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Chapter 5: Problem 5

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 \(P(X=3, Y=3)\), that is, \(p(3,3)\). b. Determine \(p(4,11)\).

Short Answer

Expert verified
a. \(p(3,3) = 0.054\); b. \(p(4,11) = 0\).

Step by step solution

01

Understanding the problem

We need to calculate the joint probabilities of two events: the number of customers waiting, given by rv \(X\), and the total number of packages to be wrapped, given by rv \(Y\). The given probabilities are for customers waiting and packages per customer.
02

Exploring Event 'X=3' and 'Y=3'

For \(X=3\), there are 3 customers, each of which can have 1, 2, or 3 packages. We need to find the probability that exactly 3 packages are being wrapped in total, which happens when one customer has 1 package and the remaining two customers have none.
03

Probability Calculation for 'Y=3' with 'X=3'

To have exactly 3 packages with 3 customers, one possible combination is one customer has all 3 packages while the others have none of their own. The probability of such a distribution is: \[ P(Y=3 | X=3) = 3! / [(1!)(0!)(0!) \cdot (2!)] \cdot (0.6)^3 = 0.216. \]
04

Joint Probability \(p(3,3)\)

The joint probability \(p(3,3)\) is the product of the probabilities: i.e., \( P(X=3) \cdot P(Y=3 | X=3) = 0.25 \times 0.216 = 0.054. \)
05

Exploring Event 'X=4' and 'Y=11'

For \(X=4\), there are 4 customers. To have exactly 11 packages, split among these customers following the given probabilities.
06

Probability Calculation for 'Y=11' with 'X=4'

To get 11 packages among 4 customers, you need to select customers so that two customers get 3 packages each, and two get 2 packages each. Calculating different possible combinations and probabilities, we note this combination is not possible due to sum restrictions.
07

Conclusion on \(p(4,11)\)

Since there is no feasible way to split 11 packages among 4 customers within the given constraints, \(p(4,11) = 0\).

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Key Concepts

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

Random Variable
A random variable, often abbreviated as rv, is a foundational concept in probability and statistics. It is a variable whose possible values are numerical outcomes of a random phenomenon. In simpler terms, a random variable quantifies the outcomes of a random process.
In the context of our exercise, we have two key random variables:
  • **X**: Represents the number of customers waiting for gift-wrap service. Its possible values are discrete, specifically, 0, 1, 2, 3, and 4.
  • **Y**: Represents the total number of packages to be wrapped. This variable also takes specific discrete values, determined by the packages each customer brings.
Understanding random variables allows us to compute probabilities and expected values, essential for analyzing real-world scenarios such as service processes in a department store.
Probability Distribution
Probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes of a random variable. For discrete random variables (like in our problem), this is often represented in tabular or graphical form where probabilities are associated with each possible value of the random variable.
In this instance:
  • For **X**, the distribution is given by the probabilities:
    • 0 customers: 0.1
    • 1 customer: 0.2
    • 2 customers: 0.3
    • 3 customers: 0.25
    • 4 customers: 0.15
  • For each **customer**, the number of packages they have is defined by the probabilities:
    • 1 package: 0.6
    • 2 packages: 0.3
    • 3 packages: 0.1
    These distributions help us understand and visually interpret the likelihood of different service scenarios.
The probability distribution is crucial for calculating outcomes like joint probabilities and expected values.
Combinatorial Probability
Combinatorial probability involves counting the number of ways certain outcomes can occur in a set of trials and then defining the probability of these events. It uses combinations and permutations to evaluate complex scenarios, especially when dealing with multiple elements that can occur in different orders or combinations.
In our task, combinatorial reasoning is applied when determining the probability that 3 customers have a total of 3 packages:
  • One possibility is where all packages are assigned to one customer, while the remaining have none.
  • The probability is calculated using the formula for permutations and the given probabilities for a single customer鈥檚 package count.
This approach of using combinatorial probability answers the feasibility of achieving certain totals under given constraints, like calculating the aforementioned scenarios.
Conditional Probability
Conditional probability refers to the likelihood of an event occurring given that another event has already occurred. It鈥檚 vital in calculating joint probabilities and understanding dependencies between events.
This concept is essential when determining joint probabilities such as:
  • The probability that there are exactly 3 packages (event Y=3) given there are 3 customers waiting (event X=3).
  • The calculation involves multiplying the probability of X by the probability of Y given X, which is evaluated in the steps provided in the problem.
      • Applying conditional probability helps to pinpoint precise outcomes like our joint probabilities \( p(3,3) \). It allows us to go beyond individual probabilities and understand how multiple random variables interact under specific conditions.

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Most popular questions from this chapter

A class has 10 mathematics majors, 6 computer science majors, and 4 statistics majors. A committee of two is selected at random to work on a problem. Let \(X\) be the number of mathematics majors and let \(Y\) be the number of computer science majors chosen. a. Determine the joint probability mass function \(p(x, y)\). This generalizes the hypergeometric distribution studied in Section 3.6. Give the joint probability table showing all nine values, of which three should be 0 . b. Determine the marginal probability mass functions by summing numerically. How could these be obtained directly? [Hint: What are the univariate distributions of \(X\) and \(Y ?]\) c. Determine the conditional probability mass function of \(Y\) given \(X=x\) for \(x=0,1,2\). Compare with the \(h(y ; 2-x, 6,10)\) distribution. Intuitively, why should this work? d. Are \(X\) and \(Y\) independent? Explain. e. Determine \(E(Y \mid X=x), x=0,1,2\). Do this numerically and then compare with the use of the formula for the hypergeometric mean, using the hypergeometric distribution given in part (c). Is \(E(Y \mid X=x)\) a linear function of \(x\) ? f. Determine \(V(Y \mid X=x), x=0,1,2\). Do this numerically and then compare with the use of the formula for the hypergeometric variance, using the hypergeometric distribution given in part (c).

The joint probability distribution of the number \(X\) of cars and the number \(Y\) of buses per signal cycle at a proposed left-turn lane is displayed in the accompanying joint probability table. $$ \begin{array}{cc|ccc} & & {y} \\ p(x, y) & & 0 & 1 & 2 \\ \hline & 0 & .025 & .015 & .010 \\ & 1 & .050 & .030 & .020 \\ & 2 & .125 & .075 & .050 \\ x & 3 & .150 & .090 & .060 \\ & 4 & .100 & .060 & .040 \\ & 5 & .050 & .030 & .020 \end{array} $$ a. What is the probability that there is exactly one car and exactly one bus during a cycle? b. What is the probability that there is at most one car and at most one bus during a cycle? c. What is the probability that there is exactly one car during a cycle? Exactly one bus? d. Suppose the left-turn lane is to have a capacity of five cars, and one bus is equivalent to three cars. What is the probability of an overflow during a cycle? e. Are \(X\) and \(Y\) independent rv's? Explain.

A restaurant serves three fixed-price dinners costing $$\$ 20, \$ 25$$, and $$\$ 30$$. For a randomly selected couple dining at this restaurant, let \(X=\) the cost of the man's dinner and \(Y=\) the cost of the woman's dinner. The joint pmf of \(X\) and \(Y\) is given in the following table: $$ \begin{array}{ll|ccc} & & {l}{y} \\ {p(x, y)} & & 20 & 25 & 30 \\ \hline {x} & 20 & .05 & .05 & .10 \\ & 25 & .05 & .10 & .35 \\ & 30 & 0 & .20 & .10 \end{array} $$ a. Compute the marginal pmf's of \(X\) and \(Y\). b. What is the probability that the man's and the woman's dinner cost at most \(\$ 25\) each? c. Are \(X\) and \(Y\) independent? Justify your answer. d. What is the expected total cost of the dinner for the two people? e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message "You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen." How much does the restaurant expect to refund?

A stockroom currently has 30 components of a certain type, of which 8 were provided by supplier 1,10 by supplier 2 , and 12 by supplier 3 . Six of these are to be randomly selected for a particular assembly. Let \(X=\) the number of supplier l's components selected, \(Y=\) the number of supplier 2 's components selected, and \(p(x, y)\) denote the joint pmf of \(X\) and \(Y\). a. What is \(p(3,2)\) ? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, \(p(3,2)\) \(=\) (number of outcomes with \(X=3\) and \(Y=2\) )/ (total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] b. Using the logic of part (a), obtain \(p(x, y)\). (This can be thought of as a multivariate hypergeometric distribution - sampling without replacement from a finite population consisting of more than two categories.)

David and Peter independently choose at random a number from \(1,2,3\), with each possibility equally likely. Let \(X\) be the larger of the two numbers, and let \(Y\) be the smaller. a. Determine \(p(x, y)\). b. Determine \(p_{X}(x), x=1,2,3\). c. Determine \(p_{Y X X}(y \mid x)\). d. Determine \(E(Y \mid X=x)\). Is this a linear function of \(x\) ? e. Determine \(V(Y \mid X=x)\).
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