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Chapter 12: Problem 3

If events \(A\) and \(B\) are given such that \(P(A)=\frac{1}{3}, P(A \cup B)=\frac{4}{9}\) and \(P(B)=\frac{2}{9},\) show that \(A\) and \(B\) are neither independent nor mutually exclusive.

Short Answer

Expert verified
A and B are neither independent nor mutually exclusive because \( P(A \cap B) \neq 0 \) and \( P(A \cap B) \neq P(A) \cdot P(B) \).

Step by step solution

01

Understanding Mutually Exclusive Events

Events A and B are mutually exclusive if they cannot occur at the same time, which implies that their intersection is empty. Mathematically, this is represented by \( P(A \cap B) = 0 \). We will use the given probabilities to check if this condition holds.
02

Verifying Mutually Exclusive Condition

We know the formula for the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Substitute the given values into the formula: \( \frac{4}{9} = \frac{1}{3} + \frac{2}{9} - P(A \cap B) \). Simplifying the expression: \( \frac{4}{9} = \frac{3}{9} + \frac{2}{9} - P(A \cap B) \), \( \frac{4}{9} = \frac{5}{9} - P(A \cap B) \). Solving for \( P(A \cap B) \), we get \( P(A \cap B) = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \). Since \( P(A \cap B) eq 0 \), A and B are not mutually exclusive.
03

Understanding Independent Events

Events A and B are independent if the probability of their intersection is the product of their individual probabilities, \( P(A \cap B) = P(A) \cdot P(B) \). We need to check if this condition holds.
04

Verifying Independence Condition

Calculate \( P(A) \cdot P(B) = \frac{1}{3} \times \frac{2}{9} = \frac{2}{27} \). We previously found \( P(A \cap B) = \frac{1}{9} \). Converting \( \frac{1}{9} \) to a denominator of 27 gives \( \frac{3}{27} \). Since \( \frac{2}{27} eq \frac{3}{27} \), \( P(A \cap B) eq P(A) \cdot P(B) \), implying A and B are not independent.

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Key Concepts

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

Mutually Exclusive Events
Events are considered mutually exclusive if they cannot happen at the same time. This means that if event A occurs, B cannot, and vice versa. For example, flipping a coin results in either heads or tails, not both at once.
If events are mutually exclusive, the probability of them happening together, known as their intersection, is zero: \[ P(A \cap B) = 0 \]
In our exercise, when we calculated the intersection, \( P(A \cap B) \), it turned out to be \( \frac{1}{9} \). Since this is not zero, A and B can happen together. This shows A and B are not mutually exclusive. Understanding this helps in problems where combinations of outcomes are defined, and it's critical in risk analysis to identify potential simultaneous events.
Independent Events
Independent events are those where the occurrence of one event does not affect the probability of the other. Imagine rolling two dice; the result of one die does not influence the other. In probability terms, two events, A and B, are independent if:\[ P(A \cap B) = P(A) \times P(B) \]For our problem, we calculated \( P(A \cap B) \) as \( \frac{1}{9} \), and the product \( P(A) \times P(B) \) was found to be \( \frac{2}{27} \). Upon converting \( \frac{1}{9} \) to the same denominator, we found it equals \( \frac{3}{27} \), which does not match the product \( \frac{2}{27} \). This mismatch means events A and B are not independent.
This concept is a cornerstone in statistics, vital for setting experiments where the assumption of independence greatly simplifies calculations.
Intersection of Events
The intersection of events in probability represents the event that both considered events happen simultaneously. It's denoted by \( A \cap B \) and plays an important role when determining if events are mutually exclusive or independent. To find the intersection probability, you might use formulas like:
  • For general probability: \( P(A \cap B) = P(A \cup B) - P(A) - P(B) \)
  • For independent events: \( P(A \cap B) = P(A) \times P(B) \)
In our exercise, through the union formula, it was calculated that \( P(A \cap B) \) was \( \frac{1}{9} \).
This value, crucially different from zero, indicates that A and B can occur together, confirming the analysis on mutually exclusive events and contributing to disproving their independence.
Knowing how to compute and interpret intersections helps in understanding overlapping probabilities and can assist in fields as diverse as finance and weather prediction.

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Most popular questions from this chapter

Police in a small town plan to enforce speed limits by installing'radar traps'at the two main town entrances: east end and west end. Traffic statistics show that \(40 \%\) of the cars entering the town use the east entrance and the rest use the west one. The east entrance traps are operated \(40 \%\) of the time and the west entrance are operated \(60 \%\) of the time. Assuming that the proportion of'speeders'is the same at both entrances, what is the probability that: a) a speeding driver is spotted passing one of the traps? b) a speeding driver who got spotted has actually come from the west end?

In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate. Give reasons for your answer. a) A die is loaded such that the probability of each face is according to the following assignment ( \(x\) is the number of spots on the upper face and \(P(x)\) is its probability.) $$\begin{array}{c|cccccc} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & 0 & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6} & 0 \\\ \hline \end{array}$$ b) A student at your school categorized in terms of gender and whether they are diploma candidates or not. P(female, diploma candidate) \(=0.57,\) P(female, not a diploma candidate) \(=0.23\) \(\mathrm{P}\) (male, diploma candidate) \(=0.43, \mathrm{P}\) (male, not a diploma candidate) \(=0.18\) c) Draw a card from a deck of 52 cards ( \(x\) is the suit of the card and \(P(x)\) is its probability). $$\begin{array}{|c|c|c|c|c|} \hline x & \text { Hearts } & \text { Spades } & \text { Diamonds } & \text { Clubs } \\ \hline P(x) & \frac{12}{52} & \frac{15}{52} & \frac{12}{52} & \frac{13}{52} \\\ \hline \end{array}$$

Five cards are chosen at random from a deck of 52 cards. Find the probability that the set contains a) 3 kings b) 4 hearts and 1 diamond.

An experiment involves rolling a pair of dice, 1 white and 1 red, and recording the numbers that come up. Find the probability a) that the sum is greater than 8 b) that a number greater than 4 appears on the white die c) that at most a total of 5 appears.

Driving tests in a certain city are relatively easy to pass the first time you take them. After going through training, the percentage of new drivers passing the test the first time is \(80 \% .\) If a driver fails the first test, there is chance of passing it on a second harder test, two weeks later. \(50 \%\) of the second-chance drivers pass the test. If the second test is unsuccessful, a third attempt, a week later, is given and \(30 \%\) of the participants pass it. Otherwise, the driver has to retrain and take the test after 1 year. a) Find the probability that a randomly chosen new driver will pass the test without having to wait one year. b) Find the probability that a randomly chosen new driver that passed the test did so on the second attempt.
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