Probability John Loses & Meghan Wins: 9/55

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Homework Help Overview

The problem involves calculating the probability that Meghan wins a game while John loses, given their individual probabilities of winning. The original poster expresses confusion regarding their calculation and the provided answer, questioning the importance of the order of winning.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the independence of winning and losing probabilities, with some questioning whether the events of John losing and Meghan winning are independent. Others suggest deriving answers from first principles rather than guessing.

Discussion Status

The discussion is active, with participants exploring different interpretations of independence in probability. Some guidance has been offered regarding the nature of the events and their independence, but no consensus has been reached on the original poster's calculation.

Contextual Notes

There is a mention of a different probability problem involving drawing balls from a bag, which highlights the difference between independent and dependent events. The original poster's approach to the problem is influenced by this example.

zorro
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Homework Statement



The probability that John wins a game is 4/5 and,
independently, the probability that Meghan wins is 9/11.
What is the probability that Meghan wins and
John loses the game?

The Attempt at a Solution



I calculated it as 18/55 but the answer given is 9/55. Why is it so? Is the order of winning not important here? Why?
 
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what is the probability that John loses the game? :)

ehild
 
1/5.
 
If the "win" probabilities are independent, are the "lose" probabilities independent? Are the events that John loses and Meghan wins independent? (Suggestion: don't _guess_. Work out the responses from first principles, using the definitions of independence, etc. This issue comes up over and over again in many places, so it would be as well to understand it.)

RGV
 
So how come that you got 18/55 for the probability that John loses the game and Meghan wins it? The events are independent, it can be any game, they are not winners and losers in the same game at the same time.

ehild
 
Last edited:
I have seen questions like ' A bag contains 4 Red and 7 Blue balls. Two balls are picked up from the bag at random without replacement. What is the probability that one ball is blue and the other is red? '

Here we compute the probability of selecting one red and one blue ball (14/55) and then multiply it by 2 because the first one can be red or blue. I applied the same logic in the question above. John can be the first or the second loser.
 
Picking out two balls from the same bag without replacement are not independent events. You can select a red one first with probability 4/11 and then a blue ones from the rest with probability 7/10 or a blue first with probability 7/11 and then a red one with probability 4/10, so the total probability is indeed 2*28/110.

The events in your problem are independent. That game is not played between Meghan and John alone as in this case the winner would determine the user so the probability that John wins and Meghan loses would be the same as the probability that John wins alone. That game can be something played by a lot of people many times. It can be, for example, writing a test successfully or failing it. The probability that John passes the tests is 4/5 and the probability that Meghan passes them is 9/11, and it is a new test now and one asks the probability that Meghan passes it and John fails. These events are totally independent, they are honest, do not look at each others tests...

ehild
 
Last edited:
I got it now. Thanks.
 

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