Different definition of the geometric distribution
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
agblt
17414
add a comment |
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
agblt
17414
add a comment |
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
agblt
17414
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
agblt
17414
asked Dec 11 '18 at 21:29
agblt
17414
edited Dec 11 '18 at 21:34
asked Dec 11 '18 at 21:29
agblt
17414
asked Dec 11 '18 at 21:29
agblt
17414
asked Dec 11 '18 at 21:29
agblt
17414
17414
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 M
44.2k31854
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035850%2fdifferent-definition-of-the-geometric-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 M
44.2k31854
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
add a comment |
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 M
44.2k31854
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
add a comment |
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 M
44.2k31854
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 M
44.2k31854
answered Dec 11 '18 at 21:42
Doug M
44.2k31854
answered Dec 11 '18 at 21:42
Doug M
44.2k31854
answered Dec 11 '18 at 21:42
Doug M
44.2k31854
44.2k31854
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
add a comment |
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
So basically, success = last visit. Thank you!
– agblt
Dec 11 '18 at 21:44
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035850%2fdifferent-definition-of-the-geometric-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
