# Exponential Distribution in place of Uniform Distribution

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 exponential distiribution does not make much sense for this problem, since it is a real-valued distribution, but seat numbers are probably integer valued. The discrete analogue of the exponential distribution is the geometric distribution. en.wikipedia.org/wiki/Geometric_distribution – Leander Tilsted Kristensen Jan 9 at 9:42
• You can get displayed equations by enclosing them in double instead of single dollar signs. – joriki Jan 9 at 9:59
• Thanks for the reply, @ Leander Tilsted Kristensen, in my case assumption is that back row, top seats have high probability of selection, so for seat number 1 to 16, N=16, the seat No.16 has high probability of selection and seat No. 0 has lowest and overall exponential distribution is followed. So as you have mentioned how can i derive expression for Pexp using geometrical distribution?. I really appreciate your interest and thank you in advance. – user3696623 Jan 9 at 20:12

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 $$P(X=0)= 1 - \sum_{k=0}^{N-1} p(1-p)^k = (1-p)^N$$

• Thanks for the reply, so it means i can write $Pexp=(1-p)^N$ and probability equation becomes $P= (1-p)^N * \sum_{i=1}^k AB_i$. Am i correct in writing this equation? – user3696623 Jan 10 at 14:21
• if you by $P_{exp}$ mean the probability mass function, then it would be $P_{exp}(N-k) = p(1-p)^k$ for $k=0,1,...,n-1$ and $P_{exp}(0)=(1-p)^N$. – Leander Tilsted Kristensen Jan 10 at 14:31
• Now if we have two variables, such as row number $R$ and seat number $S$, then we would have $P(R=r,S=s)=P(R=r | S=s)P(S=s) = P(S=s | R=r)P(R=r)$. So if $R$ follows the geometric rule described above we would have $P(R=N-k,S=s)= P(S=s | R=N-k) p(1-p)^k$ and $P(R=0,S=s)= P(S=s | R=0) (1-p)^N$ – Leander Tilsted Kristensen Jan 10 at 14:42
• Thanks for prompt reply, its really helping me out. I have just one variable i.e. seat no, ranging from 1 to N. what can i write in place of $Pexp$. – user3696623 Jan 10 at 15:32
• The equation of Probability i have needs to multiply $Pexp$ by $\sum_{i=1}^k AB_i$. So should $Pexp$ be probability mass function or just probability – user3696623 Jan 10 at 15:38