Where is my logic wrong (lottery problem)

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The discussion revolves around the misunderstanding of applying Bernoulli trials to a lottery problem involving 6 correct integers out of 40. The key point is that the probability of success changes with each selection, as you cannot pick the same number more than once, which disqualifies it from being a Bernoulli trial. Instead, the hypergeometric distribution is the appropriate model for this scenario, as it accounts for the changing probabilities when drawing without replacement. Participants clarify that order does not matter, and the focus is on selecting one correct integer and five incorrect ones. The conversation emphasizes the importance of understanding the underlying probability principles in this context.
r0bHadz
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Homework Statement
Find the probability of selecting exactly one of the correct
six integers in a lottery, where the order in which these
integers are selected does not matter, from the positive
integers not exceeding 40
Relevant Equations
In the theory of probability and statistics, a Bernoulli trial (or binomial trial) is a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted. -google
I don't understand why I can't answer this question as a bernuli trial.

There are 6 possible correct integers out of 40, and 34 incorrect integers out of 40. I'd assume it would look like this:

(6c1)(6/40)(34/40)^5

I guess, it's because when you choose and incorrect or correct integer, the probability of getting a correct/incorrect integer changes? meaning this part: "in which the probability of success is the same every time the experiment is conducted." of the equation is not satisfied? Does my reason for why this doesn't work make sense?
 
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r0bHadz said:
I don't understand why I can't answer this question as a bernuli trial.

There are 6 possible correct integers out of 40, and 34 incorrect integers out of 40. I'd assume it would look like this:

(6c1)(6/40)(34/40)^5

I guess, it's because when you choose and incorrect or correct integer, the probability of getting a correct/incorrect integer changes? meaning this part: "in which the probability of success is the same every time the experiment is conducted." of the equation is not satisfied? Does my reason for why this doesn't work make sense?

You have to pick a different number each time. There are, therefore, only 39 possibilities for your second number etc.
 
Hmm gotcha. So because the probability of success is not the same each time, this can't be a bernouli trial.
 
r0bHadz said:
Hmm gotcha. So because the probability of success is not the same each time, this can't be a bernouli trial.

Yes, your answer would be correct if You could pick the same number more than once.
 
LCKurtz said:
Order isn't really relevant in this problem. You have two subsets, the 6 winners and the 34 losers. You want 1 from the set of 6 and 5 from the set of 34. The hypergeometric distribution is appropriate. See:
http://mathworld.wolfram.com/HypergeometricDistribution.html

Nice! this problem is actually from a discrete math book, and I'm taking a probability course right now, and we haven't learned about hypergeometric distribution surprisingly.
 
You have 5 spaces for the wrong one, 34 numbers to choose those spaces.
 
r0bHadz said:
Problem Statement: Find the probability of selecting exactly one of the correct
six integers in a lottery, where the order in which these
integers are selected does not matter, from the positive
integers not exceeding 40
Relevant Equations: In the theory of probability and statistics, a Bernoulli trial (or binomial trial) is a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted. -google

I don't understand why I can't answer this question as a bernuli trial.

There are 6 possible correct integers out of 40, and 34 incorrect integers out of 40. I'd assume it would look like this:

(6c1)(6/40)(34/40)^5

I guess, it's because when you choose and incorrect or correct integer, the probability of getting a correct/incorrect integer changes? meaning this part: "in which the probability of success is the same every time the experiment is conducted." of the equation is not satisfied? Does my reason for why this doesn't work make sense?
Agree with your comparison and update. Everything is in the probability then one who know the true way always win this game
 

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