Probability of invites homework

AI Thread Summary
The discussion revolves around calculating the probability of having an equal number of men and women among a group of friends invited by a woman, with specific numbers of male and female friends. The user attempts to apply binomial coefficients to find the probability but is unsure about the calculations and the correct approach, resulting in a discrepancy between their answer and the expected result. Additionally, there is confusion regarding Bayes' theorem, particularly in determining the correct probabilities to use in various scenarios. The user expresses uncertainty about how to approach the problem and seeks clarification on the reasoning behind their calculations. Overall, the thread highlights challenges in understanding probability concepts and applying them correctly in different contexts.
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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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