Different definition of the geometric distribution
$begingroup$
From SOA Sample #199:
A company has five employees on its health insurance plan. Each year, each employee
independently has an $.80%$ probability of no hospital admissions. If an employee requires
one or more hospital admissions, the number of admissions is modeled by a geometric
distribution with a mean of $1.50$. The numbers of hospital admissions of different
employees are mutually independent.
Each hospital admission costs $20,000$.
Calculate the probability that the company’s total hospital costs in a year are less than
$50,000$.
I know how to find part of the solution, which involves finding the binomial for $0, 1$, and $2$ employees being hospitalized. I also know that the probability of the geometric distribution here is $2/3$. What I don't understand is what the geometric distribution means in this case. Until now, I understood that geometric distribution is either the number of trials to get to the first success, or the number of failures before the first success, but neither of those seem to be applicable here.
Is this some different definition of the geometric distribution, and if so what is it, or does the standard definition apply, and if so, how?
probability statistics
asked Dec 11 '18 at 21:29
agbltagblt
19814
$endgroup$
add a comment |
$begingroup$
From SOA Sample #199:
A company has five employees on its health insurance plan. Each year, each employee
independently has an $.80%$ probability of no hospital admissions. If an employee requires
one or more hospital admissions, the number of admissions is modeled by a geometric
distribution with a mean of $1.50$. The numbers of hospital admissions of different
employees are mutually independent.
Each hospital admission costs $20,000$.
Calculate the probability that the company’s total hospital costs in a year are less than
$50,000$.
I know how to find part of the solution, which involves finding the binomial for $0, 1$, and $2$ employees being hospitalized. I also know that the probability of the geometric distribution here is $2/3$. What I don't understand is what the geometric distribution means in this case. Until now, I understood that geometric distribution is either the number of trials to get to the first success, or the number of failures before the first success, but neither of those seem to be applicable here.
Is this some different definition of the geometric distribution, and if so what is it, or does the standard definition apply, and if so, how?
probability statistics
asked Dec 11 '18 at 21:29
agbltagblt
19814
$endgroup$
add a comment |
$begingroup$
From SOA Sample #199:
A company has five employees on its health insurance plan. Each year, each employee
independently has an $.80%$ probability of no hospital admissions. If an employee requires
one or more hospital admissions, the number of admissions is modeled by a geometric
distribution with a mean of $1.50$. The numbers of hospital admissions of different
employees are mutually independent.
Each hospital admission costs $20,000$.
Calculate the probability that the company’s total hospital costs in a year are less than
$50,000$.
I know how to find part of the solution, which involves finding the binomial for $0, 1$, and $2$ employees being hospitalized. I also know that the probability of the geometric distribution here is $2/3$. What I don't understand is what the geometric distribution means in this case. Until now, I understood that geometric distribution is either the number of trials to get to the first success, or the number of failures before the first success, but neither of those seem to be applicable here.
Is this some different definition of the geometric distribution, and if so what is it, or does the standard definition apply, and if so, how?
probability statistics
asked Dec 11 '18 at 21:29
agbltagblt
19814
$endgroup$
From SOA Sample #199:
A company has five employees on its health insurance plan. Each year, each employee
independently has an $.80%$ probability of no hospital admissions. If an employee requires
one or more hospital admissions, the number of admissions is modeled by a geometric
distribution with a mean of $1.50$. The numbers of hospital admissions of different
employees are mutually independent.
Each hospital admission costs $20,000$.
Calculate the probability that the company’s total hospital costs in a year are less than
$50,000$.
I know how to find part of the solution, which involves finding the binomial for $0, 1$, and $2$ employees being hospitalized. I also know that the probability of the geometric distribution here is $2/3$. What I don't understand is what the geometric distribution means in this case. Until now, I understood that geometric distribution is either the number of trials to get to the first success, or the number of failures before the first success, but neither of those seem to be applicable here.
Is this some different definition of the geometric distribution, and if so what is it, or does the standard definition apply, and if so, how?
probability statistics
probability statistics
asked Dec 11 '18 at 21:29
agbltagblt
19814
asked Dec 11 '18 at 21:29
agbltagblt
19814
edited Dec 11 '18 at 21:34
agblt
asked Dec 11 '18 at 21:29
agbltagblt
19814
asked Dec 11 '18 at 21:29
agbltagblt
19814
asked Dec 11 '18 at 21:29
agbltagblt
19814
19814
add a comment |
add a comment |
1 Answer
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$begingroup$
What does the geometric distiribution mean?
If anyone has made any number of hospital visits, they have $frac 13$ chance of making at least one more visit, and a $frac 23$ chance that this is their last visit.
Going through the cases:
No one files a claim $(0.80)^5$
One person files one claim.
1 of 5 people goes to the hospital.
They only have one visit.
$(0.80)^4(0.20)^1{5choose 1}(frac 23)$
2 claims are filed
1 claim each by 2 different people
$(0.80)^3(0.20)^2{5choose 2}(frac 23)^2$
2 claims by 1 person.
$(0.80)^4(0.20)^1{5choose 1}(frac 13)(frac 23)$
and sum them all up.
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
$endgroup$
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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$begingroup$
What does the geometric distiribution mean?
If anyone has made any number of hospital visits, they have $frac 13$ chance of making at least one more visit, and a $frac 23$ chance that this is their last visit.
Going through the cases:
No one files a claim $(0.80)^5$
One person files one claim.
1 of 5 people goes to the hospital.
They only have one visit.
$(0.80)^4(0.20)^1{5choose 1}(frac 23)$
2 claims are filed
1 claim each by 2 different people
$(0.80)^3(0.20)^2{5choose 2}(frac 23)^2$
2 claims by 1 person.
$(0.80)^4(0.20)^1{5choose 1}(frac 13)(frac 23)$
and sum them all up.
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
$endgroup$
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
add a comment |
$begingroup$
What does the geometric distiribution mean?
If anyone has made any number of hospital visits, they have $frac 13$ chance of making at least one more visit, and a $frac 23$ chance that this is their last visit.
Going through the cases:
No one files a claim $(0.80)^5$
One person files one claim.
1 of 5 people goes to the hospital.
They only have one visit.
$(0.80)^4(0.20)^1{5choose 1}(frac 23)$
2 claims are filed
1 claim each by 2 different people
$(0.80)^3(0.20)^2{5choose 2}(frac 23)^2$
2 claims by 1 person.
$(0.80)^4(0.20)^1{5choose 1}(frac 13)(frac 23)$
and sum them all up.
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
$endgroup$
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
add a comment |
$begingroup$
What does the geometric distiribution mean?
If anyone has made any number of hospital visits, they have $frac 13$ chance of making at least one more visit, and a $frac 23$ chance that this is their last visit.
Going through the cases:
No one files a claim $(0.80)^5$
One person files one claim.
1 of 5 people goes to the hospital.
They only have one visit.
$(0.80)^4(0.20)^1{5choose 1}(frac 23)$
2 claims are filed
1 claim each by 2 different people
$(0.80)^3(0.20)^2{5choose 2}(frac 23)^2$
2 claims by 1 person.
$(0.80)^4(0.20)^1{5choose 1}(frac 13)(frac 23)$
and sum them all up.
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
$endgroup$
What does the geometric distiribution mean?
If anyone has made any number of hospital visits, they have $frac 13$ chance of making at least one more visit, and a $frac 23$ chance that this is their last visit.
Going through the cases:
No one files a claim $(0.80)^5$
One person files one claim.
1 of 5 people goes to the hospital.
They only have one visit.
$(0.80)^4(0.20)^1{5choose 1}(frac 23)$
2 claims are filed
1 claim each by 2 different people
$(0.80)^3(0.20)^2{5choose 2}(frac 23)^2$
2 claims by 1 person.
$(0.80)^4(0.20)^1{5choose 1}(frac 13)(frac 23)$
and sum them all up.
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
answered Dec 11 '18 at 21:42
Doug MDoug M
45.2k31954
45.2k31954
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
add a comment |
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
$begingroup$
So basically, success = last visit. Thank you!
$endgroup$
– agblt
Dec 11 '18 at 21:44
add a comment |
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