Calculating the probability of an event

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Homework Help Overview

The problem involves calculating the probability of a football team winning all games against four opponents, given a winning probability of 0.75 for each game. The context includes concepts of independent events and the use of complements in probability calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the difference between calculating the probability of winning all games versus winning at least one game. Questions arise about the appropriate use of probabilities and the implications of independence in the context of the problem.

Discussion Status

Participants are exploring the distinctions between union and intersection in probability, with some clarifying the original poster's misunderstanding regarding the calculations. There is an ongoing examination of the correct approach to the problem, but no consensus has been reached.

Contextual Notes

There is a focus on the terminology used in probability, particularly regarding the use of complements and the conditions under which different formulas apply. The original poster's confusion highlights the need for clarity in understanding these concepts.

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Homework Statement


A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are independent, what is the probability the team wins all the games?


Homework Equations


P(w1 u w2 u w3 u w4)=1-P(w1 u w2 u w3 u w4)' the ' mark stands for complement


The Attempt at a Solution


I found that the complement w=.25, and since it says independent that 1-p(w1)'p(w2)'p(w3)'p(w4)' so 1-.25^4=.996 probability of winning however the book says .75^4 why didn't they do as I did?
 
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Like how do you know when to use .75^4 or what I did in problems what key words will tell me which one?
 
You've calculated the probability that the team wins at least one game, not the probability that it wins all four games.

To win at least one game, it has to win game 1 OR game 2 OR game 3 OR game 4.
To win all four games, it has to win game 1 AND game 2 AND game 3 AND game 4.
 
Oh ok I get it the complement bar makes the union go to intersection so therefore I calculated the or and not the and.
 
It really has nothing to do with taking the complement. You performed your calculation correctly; it just wasn't the right calculation to do. OR corresponds to the union while AND corresponds to the intersection, so ##P(w1 \cup w2 \cup w3 \cup w4)## is the probability of winning game 1 OR winning game 2 OR winning game 3 OR winning game 4.
 

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