# probability - Dependency and R.V

N people are coming to a conference. Every person comes or does not with an equal probability, independently of the others. Among the participants, randomly selected first speaker.

Suppose that N = 4, what is the probability that a person 4 will speak first?

I tried to divide the problem into cases,when in each case there are 1-4 participants, Each participant comes with a probability of 0.5, and the probability to choose the fourth person to be the first speaker is 1 divided by the number of participants.

It does not work. I get probablity bigger than 1.

Thank you.

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Do you mean that each person speaks or does not speak with equal probability? – PEV Mar 29 '11 at 20:18
no, comes to the conference. – user6163 Mar 29 '11 at 20:54

If no one comes, then there is no speaker. The chance that at least one person comes is $1-2^{-N}$. The chance that each person is the first speaker is the same for each of the $N$, so the chance that each will speak first is $(1-2^{-N})/N$.
The analogue might be to ask how often person $1$ is the speaker if person $1$ is guaranteed to show up, while everyone else shows up independently with probability $1/2$. Although there is no longer complete symmetry, the probability must be twice the probability if person $1$ were to show up with probability $1/2$. Since that is $(1-2^{-N})/N$, the conditional probability that person $1$ speaks conditioned on showing up is $(1-2^{-N})\cdot 2/N$.
The number of people who come to the conference is a binomial random variable. I think you should do conditioning (law of tatal probability). In order for person 4 to speak first, s/he must come. That's 50% probability. Given that Person 4 comes, the # of people that come among the other three is a Binomial(3, .5) random variable. Let's call it X. So the final answer should be something like this: $(.5)[1\times P(X=0)+\frac12\times P(X=1)+\frac 13\times P(X=2)+\frac14\times P(X=3)]$, and you can figure out the PMF of X.