Sum of Poisson Distribution, why is my solution incorrect
From SOA #212:
The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.
I calculated the probabilities as follows
$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$
$$=P(text{employee is sick more than two days in a three-month period})$$
This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$
$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$
$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$
$$=.8005$$
This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.
probability statistics probability-distributions
asked Nov 20 at 15:26
agblt
16214
add a comment |
From SOA #212:
The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.
I calculated the probabilities as follows
$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$
$$=P(text{employee is sick more than two days in a three-month period})$$
This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$
$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$
$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$
$$=.8005$$
This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.
probability statistics probability-distributions
asked Nov 20 at 15:26
agblt
16214
Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23
add a comment |
From SOA #212:
The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.
I calculated the probabilities as follows
$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$
$$=P(text{employee is sick more than two days in a three-month period})$$
This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$
$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$
$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$
$$=.8005$$
This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.
probability statistics probability-distributions
asked Nov 20 at 15:26
agblt
16214
From SOA #212:
The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.
I calculated the probabilities as follows
$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$
$$=P(text{employee is sick more than two days in a three-month period})$$
This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$
$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$
$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$
$$=.8005$$
This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.
probability statistics probability-distributions
probability statistics probability-distributions
asked Nov 20 at 15:26
agblt
16214
asked Nov 20 at 15:26
agblt
16214
asked Nov 20 at 15:26
agblt
16214
asked Nov 20 at 15:26
agblt
16214
asked Nov 20 at 15:26
agblt
16214
16214
Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23
add a comment |
Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23
Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23
Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23
add a comment |
3 Answers
3
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You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
add a comment |
Your calculation does not match mine. It is structurally different. We want
$1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$
For $X=2$ we have several combinations
$110 quad 200$
$011quad 020$
$101quad 002$
$X=1:$ $100,$ $ 010,$$ 001$
$X=0$: $000$
Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$
answered Nov 20 at 15:59
callculus
17.8k31427
add a comment |
We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.
answered Nov 20 at 16:47
J.G.
22.4k22035
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
add a comment |
You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
add a comment |
You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
answered Nov 20 at 15:35
Michael Lugo
17.9k33576
17.9k33576
add a comment |
add a comment |
Your calculation does not match mine. It is structurally different. We want
$1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$
For $X=2$ we have several combinations
$110 quad 200$
$011quad 020$
$101quad 002$
$X=1:$ $100,$ $ 010,$$ 001$
$X=0$: $000$
Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$
answered Nov 20 at 15:59
callculus
17.8k31427
add a comment |
Your calculation does not match mine. It is structurally different. We want
$1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$
For $X=2$ we have several combinations
$110 quad 200$
$011quad 020$
$101quad 002$
$X=1:$ $100,$ $ 010,$$ 001$
$X=0$: $000$
Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$
answered Nov 20 at 15:59
callculus
17.8k31427
add a comment |
Your calculation does not match mine. It is structurally different. We want
$1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$
For $X=2$ we have several combinations
$110 quad 200$
$011quad 020$
$101quad 002$
$X=1:$ $100,$ $ 010,$$ 001$
$X=0$: $000$
Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$
answered Nov 20 at 15:59
callculus
17.8k31427
Your calculation does not match mine. It is structurally different. We want
$1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$
For $X=2$ we have several combinations
$110 quad 200$
$011quad 020$
$101quad 002$
$X=1:$ $100,$ $ 010,$$ 001$
$X=0$: $000$
Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$
answered Nov 20 at 15:59
callculus
17.8k31427
edited Nov 20 at 16:13
answered Nov 20 at 15:59
callculus
17.8k31427
answered Nov 20 at 15:59
callculus
17.8k31427
answered Nov 20 at 15:59
callculus
17.8k31427
17.8k31427
add a comment |
add a comment |
We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.
answered Nov 20 at 16:47
J.G.
22.4k22035
add a comment |
We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.
answered Nov 20 at 16:47
J.G.
22.4k22035
add a comment |
We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.
answered Nov 20 at 16:47
J.G.
22.4k22035
We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.
answered Nov 20 at 16:47
J.G.
22.4k22035
answered Nov 20 at 16:47
J.G.
22.4k22035
answered Nov 20 at 16:47
J.G.
22.4k22035
answered Nov 20 at 16:47
J.G.
22.4k22035
22.4k22035
add a comment |
add a comment |
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Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23