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Chapter 2: Problem 58

Each of the possible five outcomes of a random experiment is equally likely. The sample space is \(\\{a, b, c, d, e\\} .\) Let \(A\) denote the event \(\\{a, b\\},\) 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) \(P(A \cap B)\)

Short Answer

Expert verified
(a) \(\frac{2}{5}\); (b) \(\frac{3}{5}\); (c) \(\frac{3}{5}\); (d) 1; (e) 0;

Step by step solution

01

Understanding the sample space and probability

The sample space has five equally likely outcomes: \(\{a, b, c, d, e\}\). Since each outcome is equally likely, the probability of any single outcome is \(\frac{1}{5}\).
02

Calculate P(A)

Event \(A\) consists of the outcomes \(\{a, b\}\). Since each outcome has a probability of \(\frac{1}{5}\), the probability of event \(A\) is the sum of the probabilities of its outcomes: \(P(A) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\).
03

Calculate P(B)

Event \(B\) consists of the outcomes \(\{c, d, e\}\). The probability of event \(B\) is the sum of the probabilities of its outcomes: \(P(B) = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{3}{5}\).
04

Calculate P(A')

\(A'\) represents the complement of event \(A\), meaning the outcomes not in \(A\), which are \(\{c, d, e\}\). Thus, \(P(A') = P(B) = \frac{3}{5}\).
05

Calculate P(A 鈭� B)

The event \(A \cup B\) represents the union of \(A\) and \(B\). Since \(A\) and \(B\) cover the entire sample space \(\{a, b, c, d, e\}\), \(P(A \cup B) = 1\).
06

Calculate P(A 鈭� B)

The event \(A \cap B\) represents the intersection of \(A\) and \(B\), which are the outcomes common to both events. Since \(A\) and \(B\) share no common outcomes, \(P(A \cap B) = 0\).

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Key Concepts

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

Sample Space
In probability theory, a sample space is the set of all possible outcomes of a random experiment. It's like the complete deck of possibilities for what can happen in your experiment. For example, consider the experiment of tossing a die. The sample space for this experiment is the set \(\{1, 2, 3, 4, 5, 6\}\), representing all possible outcomes when you roll the die.

In our exercise, the sample space is \(\{a, b, c, d, e\}\). Since each of these outcomes is equally likely, they each have a probability of \(\frac{1}{5}\). Understanding the sample space is crucial because it serves as the foundation on which we calculate probabilities of other events.
Complement of an Event
The complement of an event in probability is everything that is not part of the event within the sample space. Essentially, if you have an event, its complement is the opposite of that event happening. For any event \(A\), the complement is denoted as \(A'\). The probability of an event plus the probability of its complement always equals 1.

In our exercise, event \(A\) consists of the outcomes \(\{a, b\}\). Thus, the complement of \(A\) would be the outcomes that are not \(a\) or \(b\), which are \(\{c, d, e\}\). Consequently, \(P(A') = \frac{3}{5}\), as these are the remaining possible outcomes when \(A\) does not occur.
Union of Events
The union of two events in probability is the set of outcomes that are in either one event or the other, or in both. Think of it as combining all possible outcomes from both events. The union is denoted as \(A \cup B\). This concept is important for understanding how to calculate the probability that at least one of several events will occur.

In our problem, the union \(A \cup B\) represents all the outcomes covered by events \(A\) and \(B\). Since \(A\) and \(B\) together encompass the entire sample space \(\{a, b, c, d, e\}\), the probability \(P(A \cup B) = 1\). This indicates that by selecting an outcome at random, you will definitely select one from either \(A\) or \(B\).
Intersection of Events
The intersection of events refers to outcomes that are common to both events. Represented by \(A \cap B\), this concept narrows down to the outcomes that both events share. If the intersection is empty, it usually implies that the events are mutually exclusive, meaning they cannot both happen at the same time.

For our exercise, \(A = \{a, b\}\) and \(B = \{c, d, e\}\). Since there are no shared outcomes between \(A\) and \(B\), \(A \cap B\) is an empty set, and therefore \(P(A \cap B) = 0\). This outcome emphasizes that there is zero probability of both \(A\) and \(B\) happening together in this scenario.

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Most popular questions from this chapter

In control replication, cells are replicated over a period of two days. Not until mitosis is completed can freshly synthesized DNA be replicated again. Two control mechanisms have been identified-one positive and one negative. Suppose that a replication is observed in threc cells. Let \(A\) denote the event that all cells are identified as positive, and let \(B\) denote the event that all cells are negative. Describe the sample space graphically and display each of the following events: (a) A (b) \(B\) (c) \(A \cap B\) (d) \(A \cup B\)

Use the axioms of probability to show the following: (a) For any event \(E, P\left(E^{\prime}\right)=1-P(E)\). (b) \(P(\varnothing)=0\) (c) If \(A\) is contained in \(B,\) then \(P(A) \leq P(B)\).

A robotic insertion tool contains 10 primary components. The probability that any component fails during the warranty period is 0.01 . Assume that the components fail independently and that the tool fails if any component fails. What is the probability that the tool fails during the warranty period?

The probability that a customer's order is not shipped on time is \(0.05 .\) A particular customer places three orders, and the orders are placed far enough apart in time that they can be considered to be independent events. (a) What is the probability that all are shipped on time? (b) What is the probability that exactly one is not shipped on time? (c) What is the probability that two or more orders are not shipped on time?

A researcher receives 100 containers of oxygen. Of those containers, 20 have oxygen that is not ionized, and the rest are ionized. Two samples are randomly selected, without replacement, from the lot. (a) What is the probability that the first one selected is not ionized? (b) What is the probability that the second one selected is not ionized given that the first one was ionized? (c) What is the probability that both are ionized? (d) How does the answer in part (b) change if samples selected were replaced prior to the next selection?
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