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Given \(P(A)=0.3\) and \(P(B)=0.4\) : (a) If \(A\) and \(B\) are mutually exclusive events, compute \(P(A\) or \(B)\). (b) If \(P(A\) and \(B)=0.1\), compute \(P(A\) or \(B)\).

Short Answer

Expert verified
(a) 0.7; (b) 0.6.

Step by step solution

01

Understanding the Problem

We need to calculate the probability of the union of two events, i.e., either one or both events occur, in two different scenarios: (a) when the events are mutually exclusive, and (b) when they are not mutually exclusive as given by an intersection probability.
02

Mutually Exclusive Events

For mutually exclusive events, the probability of both events occurring together is zero. Therefore, we apply the formula: \[ P(A \cup B) = P(A) + P(B) \].Given that \( P(A) = 0.3 \) and \( P(B) = 0.4 \), we calculate:\[ P(A \cup B) = 0.3 + 0.4 = 0.7 \].
03

Events with Known Intersection Probability

When the intersection of events A and B is known, we use the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \].We know \( P(A) = 0.3 \), \( P(B) = 0.4 \), and \( P(A \cap B) = 0.1 \), so:\[ P(A \cup B) = 0.3 + 0.4 - 0.1 = 0.6 \].

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Key Concepts

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

Mutually Exclusive Events
Mutually exclusive events are a fundamental concept in probability that refer to two or more events that cannot happen at the same time. This means that if one event occurs, the other cannot. A classic example often given is the flipping of a coin, where the result can be either heads or tails, but not both.
In the context of probability theory, if events A and B are mutually exclusive, the probability of either event A or event B happening is simply the sum of the probabilities of each event happening individually. The formula used is:
  • \( P(A \cup B) = P(A) + P(B) \)
Here, \( P(A \cup B) \) represents the probability that either event A or event B, or both, will occur. Given that they cannot occur simultaneously, this is reflected by the sum without any subtraction. As in the original exercise, given \( P(A) = 0.3 \) and \( P(B) = 0.4 \), we conclude:
  • \( P(A \cup B) = 0.3 + 0.4 = 0.7 \)
This is straightforward because mutually exclusive events have no overlap.
Intersection of Events
The intersection of two events in probability theory refers to the scenario where both events occur simultaneously. This is denoted as \( A \cap B \). Visualize it as the overlapping section in a Venn diagram where both circles meet. This scenario requires us to consider how both events affect each other's occurrence.
When we know the intersection probability of events A and B, we can use it to calculate the probability that either or both events happen using the formula:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Here, \( P(A \cap B) \) is subtracted from the sum of \( P(A) \) and \( P(B) \) to avoid double-counting the overlapping section where both events occur. For example, in the given scenario with \( P(A) = 0.3 \), \( P(B) = 0.4 \), and \( P(A \cap B) = 0.1 \), the calculation becomes:
  • \( P(A \cup B) = 0.3 + 0.4 - 0.1 = 0.6 \)
This formula ensures accurate determination of the union of events, accounting for the overlap.
Union of Events
The union of two events, often denoted as \( A \cup B \), represents the probability that at least one of the events will occur. In simple terms, union means any result where event A happens, event B happens, or both happen together. This concept is significant in probability because it helps us understand the likelihood of multiple events happening.
For mutually exclusive events where \( P(A \cap B) = 0 \), the union simply sums up the probabilities. However, for events with a non-zero intersection, it's crucial to subtract the intersection probability to avoid double-counting.
The general formula to find the union probability is:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Whether events are mutually exclusive or have known intersections, this formula helps to capture the most complete picture of event occurrence. In summary, the union of events provides a comprehensive view of how events interact within a probability distribution and gives insights into the chances of various possible outcomes.

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Most popular questions from this chapter

Answer questions true or false and give a brief explanation for each answer. Hint: Review the summary of basic probability rules.$$ P\left(A \text { or } A^{c}\right)=0 $$

(a) Draw a tree diagram to display all the possible outcomes that can occur when you flip a coin and then toss a die. (b) How many outcomes contain a head and a number greater than \(4 ?\) (c) Probability Extension Assuming the outcomes displayed in the tree diagram are all equally likely, what is the probability that you will get a head and a number greater than 4 when you flip a coin and toss a die?

General: Roll a Die (a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely? (b) Assign probabilities to the outcomes of the sample space of part (a). Do the probabilities add up to 1 ? Should they add up to 1 ? Explain. (c) What is the probability of getting a number less than 5 on a single throw? (d) What is the probability of getting 5 or 6 on a single throw?

Suppose two events \(A\) and \(B\) are mutually exclusive, with \(P(A) \neq 0\) and \(P(B) \neq 0 .\) By working through the following steps, you'll see why two mutually exclusive events are not independent. (a) For mutually exclusive events, can event \(A\) occur if event \(B\) has occurred? What is the value of \(P(A \mid B) ?\) (b) Using the information from part (a), can you conclude that events \(A\) and \(B\) are not independent if they are mutually exclusive? Explain.

s The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the test is positive.) Experimentation has shown that the probability of a positive test is \(0.82\), given that a person has tuberculosis. The probability is \(0.09\) that the test registers positive, given that the person does not have tuberculosis. Assume that in the general population, the probability that a person has tuberculosis is \(0.04\). What is the probability that a person chosen at random will (a) have tuberculosis and have a positive test? (b) not have tuberculosis? (c) not have tuberculosis and have a positive test?

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