/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Let \(X\) denote the number of C... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) denote the number of Canon digital cameras sold during a particular week by a certain store. The pmf of \(X\) is $$ \begin{array}{l|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline p_{X}(x) & .1 & .2 & .3 & .25 & .15 \end{array} $$ 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. 0.34744; c. Joint pmf detailed via binomial combinations, sum provides marginal pmf for Y.

Step by step solution

01

Understanding the Problem

We are given a probability mass function (pmf) for the number of cameras sold (denoted by \(X\)). We also know that each camera sold has a 60% probability of being sold with an extended warranty. A related random variable \(Y\) represents the number of warranties out of those cameras. Our tasks include finding certain probabilities and pmfs related to \(X\) and \(Y\).
02

Calculating P(X=4, Y=2)

We need the probability that 4 cameras are sold and exactly 2 of them have extended warranties. This involves two components:1. \(P(X = 4)\), which is given in the pmf as 0.15.2. \(P(Y = 2 \mid X = 4)\). When \(X = 4\), the number of warranties \(Y\) behaves as a binomial random variable with \(n = 4\) and \(p = 0.6\), thus: \[ P(Y = 2 \mid X = 4) = \binom{4}{2} (0.6)^2 (0.4)^2 = 6 \times 0.36 \times 0.16 = 0.3456 \]Combine to find \(P(X = 4, Y = 2)\): \[ P(X = 4, Y = 2) = P(Y = 2 \mid X = 4) \times P(X = 4) = 0.3456 \times 0.15 = 0.05184 \]
03

Calculating P(X=Y)

We need to calculate the probability that the number of cameras sold equals the number with warranties, \(P(X=Y)\). We consider values of \(X\) and calculate:- When \(X = 0\), \(Y = 0\) is the only possibility: \(P(X=0) = 0.1\).- When \(X = 1\), \(Y = 1\): \(P(Y = 1 \mid X = 1) = 0.6\), so the joint probability is \(0.2 \times 0.6 = 0.12\).- When \(X = 2\), \(Y = 2\): \(P(Y = 2 \mid X = 2) = 0.36\), \(P(X=2, Y=2) = 0.3 \times 0.36 = 0.108\).- When \(X = 3\), no exact match possible since \(Y\leq X\), \(0.6\) doesn't fit .- When \(X = 4\), calculate as previous: 1 way for \(Y = 4\) using binomial, \(P(Y = 4 \mid X = 4) = 0.1296\), \(P(X=4, Y=4) = 0.15 \times 0.1296 = 0.01944\).Summing these, \(P(X=Y) = 0.1 + 0.12 + 0.108 + 0.01944 = 0.34744\).
04

Determining the Joint and Marginal PMFs

To find the joint pmf \(p_{X,Y}(x,y)\), compute \(P(Y=y | X=x)\) times \(P(X=x)\) for each \(x, y\). Given \(Y\) as binomial \(p=0.6\), calculate for each combination listed:- Use relevant binomial expressions for each \(y\) from 0 to \(x\).- Example for \(x = 3\), \(y \leq 3\): - \(y=0\), \(P = 0.064\); \(y=1\), \(P = 0.288\); \(y=2\), \(P = 0.432\); \(y=3\), \(P = 0.216\).To get marginal pmf of \(Y\), sum joint pmfs across all \(x\) for each \(y\):\(P(Y=y) = \sum_{x} P(X = x, Y = y)\) for \(y = 0, 1, 2, 3, 4\).

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Key Concepts

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

Binomial Random Variable
A binomial random variable is a type of random variable that arises in situations where there are a fixed number of trials, each with two possible outcomes: success or failure. In our context, each camera sold represents a trial, and a success is when the customer buys an extended warranty.

Key characteristics of a binomial random variable include:
  • There is a fixed number of trials (in this case, the number of cameras sold).
  • Each trial is independent of others.
  • The probability of success is the same for each trial (given as 0.6 or 60% here).
  • The random variable counts the number of successes (e.g., the number of extended warranties sold).

To calculate probabilities using a binomial distribution, you can use the formula:\[P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \(n\) is the total number of trials, \(k\) is the number of successes, and \(p\) is the probability of a success on any given trial.
Joint Probability
Joint probability provides the probability of two events happening at the same time. In our exercise, we are interested in the joint probability of the events where a certain number of cameras are sold, and a certain number of those come with an extended warranty.

The formula for joint probability can be articulated as:\[P(X = x, Y = y) = P(Y = y \mid X = x) \times P(X = x)\]This means you first determine the conditional probability \(P(Y = y \mid X = x)\), which considers the number of successful warranty sales out of the number of cameras actually sold, and then multiply it by the probability \(P(X = x)\) for that many cameras being sold, as given by the pmf.

Understanding and calculating joint probabilities allow you to assess scenarios where two interdependent outcomes occur simultaneously, giving insights into their correlated occurrences.
Marginal Probability Function
The marginal probability function is the probability of a single event without consideration of any other random variables. In the context of our exercise, we are interested in the marginal probability of events for the variable \(Y\), representing the number of extended warranty purchases irrespective of how many cameras are sold.

To calculate the marginal probability \(P(Y = y)\), you sum up the joint probabilities over all possible values of the other variable (in this case, \(X\)), using the expression:\[P(Y = y) = \sum_{x} P(X = x, Y = y)\]Where \(P(X = x, Y = y)\) is the joint probability for each particular \(x\) and \(y\), calculated previously.

Marginal probabilities enable you to focus on individual outcomes of one variable, providing a clearer picture of the likelihood of various scenarios in complex systems.
Conditional Probability
Conditional probability focuses on the likelihood of an event occurring given another event has already occurred. For example, in our exercise, we evaluated the probability that a certain number of warranties are sold given the number of cameras sold.

The concept can be expressed using the formula:\[P(Y = y \mid X = x) = \frac{P(X = x, Y = y)}{P(X = x)}\]Here, \(P(Y = y \mid X = x)\) indicates the probability of \(Y\) given \(X\), and it is crucial when events are not independent. It emphasizes how the occurrence of one event affects the probability of another.

Conditional probabilities are particularly useful in calculating scenarios where certain conditions need to be met before the occurrence of the event in question, thereby providing context-sensitive insights into probability calculations.

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