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Chapter 14: Problem 13

A check of dorm rooms on a large college campus revealed that \(38 \%\) had refrigerators, \(52 \%\) had TVs, and \(21 \%\) had both a TV and a refrigerator. What's the probability that a randomly selected dorm room has a) a TV but no refrigerator? b) a TV or a refrigerator, but not both? c) neither a TV nor a refrigerator?

Short Answer

Expert verified
a) 0.31, b) 0.48, c) 0.31

Step by step solution

01

Define the Events

Let \( R \) be the event that a dorm room has a refrigerator, and \( T \) be the event that a dorm room has a TV. We know \( P(R) = 0.38 \), \( P(T) = 0.52 \), and \( P(T \cap R) = 0.21 \), where \( T \cap R \) is the event that a dorm room has both a TV and a refrigerator.
02

Calculate TV but No Refrigerator

The probability that a dorm room has a TV but no refrigerator is given by \( P(T \cap R^c) = P(T) - P(T \cap R) \). Using the given probabilities, we find \( 0.52 - 0.21 = 0.31 \).
03

Calculate TV or Refrigerator but Not Both

This is given by the sum of the probabilities of having only a TV and only a refrigerator. Calculate \( P(R \cap T^c) = P(R) - P(T \cap R) = 0.38 - 0.21 = 0.17 \). Then, add this result to \( P(T \cap R^c) \), giving us \( 0.17 + 0.31 = 0.48 \).
04

Calculate Neither TV nor Refrigerator

This is the complement of the event of having either or both items. Thus, \( P((R \cup T)^c) = 1 - P(R \cup T) \). First, find \( P(R \cup T) = P(R) + P(T) - P(T \cap R) = 0.38 + 0.52 - 0.21 = 0.69 \). Now, \( P((R \cup T)^c) = 1 - 0.69 = 0.31 \).

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Key Concepts

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

Compound Events
Probability with compound events involves scenarios where you look at combinations of two or more single events. For our exercise:
  • The events being considered are whether a dorm room has a refrigerator or a TV, or both.
  • Compound events help us evaluate probabilities where these events interact, like having either a TV, a refrigerator, or both at once.
When examining a compound event, such as a dorm room having a TV but no refrigerator, visualize it as a subset of all possible scenarios:
  • First, calculate the likelihood of having a TV (\( P(T) = 0.52 \)).
  • Then, subtract the likelihood of having both a TV and a refrigerator to exclusively get scenarios where only a TV is present (\( P(T \cap R) = 0.21 \)).
  • This leads us to \( P(T \cap R^c) = 0.52 - 0.21 = 0.31 \), the probability of having only a TV.
Such calculations are critical as they break down complex scenarios into simpler components, making probability easier to manage.
Complement Rule
The complement rule in probability is a helpful concept for finding probabilities of events that do not occur. Mathematically, if \( A \) represents an event, the complement \( A^c \) is the probability that \( A \) does not happen. It is given by \( P(A^c) = 1 - P(A) \).

For example, if \( P(A) \) represents any rooms having either a TV or refrigerator or both, \( P((R \cup T)^c) \) will give us the chance that a room has neither:
  • First, compute \( P(R \cup T) \) which includes probabilities for having at least one or both items: \( P(R) + P(T) - P(T \cap R) = 0.38 + 0.52 - 0.21 = 0.69 \).
  • Then subtract this from 1 to find the complement鈥攔ooms lacking both items: \( P((R \cup T)^c) = 1 - 0.69 = 0.31 \).
Understanding complements helps solve problems more efficiently, especially when directly calculating a probability is complicated.
Set Operations in Probability
Set operations in probability provide a structured method to handle different combinations of events. Using operations like union and intersection, they allow precise descriptions of probabilities:
  • Intersection (\( \cap \)): It identifies the probability that two events happen at once. From our exercise, the probability of a room having both a TV and a refrigerator is an intersection, \( P(T \cap R) = 0.21 \).
  • Union (\( \cup \)): This operation finds the probability of either of the two events occurring. It includes situations where the events happen separately or together. For instance, calculating the total probability for having at least one of the two items involves \( P(R \cup T) = P(R) + P(T) - P(T \cap R) \).
Each of these operations allows us to neatly organize multi-faceted scenarios. When compounded with basic probability principles, set operations efficiently explain probabilities for complex event scenarios.

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Most popular questions from this chapter

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