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The number of customers waiting for gift-wrap service at a department store is an rv \(X\) with possible values \(0,1,2,3,4\) and corresponding probabilities \(.1, .2, .3, .25, .15\). A randomly selected customer will have 1,2 , or 3 packages for wrapping with probabilities \(.6, .3\), and .1, respectively. Let \(Y=\) the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer). a. Determine \(P(X=3, Y=3)\), i.e., \(p(3,3)\). b. Determine \(p(4,11)\).

Short Answer

Expert verified
a. 0.054; b. 0.00018.

Step by step solution

01

Understand the Joint Probability Notation

To solve this problem, recognize that the notation \( P(X = x, Y = y) \) represents the joint probability that there are \( x \) customers and \( y \) total packages. We need to find such probabilities for given \( X \) and \( Y \) values.
02

Set Up the Distribution for Number of Packages per Customer

The probability that a randomly selected customer has a certain number of packages is as follows: 1 package with probability \(0.6\), 2 packages with probability \(0.3\), and 3 packages with probability \(0.1\).
03

Calculate Probability for Part (a): \( P(X=3, Y=3) \)

For \( Y = 3 \), if \( X=3 \), to have a total of 3 packages, each of the 3 customers must have exactly 1 package (since \( 3 \times 1 = 3 \)). The probability that one customer has 1 package is \(0.6\), hence for 3 customers: \( (0.6)^3 = 0.216\). The probability for \( X=3 \) is given as 0.25. Thus, \[ P(X=3, Y=3) = 0.25 \times 0.216 = 0.054. \]
04

Calculate Probability for Part (b): \( P(X=4, Y=11) \)

To obtain a total of 11 packages from 4 customers, one possible combination is three customers with 3 packages each and one customer with 2 packages: \(3 \times 3 + 2 = 11\). The probability for one customer having 3 packages is \(0.1\) and for having 2 packages is \(0.3\). So, using the combination formula: \[ \binom{4}{3} \times (0.1)^3 \times (0.3) = 4 \times 0.001 \times 0.3 = 0.0012. \] The probability for \( X=4 \) is given as 0.15. Therefore, \[ P(X=4, Y=11) = 0.15 \times 0.0012 = 0.00018. \]

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Key Concepts

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

Random Variable Distribution
When we discuss a random variable distribution, we are dealing with a set of possible outcomes or values that a random variable can take, along with their associated probabilities. In our example, the random variable \(X\) represents the number of customers waiting for gift-wrap service, with values ranging from 0 to 4. Each of these values is associated with a specific probability: \(0.1\), \(0.2\), \(0.3\), \(0.25\), and \(0.15\). These probabilities must add up to 1, as they represent the entire sample space of possible outcomes.

Understanding the distribution of \(X\) helps us predict the likelihood of any given number of customers being in line. Similarly, another random variable \(Y\), which denotes the total number of packages, relies on the independent number of packages customers choose, distributed with probabilities \(0.6\), \(0.3\), and \(0.1\) for 1, 2, or 3 packages, respectively.

This structured distribution framework allows us to calculate joint and marginal probabilities accurately, helping us comprehend complex probability scenarios.
Probability Calculation
Probability calculation is a fundamental concept in statistics and involves determining the likelihood of an event occurring. Within the context of our problem, we are interested in calculating joint probabilities, which involve the simultaneous occurrence of two events represented by random variables \(X\) and \(Y\).

For instance, to find \(P(X=3, Y=3)\), we determine the probability that exactly 3 customers are in line and all customers together have 3 packages. We achieve this by multiplying the probability of \(X\) being 3 by the probability that those customers individually contribute to exactly 3 packages. The specific calculation involves considering how each customer might have just one package, which is carried out through \((0.6)^3\).

Similar logic applies to more complex scenarios like \(P(X=4, Y=11)\), where combinations of customer package distributions lead to specific totals. Probability calculations in these cases can be intricate but are simplified with a clear understanding of individual event probabilities and careful combination formula applications.
Independent Events
Independent events are those whose occurrence does not affect the likelihood of the other event happening. In our problem scenario, the number of packages a customer brings is independent of how many packages other customers may have selected. This assumption simplifies our calculations since the probabilities for each customer's choices remain unchanged regardless of others.

Independence allows us to multiply probabilities of individual events to find joint probabilities. Thus, when calculating joint probabilities, for example, \((0.6)^3\) for 3 customers each bringing 1 package, we can simply multiply the probabilities for each independent customer event. This enables straightforward computation without needing to adjust probabilities based on other events.

Understanding and identifying independent events is critical in probability theory, as it helps isolate and calculate individual event probabilities more efficiently, providing accurate results for more complex joint probability scenarios.

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