# Finding the given probability

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 ?

• 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? – whuber Oct 10 '16 at 15:54
• Yeah , it should be , $Pr(B)$ should be 1. Thanks for pointing out. @whuber – User9523 Oct 10 '16 at 18:51

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.