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In a town with 5000 adults , a sample of $100$ is asked about their opinion on a proposed municipal project , $60$ are found to favor it and $40$ oppose it. If in fact the adults of the town were equally divided on the proposal , what would be the probability of obtaining a majority of $60$ or more favoring it in a sample of $100$ ?

My interpretation for the problem :

Probability[60 or more favoring in a sample of 100] given the population is equally divided over the decision is asked.

which can be represented as :

P(60 or more in favor in a sample of 100 | 2500 favor it in 5000 adults )

Say , $X$ represents the number of adults in favor in a sample of 100.

$X$~Bin(100,$p$) , where $p$ = Probability of an adult in favor.

So , $P(A|B)$ = $\dfrac{P(A\&B)}{P(B)}$

Thus the solution becomes : $\dfrac{P(X \geq 60 )}{(\frac{2500}{5000})}$ , where $X$ ~ Bin(100,$\frac{2500}{5000}$)

Is the above interpretation correct ?

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  • $\begingroup$ If "$B$" stands for "2500 favor it in 5000 adults," then isn't $\Pr(B)=1$? After all, you are told this is true. And if you are viewing $B$ as a random event, then what are the alternative possible events and what would their probabilities be? That is, what justifies using $2500/5000$ as its probability? $\endgroup$ – whuber Oct 10 '16 at 15:54
  • $\begingroup$ Yeah , it should be , $Pr(B)$ should be 1. Thanks for pointing out. @whuber $\endgroup$ – User9523 Oct 10 '16 at 18:51
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While @Time Lee is technically correct that it is a hypergeometric distribution, you're gonna end up in a whole world of pain if you try to do it that way. The population and number of people who favor it are both large enough that a Binomial distribution will suffice as an approximation.

As @whuber noted in the comment above, your event $B$ is just the state of the world because it is a statement about the population, not a sample. You correctly state that this gives you $p$ for your binomial RV, but you have one more step. You should use another approximation after setting it up as a binomial otherwise you're not going to have fun computationally.

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The distribution discussed here is the hypergeometric distribution, with N=5000, K=2500, n=100, and k=60. To solve for the probability of the 60 or more, the true CDF of this function is not very easy to use, however you can use the inverse method which is easier to use. I did the work for P(X=60), and P(X=60)=.0105. You would need to sum this over P(X=61,62,...,100).

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    $\begingroup$ So it can't be approximated by a binomial distribution ? I am quite new to this hypergeometric distribution ,thus tried using binomial. $\endgroup$ – User9523 Oct 10 '16 at 17:25
  • $\begingroup$ If you are approximating it, I would solve the hypergeometric form for say, X=60,65,70,75,80,85,90,95,100, and then smooth the values between. I cannot be sure that the binomial form is a good approximation method here. The reason being that the hypergeometric form varies greatly depending on the size of the population. For example if we reduce this entire problem by an order of 10 everywhere, P(X=6)= .2. $\endgroup$ – Tim Lee Oct 10 '16 at 17:30
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    $\begingroup$ Furthermore, if this problem is being to be solved by hand, binomial distribution can be approximated by normal. This is the way this kind of problem is usually addressed in elementary courses of statistics. $\endgroup$ – Pere Oct 10 '16 at 18:20

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