Probability of invites homework

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SUMMARY

The discussion centers on calculating the probability of a woman inviting friends such that the number of women equals the number of men among the attendees. The problem involves combinatorial calculations using binomial coefficients, specifically the formula for combinations. The initial attempt yielded a probability of approximately 0.316, while the correct probability is 0.164. Additionally, the conversation touches on Bayes' theorem, particularly in the context of conditional probabilities related to health statistics.

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  • Understanding of binomial coefficients and combinations
  • Familiarity with probability theory and basic statistics
  • Knowledge of Bayes' theorem and conditional probabilities
  • Ability to perform combinatorial calculations
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  • Study combinatorial probability using binomial coefficients
  • Learn about Bayes' theorem and its applications in real-world scenarios
  • Practice problems involving conditional probabilities
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Students studying probability and statistics, educators teaching combinatorial methods, and anyone looking to deepen their understanding of Bayes' theorem and its applications.

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Sorry about not having the right writing way on the definitions, I'll update that soon.

Homework Statement



A woman takes a.. drawing among her 11 friends(ie 0 to 11 might arivve. Also 6 are women and 5 are men). What is the probability that by conincidence is as much women as men among the company?

Homework Equations



binominalkoefficient(I think that's relevant)
(p)
(r)
(p over r)

.. and then I don't know.

The Attempt at a Solution



I tried (((6 over 0)*(5 over 1))\(11 over 1)((6 over 1)*(5 over 2))\(11 over 3) ((6 over 2)*(5 over 3))\(11 over 5)((6 over 3)*(5 over 4))\(11 over 7)((6 over 4)*(5 over 5))\(11 over 9)) / 4. But that's something like 0.316. And the is 0.164.


Problem 2. I have some minor problems with Bayes sentence.

eg. 0.6 of women have aids, 0.4 of them testes positive. 0.4 have not aids, and .03 of them also get positive.

The bottom of the sentence is ok, just P(B), But when do you know it's 0.6 * P(B|A), or 0.6* (just)0.4, or just 0.6. I mess up there. (eg if you could explain it throu how the different questions relating to those differences would look)
 
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Are you saying the woman invites 11 friends, each of whom may or may not come? Seems to me you would need to know the probability each will or will not come.

"I tried (((6 over 0)*(5 over 1))\(11 over 1)((6 over 1)*(5 over 2))\(11 over 3) ((6 over 2)*(5 over 3))\(11 over 5)((6 over 3)*(5 over 4))\(11 over 7)((6 over 4)*(5 over 5))\(11 over 9)) / 4. But that's something like 0.316. And the [answer?] is 0.164."
It would help if you said WHY you tried that!
 
Yeah, you could say it like that.

I tried that answer because as you said that would be the probability for each combination. Then I added each(didn't know if to add or multiply, but neither work anyway), .. and then I guess just by intuition I divide by 4 as to get the 'line' that gives the general outcome(some would give higher probability, some lower)(don't know quite how to spell this in english, sorry!)
It doesn't feel right, but I didn't know what else to do and it was my first thought.
 

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