the timing you're likely to receive both magazines.

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