# A problem on random variable in probability

I am a starter in maths. I am doing pretty good in all other topics except for probability. I don't know why I am always confused in it. My exams are nearby and I still cant solve simple problems. Can any one help me to solve this please?

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. identify the appropriate random variable and the probability distribution for the same. hence answer the following. 1)What is the probability of no arrivals in a minute period? 2)What is the probability of no arrivals in a 15 second period? 3)What is the probability of at least one arrival in a 15 second period?

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This might help en.wikipedia.org/wiki/Poisson_process – Aang Jul 24 '12 at 6:13
I am confused on what values I should take for the variable? Mean has given. But what values I have to take? – narayanpatra Jul 24 '12 at 6:26

There are two related but very different types of random variable that could be associated with this problem. We could let $T$ be the time between consecutive arrivals. Then it is (sort of) reasonable, and certainly common, to assume that $T$ has exponential distribution.

In general, $\Pr(T \le t)=1-e^{-\lambda t}$, where $\lambda$ is a parameter that measures the arrival rate. If we measure time in minutes, then in our case $\lambda=10$.

Equivalently, $Pr(T \gt t)=e^{-\lambda t}$. So the probability that there are no arrivals in a $1$ minute interval is $\Pr(T \gt 1)$, which is $e^{-10}$.

Similarly, the probability of no arrivals in a $15$ second ($1/4$ minute) interval is $e^{-10/4}$.

The probability of at least one arrival in a specific $15$ second interval is $1$ minus the probability of no arrivals. this is $1-e^{-10/4}$.

Or let $N$ be the number of arrivals in a time interval of fixed length. It is (sort of) reasonable, and certainly standard, to assume that $X$ has Poisson distribution.

Let's go with $N$, the number of arrivals in a $1$ minute interval. Then $N$ is said to have Poisson distribution with parameter $\lambda$, where $\lambda$ is the arrival rate. In general we have $$\Pr(N=n)=e^{-\lambda}\frac{\lambda^n}{n!}.$$

We can use the Poisson distribution to solve our problems. For example, if we are measuring time in $1$ minute units, the arrival rate is $10$, and $\Pr(N=0)=e^{-10}\frac{10^0}{0!}$. This is simply $e^{-10}$.

If we measure time in $15$ second units, the arrival rate $\lambda$ is $10/4$, and we can calculate as before.

The Poisson is more complicated but more versatile than the exponential, for it lets us find quickly, for example, the probability of $7$ arrivals in a $1$ minute interval.

Remark: The exponential random variable $T$ can take on, in principle, any positive real number as a value. It is said to have continuous distribution.

By contrast, the Poisson random variable $N$ only takes on non-negative integer values. It is an example of a discrete distribution.

It is fairly likely that at this stage you have been exposed to the exponential and not to the Poisson.

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Thanks bro, but I am still confused about the probability distribution. What values the random variable take? I also have to answer " identify the appropriate random variable and the probability distribution for the same". – narayanpatra Jul 24 '12 at 6:48
It is probably the exponential distribution they are looking for. It takes on all positive (or for some people) all non-negative reals values. The probability density function (pdf) is $\lambda e^{\lambda t}$, where $\lambda=10$, for $t \gt 0$ (or $t \gt 0$) and $0$ elsewhere. The cumulative distribution function (cdf) is $0$ for $t\lt 0$ and $1-e^{-\lambda t}$ for $t \ge 0$. – André Nicolas Jul 24 '12 at 6:54