# 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 $(1-2^{-N})/N$.

This type of analysis is also useful if you are analyzing your expected share of a lottery in which the prize is split among the winning tickets. You could try to evaluate a complicated sum, but instead compute the probability that at least one person wins, and then your expected share is the same as anyone else's. This gives a way to evaluate that complicated sum, which then lets you evaluate other cases which do not have the same symmetry, such as if you buy several lottery tickets and ensure that your tickets have different numbers, while everyone else buys one random ticket.

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 $(1-2^{-N})\cdot 2/N$.

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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.

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this is exactly what I did. for example: P(x=2)- means that only 2 pepole will come along the 4th perosn, so we can choose them (3 choose 2)*0.5*0.5*1/3 1/3- for the probabilty for choosing the 4 th to talk. AM I right? – user6163 Mar 29 '11 at 21:06
That's correct, and the answer (N=4: prob=15/64) is the same as Douglas's solution, which is simpler to understand. – GWu Mar 30 '11 at 1:32