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Two cold tablets are unintentionally put in a box containing two aspirin tablets, that appear to be identical. One tablet is selected at random from the box and swallowed by the first patient. The second patient selects another tablet at random and swallows it. a. List the simple events in the sample space \(S\). b. Find the probability of event \(A\), that the first patient swallowed a cold tablet. c. Find the probability of event \(B\), that exactly one of the two patients swallowed a cold tablet. d. Find the probability of event \(C,\) that neither patient swallowed a cold tablet.

Short Answer

Expert verified
Answer: The probability of event A is 0.5.

Step by step solution

01

Identifying the simple events

Let's represent a cold tablet with the letter C and an aspirin tablet with the letter A. We have two cold tablets (CC) and two aspirin tablets (AA) in the box. Now, we need to list all possible simple events for two patients: 1. First patient takes a cold tablet (C), and the second patient takes a cold tablet (C). 2. First patient takes a cold tablet (C), and the second patient takes an aspirin tablet (A). 3. First patient takes an aspirin tablet (A), and the second patient takes a cold tablet (C). 4. First patient takes an aspirin tablet (A), and the second patient takes an aspirin tablet (A). So, the sample space S can be represented as: \(S = \{(C, C), (C, A), (A, C), (A, A)\}\). #b. Probability of event A#
02

Calculating P(A)

Event A is the event in which the first patient swallowed a cold tablet. From our sample space \(S\), we see that this happens in two cases: \((C, C)\) and \((C, A)\). There are a total of 4 possible outcomes in the sample space, and 2 of these outcomes are favorable for event A. Therefore, the probability of event A is given by: \(P(A)=\frac{\text{Number of favorable outcomes for event A}}{\text{Total number of possible outcomes}}=\frac{2}{4}=0.5\). #c. Probability of event B#
03

Calculating P(B)

Event B is the event in which exactly one of the two patients swallowed a cold tablet. From our sample space \(S\), we see that this happens in two cases: \((C, A)\) and \((A, C)\). There are a total of 4 possible outcomes in the sample space, and 2 of these outcomes are favorable for event B. Therefore, the probability of event B is given by: \(P(B)=\frac{\text{Number of favorable outcomes for event B}}{\text{Total number of possible outcomes}}=\frac{2}{4}=0.5\). #d. Probability of event C#
04

Calculating P(C)

Event C is the event in which neither patient swallowed a cold tablet. From our sample space \(S\), we see that this happens in only one case: \((A, A)\). There are a total of 4 possible outcomes in the sample space, and only 1 of these outcomes is favorable for event C. Therefore, the probability of event C is given by: \(P(C)=\frac{\text{Number of favorable outcomes for event C}}{\text{Total number of possible outcomes}}=\frac{1}{4}=0.25\).

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Key Concepts

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

Event Probability Calculation
Understanding how to calculate the probability of an event is a fundamental skill in probability theory. The basic formula is quite straightforward: the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes in the sample space. It's expressed mathematically as:

\( P(E) = \frac{\text{Number of favorable outcomes for event E}}{\text{Total number of possible outcomes}} \).

Let's relate this to the problem we're working with. When calculating the probability that the first patient swallowed a cold tablet, we identified that there were two outcomes out of four total possible outcomes from the sample space. This made it easy to calculate event A's probability as 0.5, or 50%. Similarly, we used the same method to determine the probabilities of the patients swallowing exactly one cold tablet (event B) and neither swallowing a cold tablet (event C).

Always remember that to accurately calculate probabilities, you must ensure that the sample space includes all possible outcomes and that each outcome is equally likely. If these conditions are met, simple division as shown above yields the desired probability.
Simple Events in Probability
In probability, a simple event refers to a single outcome in a sample space. It's the most basic possible outcome that cannot be broken down into simpler components. When identifying simple events, it's essential to ensure that each one is distinct and that the collection of all simple events covers the entire realm of possibility for the experiment.

In our textbook problem, we listed the simple events by considering each possibility for the patients taking a tablet. There were four distinct simple events identified, each representing a different combination of the patients taking cold and aspirin tablets. By understanding simple events, we create the foundation for determining probabilities of more complex events that may consist of multiple simple events. Recognizing and listing simple events is an invaluable step in any calculation of probability.
Probability Sample Space Outcomes
The sample space in probability is the set of all possible outcomes of an experiment. Referring to our original problem, the sample space \( S \) was identified as \( S = \{(C, C), (C, A), (A, C), (A, A)\} \), containing all possible scenarios of the two patients each taking one tablet. It's critical for the sample space to be comprehensive, meaning it should include all possible simple events without any omissions.

When dealing with sample spaces, one tip is to systematically list outcomes to ensure none are left out. For example, you might structure your list based on a pattern or use a visual representation like a tree diagram to help organize your thoughts. Moreover, understanding the sample space is integral to accurately computing event probabilities as it defines the denominator in our probability calculations—equally crucial as identifying the favorable outcomes for specific events.

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