Exponential distribution probability exercise

[archaic]

Homework Statement

In airport A, the time between the arrivals of flights coming from the UK is exponentially distributed, with $\lambda=1$ hour.
1) What is the probability that more than 5 flights arrive within 2 hours?
2) If we choose 30 separate 2 hour intervals, what is the probability that no interval is a siege of more than 5 arrivals?
3) If the probability that no flights arrive during a time interval is 0.1, what is its length?

Relevant Equations

$$f(x)=\lambda e^{-\lambda x}$$

1) Since I want at least $6$ flights to come within $2$ hours, then the time interval between each should be, at worse, $2/6=1/3$ hours, and the probability is $P(X\leq1/3)=1-e^{-1/3}$.

2) The probability that at best 5 airplanes arrive at the airport is $P(X\geq2/5)=1-P(X<2/5)=e^{-2/5}$, and since the exponential distribution is memoryless, the probability that we want is $(e^{-2/5})^{30}$.

3) I want to find $t$ such that $\int_0^tf(x)dx=0.1$.

 

[Office_Shredder]

For 1, it seems like you've missed a couple things
- that's the probability that one flight arrives within twenty minutes. You care about it happening six (or more!) times.
- even then, it's possible to construct a scenario where six flights show up in two hours, but the space between a pair of them is more than twenty minutes.

Do you know what the Poisson distribution is?

(2) has a similar issue.

I think the integral you wrote down for 3 is the correct thing to do.

 

[archaic]

For 1, it seems like you've missed a couple things
- that's the probability that one flight arrives within twenty minutes. You care about it happening six (or more!) times.
- even then, it's possible to construct a scenario where six flights show up in two hours, but the space between a pair of them is more than twenty minutes.

Thank you!

Do you know what the Poisson distribution is?

Yes.
Hm, maybe I can do this using poisson probability.
Since $\lambda=1$ hour, then I can think of the poisson parameter as being an arrival per hour, and so$$P(X>5)=1-P(X\leq5)=1-\sum_{x=0}^5\frac{2^xe^{-2}}{x!}$$am I right?

 

[Office_Shredder]

That looks like the right idea to me. The Poisson distribution was literally created to model counting the number of arrivals of an exponential distribution.

As a random side note, it looks a bit funny to use x as your summand variable. What you wrote is correct, but people generally expect letters like x and y to be real numbers, and i, j, k, l to be natural numbers.

 

[archaic]

@Office_Shredder hello! I have just thought about this.. If you think of the "units", don't you think that the problem is weird if $\lambda=1\,[T]$? If $dx$ is in probability per time, then we have a problem with the exponential, since $x$ is in units of time. Our result won't be a probability..
It looks fine, though, if $\lambda\,[T^{-1}]$.

 

[archaic]

OH! The mean of the exponential is $1/\lambda$...
It is my bad, I wrote $\lambda=1$ hour, but the question said that the mean is an hour.

 

[archaic]

I have thought of this: Let $p$ be the probability that an airplane arrives within $2$ hours, then it makes a bit of sense to look at this as a probability of success in a Bernoulli trial.$$P(X>5)=1-P(X\leq5)=1-\sum_{i=0}^5{5\choose i}p^i(1-p)^{5-i}$$with $p=\int_0^1f(x)dx$.
But this literally gives me $0$ using desmos calculator, which is not what I have found with the poisson model (either this is true, or the poisson model is).
Any guidance?

 

[Office_Shredder]

The airplane arrivals are not all with probability p. Once the first plane arrives, there are fewer than two hours left, so the second plane will arrive with probability smaller than p.

I didn't actually do the math, but the fact that you get exactly zero seems odd to me. And why is p the integral from 0 to 1 instead of the integral from 0 to 2?

 

[archaic]

The airplane arrivals are not all with probability p. Once the first plane arrives, there are fewer than two hours left, so the second plane will arrive with probability smaller than p.

I didn't actually do the math, but the fact that you get exactly zero seems odd to me. And why is p the integral from 0 to 1 instead of the integral from 0 to 2?

Right.. do you see a way of doing without recurring to the Poisson equivalent?
That was a mistake on my part, sorry!

 

[archaic]

I was mistaken about 3), it should be $\int_t^\infty f(x)dx=0.1$, since no event occurs before $t$. I want the probability that the time until the next arrival exceeds $t$ to be equal to $0.1$.
For some reason, I interpreted the distribution as one that gives us the probability of the event occurring at $t$, not before, but it actually gives us the probability that the event occurs within $t$.

 

[Office_Shredder]

You're right about 3, sorry
I think the Poisson distribution is the way to do this. Any other method is just going to rederive the distribution.

 

[haruspex]

Since λ=1 hour

It's the rate, 1 per hour.

3), it should be ∫t∞f(x)dx=0.1,

Why not just use the $\frac{(\lambda t)^k}{k!}e^{-\lambda t}$ formula you used earlier?

 

[archaic]

It's the rate, 1 per hour.

Why not just use the $\frac{(\lambda t)^k}{k!}e^{-\lambda t}$ formula you used earlier?

This is supposed to be a part of a homework sheet on continuous distributions, so I wanted to at least include something continuous in this exercise.

 

[archaic]

It's the rate, 1 per hour.

yes, thank you! i corrected myself in #6.