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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. \()=.4, P(\) 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)=0.40, P(Y=2)=0.32, P(Y=3)=0.19.

Step by step solution

01

Identify Possible Outcomes

Each magazine can arrive on one of four days: Wednesday, Thursday, Friday, or Saturday. Thus, there are a total of 16 possible outcomes for the arrival of the two magazines (4 days for magazine 1 × 4 days for magazine 2 = 16 outcomes).
02

Define the Random Variable Y

The random variable \( Y \) represents the number of days beyond Wednesday that it takes for both magazines to arrive. That means:- \( Y = 0 \) if both magazines arrive on Wednesday.- \( Y = 1 \) if the later arriving magazine arrives on Thursday.- \( Y = 2 \) if the later arriving magazine arrives on Friday.- \( Y = 3 \) if the later arriving magazine arrives on Saturday.
03

Calculate Probability for Y=0

\( Y = 0 \) occurs if both magazines arrive on Wednesday. The probability is computed as: \[ P(Y=0) = P(W, W) = 0.3 \times 0.3 = 0.09 \]
04

Calculate Probability for Y=1

\( Y = 1 \) occurs if the later magazine arrives on Thursday. This happens in the cases of (W, T), (T, W), or (T, T). The probability is \[ P(Y=1) = P(W, T) + P(T, W) + P(T, T) = (0.3 \times 0.4) + (0.4 \times 0.3) + (0.4 \times 0.4) = 0.12 + 0.12 + 0.16 = 0.40 \]
05

Calculate Probability for Y=2

\( Y = 2 \) occurs if the later magazine arrives on Friday. Possible scenarios are (W, F), (T, F), (F, W), (F, T), or (F, F). Calculate:\[ P(Y=2) = P(W, F) + P(T, F) + P(F, W) + P(F, T) + P(F, F) \] \[ = (0.3 \times 0.2) + (0.4 \times 0.2) + (0.2 \times 0.3) + (0.2 \times 0.4) + (0.2 \times 0.2) \]\[ = 0.06 + 0.08 + 0.06 + 0.08 + 0.04 = 0.32 \]
06

Calculate Probability for Y=3

\( Y = 3 \) occurs when the later magazine comes on Saturday. This happens for (W, S), (T, S), (F, S), or (S, _). Calculate:\[ P(Y=3) = P(W, S) + P(T, S) + P(F, S) + P(S, W) + P(S, T) + P(S, F) + P(S, S) \]\[ = (0.3 \times 0.1) + (0.4 \times 0.1) + (0.2 \times 0.1) + (0.1 \times 0.3) + (0.1 \times 0.4) + (0.1 \times 0.2) + (0.1 \times 0.1) \]\[ = 0.03 + 0.04 + 0.02 + 0.03 + 0.04 + 0.02 + 0.01 = 0.19 \]
07

Present the PMF

The pmf of \( Y \) for the given problem is as follows:- \( P(Y=0) = 0.09 \)- \( P(Y=1) = 0.40 \)- \( P(Y=2) = 0.32 \)- \( P(Y=3) = 0.19 \)

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Key Concepts

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

Independent Events
In probability theory, when two events are described as independent, it means that the occurrence of one event does not affect the likelihood of the other event occurring. In the given problem, the library receives two different news magazines through the mail on separate days of the week. The arrival of each magazine is an independent event. Whether one magazine arrives on a particular day does not influence the arrival of the other magazine on the same or any other day.

To better understand this concept, consider tossing two coins. The result of the first coin (heads or tails) does not affect the result of the second coin. Let's relate this to our magazine problem:
  • Each magazine has a fixed probability for arriving on any given day.
  • The probability for one magazine to arrive on a specific day doesn't change based on the arrival day of the other magazine.
This is why in the calculations provided, we multiply the individual probabilities of each magazine's arrival day to find the combined probability of both magazines arriving on particular days. The independence of these events simplifies the analysis because we can consider each magazine's arrival independently.
Random Variables
A random variable is a numerical outcome of a random process. It assigns numerical values to each event in a sample space. In the context of this exercise, the random variable designated as \( Y \) signifies the number of days beyond Wednesday that it takes for both magazines to arrive. This kind of variable is crucial for translating qualitative outcomes into quantitative data that can be analyzed using probability.

Here’s how \( Y \) is defined for the given problem:
  • \( Y = 0 \): Both magazines arrive on Wednesday.
  • \( Y = 1 \): At least one magazine arrives on Thursday, and no later arrival occurs.
  • \( Y = 2 \): At least one magazine arrives on Friday, with none arriving later.
  • \( Y = 3 \): The latest arrival is on Saturday.
Random variables provide a structured way to evaluate and predict outcomes from complex real-world scenarios like magazine deliveries.
Discrete Probability Distributions
A discrete probability distribution describes probabilities for outcomes of a discrete random variable. These distributions can take on a finite or countably infinite number of possible outcomes. In our problem, the random variable \( Y \) has four possible outcomes for the days: 0, 1, 2, and 3. The associated probabilities for these outcomes form the Probability Mass Function (PMF) of \( Y \).

The PMF gives us:
  • \( P(Y=0) = 0.09 \)
  • \( P(Y=1) = 0.40 \)
  • \( P(Y=2) = 0.32 \)
  • \( P(Y=3) = 0.19 \)
Each of these probabilities represents the likelihood that \( Y \) takes on a particular value, which corresponds to how late the second magazine arrives. Understanding discrete probability distributions is essential for making predictions and deriving insights from a set of distinct random events. This allows organizations like the library to plan and prepare for probable occurrences effectively.

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