I have basic question regarding probability distribution mathematical expressions. I need to design a model who should calculate probability of choosing seat number in the cinema theater by people and it is told that this follows exponential distribution.
Also given that this probability is also multiplied by some factor $\sum_{i=1}^k AB_i$ . Now considering N number of seats from 1 to N. I could easily get probability of particular seat selection as $Pu=1/N$ for any one of the seat, being same, if i were using uniform distribution. So the expression for probability would be
$P = Pu $ $\sum_{i=1}^k AB_i$
or
$P = \frac{1}{N} $ $\sum_{i=1}^k AB_i$
But i have to use exponential distribution for seat selection instead of uniform distribution, so how i can get this seat selection probability $Pexp$. i.e
$P= Pexp * \sum_{i=1}^k AB_i$
what would be in place of $Pexp$? Here $AB_i$ is calculated based on conditional probability.
The usual analogy for the geometric distribution is, that we flip coins untill we get a head. In this context, it means that we either pick the seat we are at with probability $p$ or we go on to the next seat with probability $1-p$.
So if we start at seat/row $N$, the probability of sitting here will be $P(X=N)=p$. Continuing on we get $$P(X=N-1) = P(\text{reject seat } N \text{ and accept seat } N-1) = p(1-p)$$ $$P(X=N-2) = P(\text{reject seat } N,N-1 \text{ and accept seat } N-2) = p(1-p)^2$$ $$\dots$$ $$P(X=N-k) = P(\text{reject seat } N,N-1,\dots,N-(k-1) \text{ and accept seat } N-k) = p(1-p)^k$$ Now it would be tempting to say $P(X=0)=p(1-p)^N$, but this introduces a problem, since $$\sum_{k=0}^N p(1-p)^k< 1.$$ A possible fix would be $$P(X=0)= 1 - \sum_{k=0}^{N-1} p(1-p)^k = (1-p)^N$$