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Chapter 2: Problem 17

Let \(A\) denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let \(B\) be the event that the next request is for help with SAS. Suppose that \(P(A)=.30\) and \(P(B)=.50\). a. Why is it not the case that \(P(A)+P(B)=1\) ? b. Calculate \(P\left(A^{\prime}\right)\). c. Calculate \(P(A \cup B)\). d. Calculate \(P\left(A^{\prime} \cap B^{\prime}\right)\).

Short Answer

Expert verified
a. P(A) and P(B) don't sum to 1 because other requests are possible. b. P(A') = 0.70. c. P(A 鈭� B) = 0.80. d. P(A' 鈭� B') = 0.20.

Step by step solution

01

Understanding Event Probabilities

Events A and B refer to different statistical software requests, SPSS and SAS respectively. The sum of their probabilities, \(P(A) + P(B) = 0.30 + 0.50 = 0.80\), does not equal 1. This is because it is possible that neither SPSS nor SAS is requested, meaning there's a probability for other events.
02

Calculating the Complement of Event A

The complement of event A, denoted as \(A'\), is the event that the next request is not for SPSS. It is calculated as \(P(A') = 1 - P(A) = 1 - 0.30 = 0.70\).
03

Calculating the Union of Events A and B

The union of two events A and B, denoted \(A \cup B\), includes any request that is for either SPSS or SAS. Using the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) and, assuming A and B are mutually exclusive (requests can't be for both softwares at once), we have \(P(A \cup B) = 0.30 + 0.50 = 0.80\).
04

Calculating the Complement of the Union

The complement of the union event \(A \cup B\), denoted as \(A' \cap B'\), is the event that the request is for neither SPSS nor SAS. It is calculated as \(P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.80 = 0.20\).

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Key Concepts

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

Event Complement
The concept of an event complement is a fundamental part of probability theory. When we talk about the complement of an event, let's say event \(A\), we refer to the scenario where event \(A\) does not occur. This is denoted as \(A'\) (read as "A prime" or "A complement"). It is often calculated using the formula:
\[ P(A') = 1 - P(A) \]
This formula is derived from the idea that the total probability of all possible outcomes of an experiment sums up to 1. Therefore, the probability of the event not happening is simply the remainder when the probability of the event occurring is subtracted from 1.
For instance, if the probability of needing assistance with SPSS is 0.30, the probability of not needing SPSS assistance is:
\[ P(A') = 1 - 0.30 = 0.70 \]
This represents all the other possibilities - requests for help with other software packages aside from SPSS.
Mutually Exclusive Events
Mutually exclusive events are a set of events that cannot occur simultaneously. This means if one event occurs, the other cannot. The probability of both events happening at the same time is zero. In probability terms, for two events \(A\) and \(B\), this can be expressed as:
\[ P(A \cap B) = 0 \]
This concept is especially relevant when calculating the probability of either of the events occurring, known as the union of events. If events \(A\) and \(B\) are mutually exclusive, then the formula simplifies to:
\[ P(A \cup B) = P(A) + P(B) \]
Take the example from the exercise about different software requests. If requests are either for SPSS or SAS and not both, these events are mutually exclusive. Thus, if \(P(A) = 0.30\) and \(P(B) = 0.50\), then the probability of either happening is:
\[ P(A \cup B) = 0.30 + 0.50 = 0.80 \]
Union of Events
The union of events is a probability operation representing the scenario where at least one of the events occurs. If we're looking at two events, \(A\) and \(B\), their union is denoted as \(A \cup B\). This includes any outcomes where event \(A\) happens, event \(B\) happens, or both happen鈥攗nless they are mutually exclusive.
The formula for the probability of the union of two events is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Here, the subtraction of \(P(A \cap B)\) ensures that any overlap between the two events isn't counted twice, except in the scenario of mutually exclusive events where \(P(A \cap B) = 0\).
For instance, using the probabilities from the exercise, where requests may relate to either SPSS or SAS, the union probability without event overlap simplifies to:
\[ P(A \cup B) = 0.30 + 0.50 = 0.80 \]
This captures all occurrences of assistance requests for these two statistical software types.

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Most popular questions from this chapter

As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com). a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.] b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.] c. Answer the questions posed in (b) for four-character sequences. d. As of April \(2006,97,786\) of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

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Show that for any three events \(A, B\), and \(C\) with \(P(C)>0\), \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\).
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