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Poisson Limit Theorem
Poisson Distribution of sum of two random independent variables $X$, $Y$Lottery probability question…Probability of winning a simple gamePoisson Processes and InsuranceProbability of earnings from lotteryUsing the Central Limit Theorem to calculate a mean from Poisson distributed random variablesA kind of Poisson random variable?Central limit theorem; Poisson distributionneed clarification on a binomial distribution questionProbability of 3 people winning 3 games out of 9 total.
$begingroup$
Using Poisson Limit Theorem derive an asymptotic probability distribution of n wins (getting 4 correct numbers out of 49) in one lottery (of course many people play the lottery). I only know how to determine the probability of getting 4 out of six, which is $$frac1binom494 sim 4,7 * 10^-6$$ but I don't know how to aplly this theorem.
probability combinatorics poisson-distribution
edited Mar 29 at 3:17
BruceET
36.2k71540
asked Mar 28 at 19:50
NotStudentNotStudent
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Using Poisson Limit Theorem derive an asymptotic probability distribution of n wins (getting 4 correct numbers out of 49) in one lottery (of course many people play the lottery). I only know how to determine the probability of getting 4 out of six, which is $$frac1binom494 sim 4,7 * 10^-6$$ but I don't know how to aplly this theorem.
probability combinatorics poisson-distribution
edited Mar 29 at 3:17
BruceET
36.2k71540
asked Mar 28 at 19:50
NotStudentNotStudent
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Suppose a quarter of a million $M = 0.25 times 10^6$ people play the lottery, then the number $X$ of winners has distribution $X sim BINOM(250.000,; p = 4.7(10^-6))$ with $E(X) = Mp = 1.175.$ This is approx $X sim POIS(lambda=1.175).$ Thus the probability of no winners is $P(X = 0) = e^-lambda = 0.3088..$ and $P(X = 1) = 0.3629.$ What is $P(X = 2),$ so that two winners split the prize?
$endgroup$
– BruceET
Mar 29 at 2:40
$begingroup$
More precisely $1/49 choose 4 = 4.719742 times 10^-6,$ so it's more like $lambda = 1.18$ and some of the other numbers above are slightly incorrect.
$endgroup$
– BruceET
Mar 29 at 3:03
add a comment |
$begingroup$
Using Poisson Limit Theorem derive an asymptotic probability distribution of n wins (getting 4 correct numbers out of 49) in one lottery (of course many people play the lottery). I only know how to determine the probability of getting 4 out of six, which is $$frac1binom494 sim 4,7 * 10^-6$$ but I don't know how to aplly this theorem.
probability combinatorics poisson-distribution
edited Mar 29 at 3:17
BruceET
36.2k71540
asked Mar 28 at 19:50
NotStudentNotStudent
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Using Poisson Limit Theorem derive an asymptotic probability distribution of n wins (getting 4 correct numbers out of 49) in one lottery (of course many people play the lottery). I only know how to determine the probability of getting 4 out of six, which is $$frac1binom494 sim 4,7 * 10^-6$$ but I don't know how to aplly this theorem.
probability combinatorics poisson-distribution
probability combinatorics poisson-distribution
edited Mar 29 at 3:17
BruceET
36.2k71540
asked Mar 28 at 19:50
NotStudentNotStudent
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 29 at 3:17
BruceET
36.2k71540
asked Mar 28 at 19:50
NotStudentNotStudent
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 29 at 3:17
BruceET
36.2k71540
edited Mar 29 at 3:17
BruceET
36.2k71540
edited Mar 29 at 3:17
BruceET
36.2k71540
36.2k71540
asked Mar 28 at 19:50
NotStudentNotStudent
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 19:50
NotStudentNotStudent
63
asked Mar 28 at 19:50
NotStudentNotStudent
63
63
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
NotStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Suppose a quarter of a million $M = 0.25 times 10^6$ people play the lottery, then the number $X$ of winners has distribution $X sim BINOM(250.000,; p = 4.7(10^-6))$ with $E(X) = Mp = 1.175.$ This is approx $X sim POIS(lambda=1.175).$ Thus the probability of no winners is $P(X = 0) = e^-lambda = 0.3088..$ and $P(X = 1) = 0.3629.$ What is $P(X = 2),$ so that two winners split the prize?
$endgroup$
– BruceET
Mar 29 at 2:40
$begingroup$
More precisely $1/49 choose 4 = 4.719742 times 10^-6,$ so it's more like $lambda = 1.18$ and some of the other numbers above are slightly incorrect.
$endgroup$
– BruceET
Mar 29 at 3:03
add a comment |
$begingroup$
Suppose a quarter of a million $M = 0.25 times 10^6$ people play the lottery, then the number $X$ of winners has distribution $X sim BINOM(250.000,; p = 4.7(10^-6))$ with $E(X) = Mp = 1.175.$ This is approx $X sim POIS(lambda=1.175).$ Thus the probability of no winners is $P(X = 0) = e^-lambda = 0.3088..$ and $P(X = 1) = 0.3629.$ What is $P(X = 2),$ so that two winners split the prize?
$endgroup$
– BruceET
Mar 29 at 2:40
$begingroup$
More precisely $1/49 choose 4 = 4.719742 times 10^-6,$ so it's more like $lambda = 1.18$ and some of the other numbers above are slightly incorrect.
$endgroup$
– BruceET
Mar 29 at 3:03
$begingroup$
Suppose a quarter of a million $M = 0.25 times 10^6$ people play the lottery, then the number $X$ of winners has distribution $X sim BINOM(250.000,; p = 4.7(10^-6))$ with $E(X) = Mp = 1.175.$ This is approx $X sim POIS(lambda=1.175).$ Thus the probability of no winners is $P(X = 0) = e^-lambda = 0.3088..$ and $P(X = 1) = 0.3629.$ What is $P(X = 2),$ so that two winners split the prize?
$endgroup$
– BruceET
Mar 29 at 2:40
$begingroup$
Suppose a quarter of a million $M = 0.25 times 10^6$ people play the lottery, then the number $X$ of winners has distribution $X sim BINOM(250.000,; p = 4.7(10^-6))$ with $E(X) = Mp = 1.175.$ This is approx $X sim POIS(lambda=1.175).$ Thus the probability of no winners is $P(X = 0) = e^-lambda = 0.3088..$ and $P(X = 1) = 0.3629.$ What is $P(X = 2),$ so that two winners split the prize?
$endgroup$
– BruceET
Mar 29 at 2:40
$begingroup$
More precisely $1/49 choose 4 = 4.719742 times 10^-6,$ so it's more like $lambda = 1.18$ and some of the other numbers above are slightly incorrect.
$endgroup$
– BruceET
Mar 29 at 3:03
$begingroup$
More precisely $1/49 choose 4 = 4.719742 times 10^-6,$ so it's more like $lambda = 1.18$ and some of the other numbers above are slightly incorrect.
$endgroup$
– BruceET
Mar 29 at 3:03
add a comment |
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$begingroup$
Suppose a quarter of a million $M = 0.25 times 10^6$ people play the lottery, then the number $X$ of winners has distribution $X sim BINOM(250.000,; p = 4.7(10^-6))$ with $E(X) = Mp = 1.175.$ This is approx $X sim POIS(lambda=1.175).$ Thus the probability of no winners is $P(X = 0) = e^-lambda = 0.3088..$ and $P(X = 1) = 0.3629.$ What is $P(X = 2),$ so that two winners split the prize?
$endgroup$
– BruceET
Mar 29 at 2:40
$begingroup$
More precisely $1/49 choose 4 = 4.719742 times 10^-6,$ so it's more like $lambda = 1.18$ and some of the other numbers above are slightly incorrect.
$endgroup$
– BruceET
Mar 29 at 3:03