Mathematics Asked by Soid on January 21, 2021
I’m working on the following problem from Ross "A First Course in Probability" (9th edition):
People enter a gambling casino at a rate of 1 every 2 minutes. (a)
What is the probability that no one enters between 12:00 and 12:05?
(b) What is the probability that at least 4 people enter the casino
during that time?
(problem 63, chapter 4, page 178)
The official solution manual approaches the problem as a Poisson process, but why is it appropriate? I feel like for a small number of trials (n = 5) and high probability (p = 0.5) the Binomial distribution would be more fit. The answer is quite different depending on what distribution we use:
Binomial(n = 5, p=0.5): $ P(X=0) = 0.03125; P(X geq 4) = 0.1875 $
Poisson($lambda = 2.5$): $ P(X = 0) = 0.082; P(X geq 4) = 0.242 $
It's not just five trials; it's infinitely many trials with an infinitely small probability of success on each trial. Or, if you like, one can approximate it by dividing the dividing every minute into $10,000$ small intervals, each with probability $1/20,000$ of someone entering the casino, so it's $operatorname{Binomial}(2.5times10,000,,,, 1/20,000).$ And an even closer approximation to the Poisson distribution (but not very much closer since you're extremely close already) is achieved by dividing each minute into $100,000$ parts, each with probability $1/200,000$ of a customer entering.
Correct answer by Michael Hardy on January 21, 2021
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