Problem 59
Question
The sample space of a random experiment is \(\\{a, b,\) \(c, d, e\\}\) with probabilities \(0.1,0.1,0.2,0.4,\) and \(0.2,\) respectively. Let \(A\) denote the event \(\\{a, b, c\\},\) and let \(B\) denote the event \(\\{c, d, e\\} .\) Determine the following: (a) \(P(A)\) (b) \(P(B)\) (c) \(P\left(A^{\prime}\right)\) (d) \(P(A \cup B)\) (e) \(P(A \cap B)\)
Step-by-Step Solution
Verified Answer
(a) 0.4; (b) 0.8; (c) 0.6; (d) 1.0; (e) 0.2
1Step 1: Identify Probabilities of Each Event
The sample space is \{a, b, c, d, e\} with probabilities associated as follows: \( P(a) = 0.1 \), \( P(b) = 0.1 \), \( P(c) = 0.2 \), \( P(d) = 0.4 \), \( P(e) = 0.2 \). Event \( A \) includes \{a, b, c\}, and event \( B \) includes \{c, d, e\}.
2Step 1: Calculate P(A)
The probability of event \( A \) is the sum of the probabilities of \( a \), \( b \), and \( c \). Thus, \( P(A) = P(a) + P(b) + P(c) = 0.1 + 0.1 + 0.2 = 0.4 \).
3Step 2: Calculate P(B)
Event \( B \) consists of \{c, d, e\}. So, the probability of event \( B \) is \( P(B) = P(c) + P(d) + P(e) = 0.2 + 0.4 + 0.2 = 0.8 \).
4Step 3: Calculate P(A')
The complement of \( A \) includes elements in the sample space which are not in \( A \), namely \{d, e\}. Therefore, \( P(A') = P(d) + P(e) = 0.4 + 0.2 = 0.6 \).
5Step 4: Calculate P(A ∪ B)
The union \( A \cup B \) includes elements in \{a, b, c, d, e\}. Therefore, \( P(A \cup B) = P(a) + P(b) + P(c) + P(d) + P(e) = 0.1 + 0.1 + 0.2 + 0.4 + 0.2 = 1.0 \).
6Step 5: Calculate P(A ∩ B)
The intersection \( A \cap B \) includes only the element \{c\}. Thus, \( P(A \cap B) = P(c) = 0.2 \).
Key Concepts
Sample SpaceProbability of EventsSet Operations in ProbabilityComplementary Events
Sample Space
In probability theory, the sample space is a fundamental concept. It refers to the set of all possible outcomes of a random experiment. Imagine it as the complete list of every potential result that could occur.
For instance, if you are throwing a six-sided die, the sample space consists of the numbers \{1, 2, 3, 4, 5, 6\}. In our original exercise, the sample space is given as \{a, b, c, d, e\}.
Each of these elements in the sample space has a certain probability associated with it. In this case, the probabilities are: 0.1 for \(a\), 0.1 for \(b\), 0.2 for \(c\), 0.4 for \(d\), and 0.2 for \(e\).
The sum of the probabilities in a sample space should always be 1, meaning that one of the outcomes must occur. Let's check: \(0.1 + 0.1 + 0.2 + 0.4 + 0.2 = 1.0\). This confirms that our set is indeed a complete sample space.
For instance, if you are throwing a six-sided die, the sample space consists of the numbers \{1, 2, 3, 4, 5, 6\}. In our original exercise, the sample space is given as \{a, b, c, d, e\}.
Each of these elements in the sample space has a certain probability associated with it. In this case, the probabilities are: 0.1 for \(a\), 0.1 for \(b\), 0.2 for \(c\), 0.4 for \(d\), and 0.2 for \(e\).
The sum of the probabilities in a sample space should always be 1, meaning that one of the outcomes must occur. Let's check: \(0.1 + 0.1 + 0.2 + 0.4 + 0.2 = 1.0\). This confirms that our set is indeed a complete sample space.
Probability of Events
An event in probability is any subset of the sample space. Events can include one or more of the outcomes listed in the sample space.
For the original exercise, two events are defined: \(A = \{a, b, c\}\) and \(B = \{c, d, e\}\).
The probability of an event is found by summing the probabilities of the individual outcomes that make up the event. This basic rule stems from the additivity of probabilities.
For the original exercise, two events are defined: \(A = \{a, b, c\}\) and \(B = \{c, d, e\}\).
The probability of an event is found by summing the probabilities of the individual outcomes that make up the event. This basic rule stems from the additivity of probabilities.
- To find \(P(A)\), sum the probabilities of the outcomes in \(A\): \(0.1 + 0.1 + 0.2 = 0.4\).
- For \(P(B)\), you sum the probabilities of the outcomes in \(B\): \(0.2 + 0.4 + 0.2 = 0.8\).
Set Operations in Probability
In probability theory, set operations such as union and intersection play a key role. They help in understanding how different events relate to one another.
The **union** of two events \(A\) and \(B\), represented as \(A \cup B\), includes outcomes that are in either \(A\) or \(B\), or in both. In our exercise, the union of \(A\) and \(B\) comprises all elements in \{a, b, c, d, e\}, which encompasses the entire sample space. Thus, the probability is \(P(A \cup B) = 1.0\).
The **intersection** of \(A\) and \(B\), denoted by \(A \cap B\), consists of outcomes that are in both \(A\) and \(B\). The only outcome satisfying this condition is \(c\). Therefore, \(P(A \cap B) = 0.2\).
Understanding these operations is crucial for solving problems involving combined events, such as mutually exclusive or overlapping events.
The **union** of two events \(A\) and \(B\), represented as \(A \cup B\), includes outcomes that are in either \(A\) or \(B\), or in both. In our exercise, the union of \(A\) and \(B\) comprises all elements in \{a, b, c, d, e\}, which encompasses the entire sample space. Thus, the probability is \(P(A \cup B) = 1.0\).
The **intersection** of \(A\) and \(B\), denoted by \(A \cap B\), consists of outcomes that are in both \(A\) and \(B\). The only outcome satisfying this condition is \(c\). Therefore, \(P(A \cap B) = 0.2\).
Understanding these operations is crucial for solving problems involving combined events, such as mutually exclusive or overlapping events.
Complementary Events
Complementary events refer to the outcomes that do not belong to a particular event within the sample space.
If \(A\) is an event, then the complement of \(A\), denoted as \(A'\), includes everything in the sample space that is not in \(A\). Essentially, \(A'\) contains all elements that are excluded from \(A\).
To find the probability of \(A'\), sum the probabilities of the elements that are in \(A'\) rather than in \(A\).
From our exercise, if \(A = \{a, b, c\}\), then \(A' = \{d, e\}\). Therefore, \(P(A') = P(d) + P(e) = 0.4 + 0.2 = 0.6\).
The concept of complementary events is important, as the probability of an event plus the probability of its complement always equals 1. This relationship can simplify complex probability problems by allowing the calculation of one from the other.
If \(A\) is an event, then the complement of \(A\), denoted as \(A'\), includes everything in the sample space that is not in \(A\). Essentially, \(A'\) contains all elements that are excluded from \(A\).
To find the probability of \(A'\), sum the probabilities of the elements that are in \(A'\) rather than in \(A\).
From our exercise, if \(A = \{a, b, c\}\), then \(A' = \{d, e\}\). Therefore, \(P(A') = P(d) + P(e) = 0.4 + 0.2 = 0.6\).
The concept of complementary events is important, as the probability of an event plus the probability of its complement always equals 1. This relationship can simplify complex probability problems by allowing the calculation of one from the other.
Other exercises in this chapter
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