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91Ó°ÊÓ

Five percent of all cars manufactured at a large auto company are lemons. Suppose two cars are selected at random from the production line of this company. Let \(x\) denote the number of lemons in this sample. Write the probability distribution of \(x .\) Draw a tree diagram for this problem.

Short Answer

Expert verified
The probability distribution of \(x\) (number of lemons) is \[\{ (0, 0.9025), (1, 0.095), (2, 0.0025) \]\]. Hence, the chances are that there won't be a lemon in the selected cars. A tree diagram further elaborates all possible outcomes.

Step by step solution

01

Define the Random Variable and Probabilities

Let's denote \(x\) to be the number of lemons. The number of lemons \(x\) can be 0, 1, or 2(if both the selected cars are lemons). Let's denote the probability of selecting a lemon car as \(P(L) = 0.05\) and selecting a good car as \(P(G) = 0.95\).
02

Calculate Probabilities

Now, we calculate the probabilities for each possible value of \(x\). The probability of \(x = 0\) is the probability that both the selected cars are good, which is \(P(GG) = P(G) \times P(G) = 0.95^2 = 0.9025\). The probability of \(x = 1\) is the probability that one car is a lemon and the other is good. There are two outcomes that result to this - a good car followed by a lemon car \(GL\) and a lemon car followed by a good car \(LG\). Hence, \(P(x = 1) = P(GL) + P(LG) = P(G) \times P(L) + P(L) \times P(G) = 2 \times 0.05 \times 0.95 = 0.095\). Finally, the probability of \(x = 2\), which corresponds to both cars being lemons, is \(P(LL) = P(L) \times P(L) = 0.05 \times 0.05 = 0.0025\).
03

Write the Probability Distribution

In the table of probability, each possible value of \(x\) is listed along with the probabilities calculated in Step 2. Hence, the probability distribution of \(x\) is: \[\{ (0, 0.9025), (1, 0.095), (2, 0.0025) \]\}
04

Draw a Tree Diagram

Starting from a single point, draw two branches to represent the first selection: one for a good car (G) and another for a lemon car (L). Each of these branches is then split into two smaller branches representing the second selection. Label each branch with the selection it represents and the respective probability. Be sure to multiply the probabilities along the branches to get the combined probabilities. This will show all possible outcomes and their probabilities.

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Key Concepts

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

Understanding Random Variables
A random variable is a concept in probability that helps us to quantify outcomes. In this exercise, our random variable, denoted as \(x\), represents the number of lemons selected from a sample of two cars.

Random variables can be discrete or continuous. Here, \(x\) is a discrete random variable because it can take on only specific values, namely 0, 1, or 2. Each of these values represents how many of the selected cars are lemons.

By defining \(x\), we can write its probability distribution, which outlines the likelihood of each possible outcome for \(x\). This helps us understand the probabilities associated with different numbers of defective cars (lemons) from a random selection.
Steps in Probability Calculation
Probability calculation is central to determining how likely different outcomes are.

To find the probability distribution of \(x\) in this scenario, we start by calculating the likelihood of each possible value of \(x\):
  • \(P(x=0)\) is the chance that both cars are not lemons. This means both cars are good, calculated as the product of their individual probabilities: \(P(GG) = 0.95 \times 0.95 = 0.9025\).

  • \(P(x=1)\) accounts for exactly one car being a lemon, with two possible sequences: a good car first, followed by a lemon car or vice-versa. Hence, \(P(GL) + P(LG) = 2 \times 0.05 \times 0.95 = 0.095\).

  • \(P(x=2)\) indicates both selected cars are lemons, calculated by \(P(LL) = 0.05 \times 0.05 = 0.0025\).
These values form the probability distribution of the random variable \(x\).
Visualizing with a Tree Diagram
Tree diagrams are a wonderful visual tool in probability theory. They help us systematically map out all possible outcomes and their corresponding probabilities.

In our exercise, the tree starts at a single point (the first selection). This splits into two branches, representing the choice between a good car \(G\) and a lemon car \(L\). Each of these branches then splits again, capturing the outcome of the second selection.

Each branch is labeled with the probability of its respective event occurring. To find the probability of a complete sequence (like \(GL\) or \(LL\)), you multiply the probabilities along the branches. This structure clarifies how different combinations of events relate and ensures all possibilities are considered.
Basics of Probability Theory
Probability theory forms the mathematical foundation for studying random events. It guides us in calculating the likelihood of different events through defined rules and formulas.

Key principles here include the multiplication rule, which helps compute the probability of sequential independent events (like our car selections). By multiplying individual event probabilities, we capture the combined likelihood of a series of events.

This foundational understanding of probability theory helps solve larger, more complex problems. It allows you to explore various configurations of events and see how likely each scenario is, fortifying your grasp of how randomness behaves in practical situations.

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Most popular questions from this chapter

$$ \begin{aligned} &\text { The following table gives the probability distribution of a discrete random variable } x \text { . }\\\ &\begin{array}{l|llllll} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(x) & .03 & .17 & .22 & 31 & .15 & .12 \\ \hline \end{array} \end{aligned} $$ Find the following probabilities. a. \(P(x=1)\) b. \(P(x \leq 1)\) c. \(P(x \geq 3)\) d. \(P(0 \leq x \leq 2)\) e. Probability that \(x\) assumes a value less than 3 f. Probability that \(x\) assumes a value greater than 3 g. Probability that \(x\) assumes a value in the interval 2 to 4

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A veterinarian assigned to a racetrack has received a tip that one or more of the 12 horses in the third race have been doped. She has time to test only 3 horses. How many ways are there to randomly select 3 horses from these 12 horses? How many permutations are possible?

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