Find a counterexample for a false statment about independent events.

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SUMMARY

The discussion centers on demonstrating that the equation P(A∩B∩C)=P(A)×P(B)×P(C) does not guarantee mutual independence among events A, B, and C. A valid counterexample involves a sample space consisting of the numbers 1 through 8, where A and B are defined as {1,2,3,4} and C as {1,5,6,7}. In this scenario, P(A)=P(B)=1/2 and P(A∩B∩C)=1/8, yet A and B are not independent, nor are A and C. The discussion highlights the importance of understanding the distinction between independence and the multiplication rule of probabilities.

PREREQUISITES
  • Understanding of probability theory, specifically independent events.
  • Familiarity with sample spaces and event definitions.
  • Knowledge of basic probability calculations, including intersection and multiplication of probabilities.
  • Experience with finite sample spaces and equally likely outcomes.
NEXT STEPS
  • Explore the concept of conditional probability and its relation to independence.
  • Study examples of dependent events in finite sample spaces.
  • Learn about the implications of the empty set in probability theory.
  • Investigate advanced probability topics, such as the law of total probability and Bayes' theorem.
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Students of probability theory, mathematicians, and educators seeking to deepen their understanding of event independence and probability calculations.

spyrustheviru
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"Construct a sample space to show that the truth of this statement P(A\bigcapB\bigcapC)=P(A)*P(B)*P(C) is not enough for the events A,B,C to be mutually independent.

Hint: Try finite sample spaces with equally likely simple events."

So, my though is that I need to find a sample space with 3 events A, B, C, that are not independent, yet P(A\bigcapB\bigcapC)=P(A)*P(B)*P(C) is true for them.
But I have already tried a couple of simple things and I can't seem to find a proper one. My problem propably lies in the events I take, not the spaces. I tried to use 2 tosses of fair, 6 sided dice, but my events were independent, 10 cards with the numbers 1-10 on them, 2 draws, and the card does not return to the deck. That one had dependent events, but the above statement was not true, and lastly, 2 coin tosses, but again, dependent events, untrue statement.

Any ideas?
 
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Well, the easiest way to do this would be to let one of the events be the empty set -- then the equation P(A∩B∩C)=P(A)*P(B)*P(C) is automatically true.
 
Last edited:
Useful example. Whole space has numbers 1 through 8, each with probability = 1/8.
Let A = B = {1,2,3,4}
Let C = {1,5,6,7}
P(A)=P(B)=P(C)=1/2
P(A and B and C)=P({1})=1/8.

However A and B are obviously not independent. Also A and C are not independent.
 

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