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Chapter 4: Problem 93

Five percent of all items sold by a mail-order company are returned by customers for a refund. Find the probability that, of two items sold during a given hour by this company, a. both will be returned for a refund b. neither will be returned for a refund Draw a tree diagram for this problem.

Short Answer

Expert verified
The probability that both items will be returned for a refund is \(0.0025\) and the probability that neither will be returned for refund is \(0.9025\).

Step by step solution

01

Defining the probabilities

Define the probability of an item being returned for a refund as \(p = 0.05\). Thus the probability of an item not being returned will be \(q = 1 - p = 0.95\).
02

Probability of both items being returned

To find the probability of both items being returned, multiply the probabilities of each individual event, as they occur independently. \(P(\text{both returned}) = p \times p = (0.05)^2 = 0.0025\).
03

Probability of neither items being returned

Similarly, to find the probability of neither item being returned, multiply the probabilities of each individual event. \(P(\text{neither returned}) = q \times q = (0.95)^2 = 0.9025\).
04

Tree diagram

A tree diagram would have two layers, the first layer representing the outcome of the first item (returned or not), and the second layer representing the outcome of the second item based on the outcome of the first one. Each branch would indicate the probability; \(0.05\) for returned and \(0.95\) for not returned, leading to 4 outcomes: both returned with a joint probability of \(0.0025\), both not returned with a joint probability of \(0.9025\), one returned and one not with a joint probability of \(0.0475\), and vice versa - same joint probability of \(0.0475\).

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Key Concepts

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

Tree Diagram
A tree diagram is a visual representation that helps us lay out all possible outcomes of an experiment by branching out each scenario. It is especially useful in probability theory when analyzing problems with multiple events, as it helps to visualize all possible combinations that can occur.

For this exercise, we start with two main branches for the first item:
  • One branch for the item being returned
  • One branch for the item not being returned
At each of these branches, further branches appear for the second item鈥攁gain representing returned or not. This makes the tree have four outcomes, each a combination of the following:
  • Both returned
  • First returned, second not
  • First not returned, second returned
  • Neither returned
The probability associated with each branch is indicated as you go further down the tree, making it easier to view the probability of combined events.
Independent Events
In probability theory, events are called independent if the outcome of one event does not affect the outcome of another. This principle is critical when considering scenarios like our exercise involving the return of items.

The decision to return the first item has no influence on whether the second item is returned. This is what characterizes them as independent events. Because of this independence, to find the probability of combined events, you multiply the probability of the two individual outcomes.

For example:
  • The probability of both items being returned is the product of the probability of each being returned.
  • The probability of either just the first or just the second being returned is gained in the same way.
Understanding independent events helps simplify the calculation process significantly.
Probability Calculation
Calculating probability involves determining how likely an event is to occur out of all possible outcomes. In our example scenario, we calculate the probability of different outcomes:
  • Find the probability of both items being returned: Since each return is independent with a probability of 0.05, we multiply these probabilities: \( P(\text{both returned}) = 0.05 \times 0.05 = 0.0025 \).
  • Next, calculate the probability where neither is returned. Since the chance of not being returned is 0.95, \( P(\text{neither returned}) = 0.95 \times 0.95 = 0.9025 \).
These calculations illustrate the core concept of probability calculation in independent events, allowing us to systematically evaluate each possible outcome and its likelihood.

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Most popular questions from this chapter

Two thousand randomly selected adults were asked whether or not they have ever shopped on the Internet. The following table gives a two-way classification of the responses obtained $$\begin{array}{lcc} \hline & \text { Have Shopped } & \text { Have Never Shopped } \\ \hline \text { Male } & 500 & 700 \\ \text { Female } & 300 & 500 \\ \hline \end{array}$$ a. Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. i. \(P(\) has never shopped on the Internet and is a male) ii. \(P(\) has shopped on the Internet \(a n d\) is a female) b. Mention what other joint probabilities you can calculate for this table and then find those. You may draw a tree diagram to find these probabilities.

Suppose that \(20 \%\) of all adults in a small town live alone, and \(8 \%\) of the adults live alone and have at least one pet. What is the probability that a randomly selected adult from this town has at least one pet given that this adult lives alone?

The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduated \(\begin{array}{lcc} \hline & \text { Graduated } & \text { Did Not Graduate } \\ \hline \text { Male } & 126 & 55 \\ \text { Female } & 133 & 32 \\ \hline \end{array}\) a. If one of these players is selected at random, find the following probabilities. i. \(P(\) female and graduated \()\) ii. \(P(\) male and did not graduate \()\) b. Find \(P\) (graduated and did not graduate). Is this probability zero? If yes, why?

Let \(A\) be the event that a number less than 3 is obtained if we roll a die once. What is the probability of \(A ?\) What is the complementary event of \(A\), and what is its probability?

A consumer agency randomly selected 1700 flights for two major airlines, \(\mathrm{A}\) and \(\mathrm{B}\). The following table gives the two-way classification of these flights based on airline and arrival time. Note that "less than 30 minutes late" includes flights that arrived early or on time. $$\begin{array}{cccc} \hline & \begin{array}{c} \text { Less Than 30 } \\ \text { Minutes Late } \end{array} & \begin{array}{c} \mathbf{3 0} \text { Minutes to } \\ \text { 1 Hour Late } \end{array} & \begin{array}{c} \text { More Than } \\ \text { 1 Hour Late } \end{array} \\ \hline \text { Airline A } & 429 & 390 & 92 \\ \text { Airline B } & 393 & 316 & 80 \\ \hline \end{array}$$ a. Suppose one flight is selected at random from these 1700 flights. Find the following probabilities. i. \(P(\) more than 1 hour late and airline \(\mathrm{A}\) ) ii. \(P(\) airline \(\mathrm{B}\) and less than 30 minutes late) b. Find the joint probability of events " 30 minutes to 1 hour late" and "more than 1 hour late." Is this probability zero? Explain why or why not.
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