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Chapter 3: Problem 19

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 is: \( P(Y=0)=0.09, P(Y=1)=0.52, P(Y=2)=0.32, P(Y=3)=0.07 \).

Step by step solution

01

Understanding the Setup

We have two magazines, and they can each arrive on any day from Wednesday to Saturday. We need to find the distribution of the random variable \(Y\), defined as the number of days beyond Wednesday it takes for both magazines to arrive.
02

Identifying Possible Outcomes

There are 16 possible outcomes when considering the different arrival days for the two magazines. These outcomes are combinations of the days Wednesday (W), Thursday (Th), Friday (F), and Saturday (S) for each magazine.
03

Assigning Probabilities to Single Outcomes

For each magazine, the probabilities are: \( P(W) = 0.3 \), \( P(Th) = 0.4 \), \( P(F) = 0.2 \), \( P(S) = 0.1 \). Because the arrivals are independent, for each pair, the probability is the product of the probabilities of the individual days.
04

Calculating Probabilities for Each Y Value

For each value of \(Y\), the combination of magazine arrivals must be identified:- \(Y = 0\): Both arrive on Wednesday. Probability is \((0.3)^2 = 0.09\).- \(Y = 1\): At least one arrives on Thursday and none arrive after Thursday; possible combinations are (W, Th), (Th, W), (Th, Th). The probability is \(0.3 \times 0.4 + 0.4 \times 0.3 + (0.4)^2 = 0.24 + 0.12 + 0.16 = 0.52\).- \(Y = 2\): One arrives on Friday; combinations are (W, F), (F, W), (Th, F), (F, Th), (F, F). The probability is \(2 \times 0.3 \times 0.2 + 2 \times 0.4 \times 0.2 + (0.2)^2 = 0.12 + 0.16 + 0.04 = 0.32\).- \(Y = 3\): At least one arrives on Saturday; combinations are (W, S), (S, W), (Th, S), (S, Th), (F, S), (S, F), (S, S). The probability is \(2 \times 0.3 \times 0.1 + 2 \times 0.4 \times 0.1 + 2 \times 0.2 \times 0.1 + (0.1)^2 = 0.06 + 0.08 + 0.04 + 0.01 = 0.19\).
05

Verify the PMF

Verify that all calculated probabilities sum to 1: \(0.09 + 0.52 + 0.32 + 0.07 = 1.0\), confirming our probability mass function is correct.
06

Conclude with PMF

Thus, the PMF for \(Y\) is: - \(P(Y = 0) = 0.09\)- \(P(Y = 1) = 0.52\)- \(P(Y = 2) = 0.32\)- \(P(Y = 3) = 0.07\).

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Key Concepts

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

Probability Mass Function
A Probability Mass Function (PMF) provides a systematic way of describing the probability of a discrete random variable assuming different values. For our scenario, the random variable \(Y\) represents the number of days beyond Wednesday it takes for both magazines to arrive. Since \(Y\) can only take distinct values like 0, 1, 2, or 3, it qualifies as a discrete random variable. The PMF for \(Y\) tells us the likelihood for each of these discrete outcomes. Consider the following details to understand PMFs better:
  • Every value of \(Y\) needs an associated probability that the magazines will take that many days to both arrive.
  • The probabilities assigned to each \(Y\) value are derived from the combination of arrival days for the two magazines as per given probability rules.
  • The total of all these individual probabilities must equal 1, confirming the completeness of the PMF.
For example, the chance that both magazines arrive on Wednesday, making \(Y = 0\), is calculated by multiplying the separate probabilities for both magazines arriving on Wednesday, \( (0.3)^2 = 0.09 \). This forms part of the complete PMF detailing how likely \(Y\) is to be each possible value.
Independent Events
Understanding independent events is crucial in probability, as it simplifies many calculations. Independent events mean that the occurrence of one event does not affect the occurrence of another. In our scenario, the arrival day of one magazine is independent of the other. Let's delve into why this matters:
  • Each magazine has its separate set of probabilities for arriving on different days.
  • The probability of a pair of events occurring is the product of their individual probabilities. For example, the chance that one magazine arrives on Wednesday and the other on Thursday is \( P(W) \times P(Th) = 0.3 \times 0.4 = 0.12 \).
  • This lack of dependency allows us to compute the probability of each combination of arrivals independently and then sum these to understand the likelihood of each \(Y\) value.
This principle makes calculating probabilities for \(Y = 0\), \(Y = 1\), etc., straightforward because each combination of magazine arrivals is treated as a separate event whose probability is determined by multiplying the individual probabilities.
Random Variables
Random variables are foundational in understanding probability distributions. In this exercise, our random variable, \(Y\), represents a clear definition of an outcome: how many days beyond Wednesday it takes for both magazines to arrive. Here's what you need to grasp about random variables:
  • Random variables can be discrete or continuous, and \(Y\) is a discrete random variable since it only takes specific values (0, 1, 2, or 3).
  • They serve as a bridge between probabilistic concepts and real-world phenomena鈥攊n this case, capturing the lag days for weekly magazine arrival.
  • A key part of working with random variables is defining the outcomes they describe, and then using probability measures, such as PMFs, to articulate the likelihood of these outcomes.
Thus, understanding \(Y\) as a random variable not only helps in organizing our approach to the magazine delivery problem; it also clarifies how different statistical concepts integrate to provide insights into real-world situations.

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Most popular questions from this chapter

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