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If \(P(A)=0.2, P(B)=0.2,\) and \(A\) and \(B\) are mutually exclusive, are they independent?

Short Answer

Expert verified
No, mutually exclusive events cannot be independent.

Step by step solution

01

Understand Key Concepts

Mutually exclusive events mean that if one event occurs, the other cannot. For two events \(A\) and \(B\) to be independent, the occurrence of one should not affect the occurrence of the other.
02

Check Mutually Exclusive Condition

Since \(A\) and \(B\) are mutually exclusive, we have \(P(A \cap B) = 0\).
03

Calculate Independence Condition

For events \(A\) and \(B\) to be independent, \(P(A \cap B)\) should be equal to \(P(A) \times P(B)\). This would be \(0.2 \times 0.2 = 0.04\).
04

Compare Results

Compare \(P(A \cap B) = 0\) with \(P(A) \times P(B) = 0.04\). Since these are not equal, \(A\) 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
Mutually exclusive events are two or more events that cannot happen at the same time. If one event occurs, the others cannot occur simultaneously. A classic example is flipping a fair coin. You cannot get both 'heads' and 'tails' at the same time.

In the exercise, we are given that events \(A\) and \(B\) are mutually exclusive. This implies that the probability of both events occurring together is zero, or mathematically, \(P(A \cap B) = 0\).
  • Mutually exclusive events have no overlap in possible outcomes.
  • If one event happens, the other does not.
Understanding this concept is crucial because it sets the groundwork for understanding how different events interact or fail to interact with each other in probability theory.
Independent Events
Independent events are events where the occurrence of one does not affect the probability of the other occurring. These events are the opposite of mutually exclusive events, where one event's outcome does not change the likelihood of the other event occurring.

For instance, rolling a die and flipping a coin are independent events. The result of the coin has no effect on the outcome of the die roll and vice versa.
  • Independence implies that the knowledge of one event's outcome provides no information about the other.
  • The probability of both independent events occurring is the product of their individual probabilities.
In our exercise, if \(A\) and \(B\) were to be considered independent, their joint probability \(P(A \cap B)\) would be \(P(A) \times P(B)\). However, because \(A\) and \(B\) are mutually exclusive with \(P(A \cap B) = 0\), they cannot be independent.
Probability Calculation
Probability calculation involves determining the likelihood of events occurring. This can be achieved using various principles of probability theory, including the properties of mutually exclusive and independent events.

In the problem given, we use the fact that the probability of both mutually exclusive events \(A\) and \(B\) happening together, \(P(A \cap B)\), is zero. This is a direct consequence of their mutually exclusive nature.
  • For mutually exclusive events, \(P(A \cap B) = 0\).
  • For independent events, you would expect \(P(A \cap B) = P(A) \times P(B)\).
Here, the calculation \(0.2 \times 0.2 = 0.04\) does not match \(P(A \cap B) = 0\), confirming that \(A\) and \(B\) cannot be independent. The calculation underlines the importance of understanding event relationships when determining probabilities.

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Most popular questions from this chapter

In the manufacturing of a chemical adhesive, \(3 \%\) of all batches have raw materials from two different lots. This occurs when holding tanks are replenished and the remaining portion of a lot is insufficient to fill the tanks. Only \(5 \%\) of batches with material from a single lot require reprocessing. However, the viscosity of batches consisting of two or more lots of material is more difficult to control, and \(40 \%\) of such batches require additional processing to achieve the required viscosity. Let \(A\) denote the event that a batch is formed from two dif- ferent lots, and let \(B\) denote the event that a lot requires additional processing. Determine the following probabilities: (a) \(P(A)\) (b) \(P\left(A^{\prime}\right)\) (c) \(P(B \mid A)\) (d) \(P\left(B \mid A^{\prime}\right)\) (e) \(P(A \cap B)\) (f) \(P\left(A \cap B^{\prime}\right)\) (g) \(P(B)\)

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A computer system uses passwords that are six characters, and each character is one of the 26 letters \((a-z)\) or 10 integers \((0-9)\). Uppercase letters are not used. Let \(A\) denote the event that a password begins with a vowel (either \(a, e, i, o\), or \(u\) ), and let \(B\) denote the event that a password ends with an even number (either \(0,2,4,6,\) or 8 ). Suppose a hacker selects a password at random. Determine the following probabilities: (a) \(P(A)\) (b) \(P(B)\) (c) \(P(A \cap B)\) (d) \(P(A \cup B)\)

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