Question

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

31%, 48%, and 31% for parts a, b, and c respectively.

Step-by-step solution

Understanding the given statistics

We are given three key pieces of information: 38% of rooms have refrigerators, 52% have TVs, and 21% have both. Let R be the event that a room has a refrigerator, and T be the event that a room has a TV. The problem asks us to find probabilities using these percentages.

Probability of TV but no refrigerator (a)

To find the probability of a room having a TV but no refrigerator, we calculate the probability of just having a TV, and subtract the probability of having both a TV and a refrigerator: \( P(T \text{ and not } R) = P(T) - P(T \text{ and } R) = 0.52 - 0.21 = 0.31 \). Thus, the answer for part (a) is 31%.

Probability of either TV or refrigerator but not both (b)

Here we need the probability of having a TV only or a refrigerator only. Use the formula: \( P(T \text{ xor } R) = P(T \text{ and not } R) + P(R \text{ and not } T) \). We already found \( P(T \text{ and not } R) = 0.31 \). For \( P(R \text{ and not } T) \), calculate as \( P(R \text{ and not } T) = P(R) - P(T \text{ and } R) = 0.38 - 0.21 = 0.17 \). Then, \( P(T \text{ xor } R) = 0.31 + 0.17 = 0.48 \). Hence, for part (b), it is 48%.

Probability of neither TV nor refrigerator (c)

Use the formula \( P(\text{neither } T \text{ nor } R) = 1 - P(T \text{ or } R) \). First, find \( P(T \text{ or } R) \) using \( P(T \text{ or } R) = P(T) + P(R) - P(T \text{ and } R) = 0.52 + 0.38 - 0.21 = 0.69 \). Then, \( P(\text{neither } T \text{ nor } R) = 1 - 0.69 = 0.31 \). The final answer for part (c) is 31%.

Key concepts

Set Theory

Set theory is a fundamental concept in mathematics that deals with collections of objects, known as sets. In probability, sets are used to represent events. For instance, in the problem, the events are defined by whether a dorm room has a refrigerator or a TV.

To visualize this using set theory, consider the following:

• The set \( R \) represents rooms with a refrigerator.
• The set \( T \) represents rooms with a TV.
• The intersection \( T \cap R \) encompasses rooms that have both a TV and a refrigerator, which is 21% of the sample.
• The union \( T \cup R \) represents rooms that have either a TV, a refrigerator, or both.

You can calculate probabilities involving the combinations of these sets, aiding in understanding overlaps and exclusive occurrences. This framework helps in solving various probability problems using principles such as union, intersection, and differences between sets.

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already occurred. In simpler terms, it helps answer questions like, "What is the probability of event A occurring if event B is already true?"

In our case, though we calculated probabilities for exclusive events, understanding conditional probability is foundational. For example, suppose we wanted to know the probability that a room with a TV also has a refrigerator. This would be expressed as:

• \( P(R|T) = \frac{P(T \text{ and } R)}{P(T)} \)
• Here, \( P(R|T) \) represents the probability of having a refrigerator given there is a TV.

Although this is not directly in the solution steps, grasping this part of probability theory enriches your understanding of probabilistic relationships between events.

Complement Rule

The complement rule in probability simplifies the calculation of an event's probability by dealing with its complement. The complement of an event \( A \), denoted \( A^c \), includes all outcomes not in \( A \).

A crucial application of the complement rule is demonstrated in step (c) of the original solution. We are asked to calculate the probability that a room does not have either a TV or a refrigerator. Using the complement rule:

• First, compute \( P(T \cup R) \), the probability of having at least a TV or a refrigerator.
• Then, logically deduce \( P(T^c \cap R^c) \), which is really \( 1 - P(T \cup R) \).
• This results in \( P(T^c \cap R^c) = 1 - 0.69 = 0.31 \).

The complement rule provides an efficient way to solve problems that involve calculating the probability of 'none' or 'complementary' events.