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A positive integer from one to six is to be chosen by casting a die. Thus the elements c of the sample space C are 1,2,3,4,5,6. Suppose C1=1,2,3,4 and C2=3,4,5,6. If the probability set function P assigns a probability of 16 to each of the elements of C, compute P(C1),P(C2),P(C1C2), and P(C1C2).

Short Answer

Expert verified
The probability of event C1 is 23, for event C2 is 23, for the intersection of C1 and C2 is 13, and for the union of C1 and C2 is 1.

Step by step solution

01

Determine the Probability of Event C1

The event C1 contains four outcomes: 1,2,3, and 4. Since each outcome has a 16 probability, the probability of event C1 is 4×16=23.
02

Determine the Probability of Event C2

Similarly, the event C2 contains four outcomes: 3,4,5, and 6. As in step 1, using the probability of each outcome, we find the probability of event C2 is also 4×16=23.
03

Determine the Probability of the Intersection of C1 and C2

The intersection of C1 and C2, written as C1C2, contains the outcomes that are common to both C1 and C2, which are 3 and 4. Therefore, applying the rule of product in probability, the probability of the intersection of C1 and C2 is 2×16=13.
04

Determine the Probability of the Union of C1 and C2

The union of C1 and C2, written as C1C2, contains all the outcomes that are in either C1, C2, or both. Since the sample space includes numbers from 1 to 6, the union of C1 and C2 is the same as the sample space. Therefore, the sum of all six equally likely outcomes gives the probability of the union of C1 and C2, which is 6×16=1.

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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
In probability theory, the sample space, often denoted as S or Ω is a fundamental concept. It's the set that includes all the possible outcomes of a random experiment. For example, if you roll a six-sided die, the sample space is C={1,2,3,4,5,6} as these are all possible outcomes. Each outcome in the sample space is equally likely if the die is fair.

The size of a sample space is essential since it affects the computation of probabilities for events. In our example, since there are six outcomes and the die is fair, each outcome's probability is 16. Understanding the sample space is crucial before you can delve into more complicated concepts in probability.
Probability Set Function
The probability set function, usually denoted as P is a way to assign a probability to events in a sample space. It maps subsets of the sample space to numbers between 0 and 1—where 0 signifies the event is impossible, and 1 indicates it is certain.

In our die example, the function P assigns a probability of 16 to each of the elements in the sample space. This means that for any event E consisting of outcomes from the sample space, the probability P(E) is calculated by summing the probabilities of each outcome within that event.

For instance, for an event C1={1,2,3,4} with four outcomes, P(C1) would be the sum of probabilities of outcomes 1 through 4— giving us P(C1)=4×16=23 as seen in the solution.
Intersection of Events
The intersection of events, denoted as AB where A and B are events, refers to a new event that includes all the outcomes common to both A and B only. It is the equivalent of the 'and' condition in probability.

Considering the previous events C1 and C2 from our die example, the intersection C1C2 would consist of outcomes that appear in both sets— which are 3 and 4 only.

The probability of the intersection P(C1C2) is then the sum of the probabilities of these common outcomes. Since 3 and 4 each have a probability of 16, P(C1C2) is 2×16=13 as previously calculated in the solutions.
Union of Events
The union of events represents the combination of all outcomes from multiple events, denoted as AB where A and B are two events. It encompasses any outcome that is in either A or B or in both, aligning with the 'or' condition in probability.

For instance, if you look at events C1 and C2 like in our die-casting example, the union C1C2 would include all the numbers from 1 to 6 since these are the numbers in at least one of the sets. Here, the union comprises the entire sample space, indicating that one of these outcomes will certainly occur, which gives us P(C1C2)=1—matching the sample space's total probability.

Understanding the union of events is crucial in scenarios where multiple outcomes lead to a successful or desired result, as it helps to calculate the overall chances of success.

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