91Ó°ÊÓ

To help ensure the safety of school classrooms, the local fire marshal does an inspection at Thomas Jefferson High School each month to check for faulty wiring, overloaded circuits, and other fire code violations. Each month, one room is selected for inspection. Suppose that the probability that the selected room is a science classroom (biology, chemistry, or physics) is 0.6 and the probability that the selected room is a chemistry room is 0.4 . Use probability formulas to find the following probabilities. a. The probability that the selected room is not a science room. b. The probability that the selected room is a chemistry room and a science room. c. The probability that the selected room is a chemistry room given that the room selected was a science room. d. The probability that the selected room was a chemistry room or a science room.

Step by step solution

01

Calculate \(P(A')\)

\(P(A') = 1 - P(A) = 1 - 0.6 = 0.4\) So, the probability that the selected room is not a science room is 0.4. #b. The probability that the selected room is a chemistry room and a science room.# In order to compute the intersection, we need to consider that every chemistry classroom is a science classroom.

02

Calculate \(P(A \cap B)\)

Since every chemistry classroom is a science classroom, we can deduce that \(P(A \cap B) = P(B) = 0.4\) Hence, the probability that the selected room is a chemistry room and a science room is 0.4. #c. The probability that the selected room is a chemistry room given that the room selected was a science room.# Now, we find the conditional probability \(P(B|A)\), the probability that the selected room is a chemistry room given that the room selected was a science room. We use the formula \(P(B|A) = \frac{P(A \cap B)}{P(A)}\).

03

Calculate \(P(B|A)\)

Using values from before, we have \(P(B|A) =\frac{P(A \cap B)}{P(A)} =\frac{0.4}{0.6} =\frac{2}{3}\). So, the probability that the selected room is a chemistry room given that the room selected was a science room is \(\frac{2}{3}\). #d. The probability that the selected room was a chemistry room or a science room.# Finally, we need to find the probability of the union of events \(A\) and \(B\), which is denoted by \(P(A \cup B)\). We will use the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

04

Calculate \(P(A \cup B)\)

Using the values from before, we have \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 - 0.4 = 0.6\). So, the probability that the selected room was a chemistry room or a science room is 0.6.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conditional Probability

Conditional probability helps us understand the likelihood of an event occurring given that another event has already happened. In the context of the exercise, we are interested in the probability that a room is a chemistry room given that it is already known to be a science room.
To calculate conditional probability, we use the formula:

• \(P(B|A) = \frac{P(A \cap B)}{P(A)}\)

Here, \(P(B|A)\) represents the probability of event B occurring given that A has occurred. Using this formula, we can focus only on the subset of science rooms and find how likely it is to pick a chemistry room from this subset. In our exercise, this gave us \(\frac{2}{3}\), meaning there's a two-thirds chance that a science room selected at random is specifically a chemistry room.

Intersection of Events

The intersection of events refers to situations where two or more events happen at the same time. In probability terms, this is often denoted as \(P(A \cap B)\), which stands for the probability that both events A and B occur simultaneously.
In the exercise, when asked for the intersection of chemistry rooms and science rooms, we already know every chemistry room is also a science room. Therefore, the probability \(P(A \cap B)\) equals the probability of the selected room being a chemistry room, which is \(0.4\). This showcases that the intersection can be straightforward when one event is a subset of another.

Union of Events

The union of events is about understanding the probability that at least one of these events occurs. It's written as \(P(A \cup B)\), which calculates the probability that either event A occurs, event B occurs, or both occur.
To find this union probability, use the formula:

• \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

The formula accounts for subtracting the intersection because things counted double when they overlap. In our situation, it turned out to be \(0.6\), meaning a 60% chance of selecting either a science room or a chemistry room.

Complementary Probability

Complementary probability deals with the likelihood of an event not happening. If you know the probability of an event occurring, finding the probability it won't occur is as simple as subtracting that probability from 1.
For example, if the probability of selecting a science room is \(0.6\), then the probability of not selecting a science room is:

• \(P(A') = 1 - P(A)\)

Here, this gave us \(0.4\), meaning there's a 40% chance that a randomly selected room is not a science room. Complementary probability is a helpful tool, often simplifying problems by switching perspectives from what does happen to what does not.