# Probability question with a definite outcome

1. Jul 16, 2012

### jodave614

So I'm having trouble calculating the probability of the following scenario:

I have 4 possible locations for 4 boxes. Each of these boxes may exist with a probability, β. There is also a circle, this circle definitely exists.There can only be one connection between the circle and the the group of boxes (i.e. the circle can only be connected to one boxes). What is the probability that the circle is connected to a spot with a box which does not exist?

Attempt at solution:

So I figure that i would need to multiply the probability that there is only one connection by the probability that that box doesn't exist. The only problem is that I couldn't figure out which probability I am supposed to put in the binomial equation, because I can't put in (1-β) as p because then q would be β, but it doesn't matter to me that all the other boxes exist, all I care about is the box with the connection exists. And then I thought to put in the probability that there is one connection in for p, but then I got 1 and q is 0. So I am really not sure what to do...

Last edited: Jul 16, 2012
2. Jul 17, 2012

### haruspex

Are you saying the circle can be connected to any number of spots, but to at most one that contains a box? If so, there's missing information: what controls the tendency for the connections to exist?
E.g. suppose the circle connects to each spot independently with probability p, except that you happen to know it has connected to at most one box. A priori, the probability of connecting to no boxes was P0 = Ʃpi(1-p)4-i 4Ci (1-β)i (i=0..4). The a priori probability of connecting to exactly one box was P1 = Ʃpi(1-p)4-i 4Ci iβ(1-β)i-1 (i=1..4). Now, armed with the knowledge that only one of these two possibilities occurred, the probability of not connecting to a box is P0/(P0+P1).
If there is always exactly one connection then it's easy: 1-β.