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Chapter 5: Problem 6

Let \(X\) denote the number of Canon digital cameras sold during a particular week by a certain store. The pmf of \(X\) is \begin{tabular}{l|ccccc} \(x\) & 0 & 1 & 2 & 3 & 4 \\ \hline\(p_{\lambda}(x)\) & \(.1\) & \(.2\) & \(.3\) & \(.25\) & \(.15\) \end{tabular} Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let \(Y\) denote the number of purchasers during this week who buy an extended warranty. a. What is \(P(X=4, Y=2)\) ? [Hint: This probability equals \(P(Y=2 \mid X=4) \cdot P(X=4)\); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty. b. Calculate \(P(X=Y)\). c. Determine the joint pmf of \(X\) and \(Y\) and then the marginal pmf of \(Y\).

Short Answer

Expert verified
a) 0.05184, b) 0.40144, c) Construct joint pmf using conditional and marginal formulas.

Step by step solution

01

Probability P(X = 4, Y = 2)

First, we need to calculate \(P(X=4)\). From the given pmf, \(P(X=4) = 0.15\). Next, we calculate \(P(Y=2 \mid X=4)\). If \(X=4\), \(Y\) follows a binomial distribution with parameters \(n=4\) and \(p=0.6\) (since 60% of customers buy the warranty). \(P(Y=2 \mid X=4)\) is calculated using the binomial probability formula: \[ P(Y=2 \mid X=4) = \binom{4}{2} (0.6)^2 (0.4)^2 = 6 \times 0.36 \times 0.16 = 0.3456 \] So, \(P(X=4, Y=2) = P(Y=2 \mid X=4) \cdot P(X=4) = 0.3456 \times 0.15 = 0.05184\).
02

Probability P(X = Y)

To calculate \(P(X=Y)\), we need to sum the probabilities \(P(X=k, Y=k)\) for \(k = 0, 1, 2, 3, 4\). For each \(k\), calculate \(P(Y=k \mid X=k)\) using the binomial formula with parameters \(n=k\) and \(p=0.6\). Then multiply by \(P(X=k)\) from the pmf. - \(k=0\): \(P(X=0, Y=0) = 1 \cdot 0.1 = 0.1\) - \(k=1\): \(P(X=1, Y=1) = 0.6 \cdot 0.2 = 0.12\) - \(k=2\): \(P(X=2, Y=2) = (0.6)^2 \cdot 0.3 = 0.108\) - \(k=3\): \(P(X=3, Y=3) = (0.6)^3 \cdot 0.25 = 0.054\) - \(k=4\): \(P(X=4, Y=4) = (0.6)^4 \cdot 0.15 = 0.01944\) Finally, \(P(X=Y) = 0.1 + 0.12 + 0.108 + 0.054 + 0.01944 = 0.40144\).
03

Joint pmf of X and Y

The joint pmf of \(X\) and \(Y\), \(p(x,y) = P(X=x, Y=y)\), can be determined by using \( P(Y=y \mid X=x) \cdot P(X=x) \) for each combination of \(x\) and \(y\). For each \(x\):- Use the binomial distribution \(P(Y=y \mid X=x)\) for 0 to \(x\).- Multiply by \(P(X=x)\). For example, for \(x=4\), compute all \(y = 0, 1, 2, 3, 4\) using binomial: \[ P(Y=0 \mid X=4) = (0.4)^4, P(Y=1 \mid X=4) = 4 \times (0.6)(0.4)^3, \ldots \] and so on for other \(x\).
04

Marginal pmf of Y

To find the marginal pmf \(p_Y(y)\), sum over all \(x\) for \(P(X=x, Y=y)\) using values from the joint pmf. For each \(y\), compute: \[ p_Y(y) = \sum_{x} P(X=x, Y=y) \] Summing contributions of each \(x\) for given \(y\) completes the calculation.

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Key Concepts

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

Joint Probability Distribution
A joint probability distribution provides a way to study the relationship between two random variables by specifying the probability of different combinations of outcomes. In this exercise, we are dealing with random variables \(X\), the number of digital cameras sold in a week, and \(Y\), the number of customers buying an extended warranty. To comprehend their relationship, we need to establish their joint probability mass function (pmf).

The joint pmf \(p(x, y) = P(X = x, Y = y)\) represents the probability that \(X\) takes a particular value \(x\) while \(Y\) takes a particular value \(y\). To calculate this, given the problem, we use the formula:
  • \(P(Y = y \mid X = x) \cdot P(X = x)\).
Let's consider \(x = 4\), the joint pmf then is calculated for each possible value of \(y\) using the binomial distribution. This gives us probabilities for combinations based on the number of sold cameras and purchased warranties.

Calculating the joint pmf for all combinations of \(x\) and \(y\) builds a comprehensive picture of their interaction, which serves as a cornerstone for both expectation and covariance calculations as well as further inference about these variables.
Marginal Probability Distribution
The marginal probability distribution of a variable is derived from its joint distribution with another variable, essentially focusing on one variable by summing or integrating out the others. Here, we are interested in the marginal probability distribution of \(Y\), the number of customers who buy an extended warranty.

To find this, the formula is:
  • \(p_Y(y) = \sum_{x} P(X = x, Y = y)\).
This sum is over all possible values of \(X\). It provides the probability of each value of \(Y\) regardless of any particular value \(X\) might take. This is crucial, as it informs us solely about the behavior of warranty purchases, independent of the number of cameras sold.

Understanding the marginal distribution allows retailers to predict sales strategies for warranties, directly related to customer satisfaction and support prevalent business goals.
Binomial Distribution
The binomial distribution is fundamental when the outcome of a series of trials can be classified into two distinct categories: success and failure. Here, buying an extended warranty is a "success."

The binomial distribution is characterized by two parameters, \(n\), the number of trials, and \(p\), the probability of success in each trial. It helps calculate the likelihood of achieving a certain number of successes across these trials. In our case, when \(X = 4\), there are 4 camera purchases (trials), with each having a 0.6 probability of "success," i.e., buying an extended warranty.

The probability of \(Y = 2\), or exactly two customers purchasing the warranty out of four camera buyers, uses the formula:
  • \[\binom{n}{k} p^k (1-p)^{n-k}\]
This formula evaluates how often \(Y = k\) successes appear among the \(n\) trials, lending significant tools for the data-driven business world to model real-world scenarios accurately.

Through this understanding, organizations create potential strategies and make informed decisions, from stocking the correct number of warranty packages to targeting clients more efficiently.

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Most popular questions from this chapter

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter \(\lambda\). a. What is the probability that all ten bulbs fail before time \(r\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(t\) ? c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \(\lambda\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \(\theta\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time \(t\) ?

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A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let \(X\) denote the number of hoses being used on the self-service island at a particular time, and let \(Y\) denote the number of hoses on the full-service island in use at that time. The joint pmf of \(X\) and \(Y\) appears in the accompanying tabulation. \begin{tabular}{ll|ccc} \(p(x, y)\) & & 0 & 1 & 2 \\ \hline & 0 & \(.10\) & 04 & \(.02\) \\ \(x\) & 1 & \(.08\) & \(.20\) & \(.06\) \\ & 2 & \(.06\) & \(.14\) & \(.30\) \end{tabular} a. What is \(P(X=1\) and \(Y=1)\) ? b. Compute \(P(X \leq 1\) and \(Y \leq 1)\). c. Give a word description of the event \(\\{X \neq 0\) and \(Y \neq 0\\}\), and compute the probability of this event. d. Compute the marginal pmf of \(X\) and of \(Y\). Using \(p_{X}(x)\), what is \(P(X \leq 1)\) ? e. Are \(X\) and \(Y\) independent rv's? Explain.
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