Shooting Star Problem — Simple Solution and Poisson Process Demonstration
See the usage of the Poisson process with a mathematical problem
It has been a while since I wrote. I don’t write as a routine phenomenon. Writing requires motivation and yearning for me. Thankfully, I found something when going through some of the probability interview questions for data science i.e. Shooting Star Problem. The name is not given as “Shooting Star Problem” per se but I coined the name myself. The question tests interview candidates' knowledge of basic mathematics as well as advanced mathematics.
The problem states as:
In any 15-minute interval, there is a 20% probability that you will see at least one shooting star. What is the probability that you see at least one shooting star in the period of an hour?
Simple Solution
I call this a simple solution because it does not involve probability distributions. It just requires simple knowledge of probability, that’s it!
The question says any 15-minute interval, so one hour can be thought of as four 15-minute intervals.
Now, the question asks the probability of seeing at least one shooting star in those four 15-minute intervals.
Such a probability is the same as the complement of the probability of not seeing any shooting star in those four 15-minute intervals.
All of those four 15-minute intervals are independent of each other. So, their combined probability is the product of the individual probabilities.
The individual probability of not seeing any shooting star in an interval of 15 minutes is 1 - 0.2 = 0.8.
So, the combined probability of not seeing any of the shooting stars in the four intervals is: 0.8 * 0.8 * 0.8 * 0.8 = 0.8⁴ = 0.4096
Now, the complement of its probability is 1 - 0.4096 = 0.5904
So, that is the answer or the probability of seeing at least one shooting star in an interval of an hour.
Solution using Poisson Process
Why?
Before diving into the Poisson process, let me explain why such a demonstration is helpful even if the problem can be solved simply.
The problems we encounter in the real-life can’t be solved by mere simple probability formulas. That is the reason more complex concepts are developed in mathematics. They help model real-life scenarios. When the scenarios are not exactly the same as the mathematical model, we do some assumptions and approximate the scenarios with mathematical models. Then, we do the modeling.
Mathematical modeling requires prior practice. To visualize the mathematical models and naturalize ourselves to such (abstract) models, good practice with the problems always helps. So, even if these problems can be solved easily using simpler methods, such problems provide us a good opportunity to practice mathematics and modeling.
Assumptions
At first, let’s see if the assumptions of the Poisson process hold here. The number of events can be counted. The occurrences of the events are independent of each other. The average rate at which the events occur can be calculated and let’s assume two events can not occur exactly at the same instant in time.
So, all the assumptions of the Poisson process hold here. That gives us the green light to move forward.
Mathematical Work
Let’s assume the rate of λ per minute. Then, the rate corresponds to 15λ per quarter.
Now, the formula for the probability distribution is:
P(X = x) = μ^x * e^-μ / x!
Before we go further, let’s understand the mathematical statement. The mathematical statement expresses the probability of seeing x events in a time period. μ is the average number of events seen in the time period. The time period can be a minute, an hour, or even a day. It can be any time period.
At first, let’s see the first statement — In any 15-minute interval, there is a 20% probability that you will see at least one shooting star.
Here, the interval is of 15 minutes. So, the average number of events in the interval is 15λ. The probability of seeing 0 events (i.e X = 0) is:
P(X = 0) = μ⁰ * e^-(15λ) / 0! = 1 / e^(15λ) — — (1)
(1) is equal to 80% as the probability of at least one event happening is 20%.
So, 1 / e^(15λ) = 0.8
or, e^(15λ) = 1.25
or, 15λ = ln(1.25) = 0.2231
=> λ = 0.0149 / min
So, the average number of events (shooting stars) is 0.0149 per minute.
Now, the average number of events in an hour is:
0.0149 * 60 = 0.8926 / hour
So, the Poisson distribution expression becomes:
P(X = x) = 0.8926^x * e^-0.8926 / x! — — (2)
Here, the time period is an hour.
Now, let’s find the probability of not having even a single event in an hour:
P(X = 0) = 0.8926⁰ * e^-0.8926 / 0! = 0.4096
So, the probability of having at least one event is the complement of the above probability.
So, P(X≥1) = 1 - P(X=0) = 1 - 0.4096 = 0.5904
Did not we have the same probability when working the problem using the simple formulas?
It is beautiful how both methods — one very simple and involving basic definitions of probability and another more complex and involving probability distributions — converge to the same result. Mathematics is not a coincidence, it is a science — a science that is well defined and beautiful!
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