Probability of getting at least one job offer based on interview odds
$begingroup$
I have recently interviewed for a number of jobs, and am wondering what the odds are of getting accepted for one based on the odds of each interview and the total number of interviews.
- I had $7$ interviews which had a $1$ in $10$ chance of securing a job ($10$
interviewees for every applicant). - I had another interview with a $6%$ chance, another with a $5%$ chance, and another with a $14%$ chance.
So in total, $10$ places with odds of $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $14%$, $6%$, $5%$.
I know that I cannot simply add the numbers, but I am also stumped in that I am trying to find the odds of not one job offer, but a minimum of one offer....so the odds of either one positive return, two positive return, etc. vs the odds of all $10$ interviews coming up negative.
Anyone know how to calculate this problem?
probability
edited 4 hours ago
Eevee Trainer
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asked 4 hours ago
Yi CalvertYi Calvert
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$endgroup$
add a comment |
$begingroup$
I have recently interviewed for a number of jobs, and am wondering what the odds are of getting accepted for one based on the odds of each interview and the total number of interviews.
- I had $7$ interviews which had a $1$ in $10$ chance of securing a job ($10$
interviewees for every applicant). - I had another interview with a $6%$ chance, another with a $5%$ chance, and another with a $14%$ chance.
So in total, $10$ places with odds of $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $14%$, $6%$, $5%$.
I know that I cannot simply add the numbers, but I am also stumped in that I am trying to find the odds of not one job offer, but a minimum of one offer....so the odds of either one positive return, two positive return, etc. vs the odds of all $10$ interviews coming up negative.
Anyone know how to calculate this problem?
probability
edited 4 hours ago
Eevee Trainer
5,9021936
asked 4 hours ago
Yi CalvertYi Calvert
261
New contributor
Yi Calvert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
So the assumption is that in the first 7 interviews, each interviewee has an equal chance to get the offer? ... not very realistic IMHO.... Anyway, it sounds like you're not looking for the chance of getting exactly one job offer, but of getting at least one job offer, right? OK, then here's a HINT: Calculate the chance of not getting any job offer at all! Do you see how to do that? And how, if you have that, you can calculate the chance of getting at least one job offer on the basis of that?
$endgroup$
– Bram28
4 hours ago
add a comment |
$begingroup$
I have recently interviewed for a number of jobs, and am wondering what the odds are of getting accepted for one based on the odds of each interview and the total number of interviews.
- I had $7$ interviews which had a $1$ in $10$ chance of securing a job ($10$
interviewees for every applicant). - I had another interview with a $6%$ chance, another with a $5%$ chance, and another with a $14%$ chance.
So in total, $10$ places with odds of $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $14%$, $6%$, $5%$.
I know that I cannot simply add the numbers, but I am also stumped in that I am trying to find the odds of not one job offer, but a minimum of one offer....so the odds of either one positive return, two positive return, etc. vs the odds of all $10$ interviews coming up negative.
Anyone know how to calculate this problem?
probability
edited 4 hours ago
Eevee Trainer
5,9021936
asked 4 hours ago
Yi CalvertYi Calvert
261
New contributor
Yi Calvert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have recently interviewed for a number of jobs, and am wondering what the odds are of getting accepted for one based on the odds of each interview and the total number of interviews.
- I had $7$ interviews which had a $1$ in $10$ chance of securing a job ($10$
interviewees for every applicant). - I had another interview with a $6%$ chance, another with a $5%$ chance, and another with a $14%$ chance.
So in total, $10$ places with odds of $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $10%$, $14%$, $6%$, $5%$.
I know that I cannot simply add the numbers, but I am also stumped in that I am trying to find the odds of not one job offer, but a minimum of one offer....so the odds of either one positive return, two positive return, etc. vs the odds of all $10$ interviews coming up negative.
Anyone know how to calculate this problem?
probability
probability
edited 4 hours ago
Eevee Trainer
5,9021936
asked 4 hours ago
Yi CalvertYi Calvert
261
New contributor
Yi Calvert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Eevee Trainer
5,9021936
asked 4 hours ago
Yi CalvertYi Calvert
261
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Yi Calvert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 4 hours ago
Eevee Trainer
5,9021936
edited 4 hours ago
Eevee Trainer
5,9021936
edited 4 hours ago
Eevee Trainer
5,9021936
5,9021936
asked 4 hours ago
Yi CalvertYi Calvert
261
New contributor
Yi Calvert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
Yi CalvertYi Calvert
261
asked 4 hours ago
Yi CalvertYi Calvert
261
261
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Yi Calvert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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$begingroup$
So the assumption is that in the first 7 interviews, each interviewee has an equal chance to get the offer? ... not very realistic IMHO.... Anyway, it sounds like you're not looking for the chance of getting exactly one job offer, but of getting at least one job offer, right? OK, then here's a HINT: Calculate the chance of not getting any job offer at all! Do you see how to do that? And how, if you have that, you can calculate the chance of getting at least one job offer on the basis of that?
$endgroup$
– Bram28
4 hours ago
add a comment |
$begingroup$
So the assumption is that in the first 7 interviews, each interviewee has an equal chance to get the offer? ... not very realistic IMHO.... Anyway, it sounds like you're not looking for the chance of getting exactly one job offer, but of getting at least one job offer, right? OK, then here's a HINT: Calculate the chance of not getting any job offer at all! Do you see how to do that? And how, if you have that, you can calculate the chance of getting at least one job offer on the basis of that?
$endgroup$
– Bram28
4 hours ago
$begingroup$
So the assumption is that in the first 7 interviews, each interviewee has an equal chance to get the offer? ... not very realistic IMHO.... Anyway, it sounds like you're not looking for the chance of getting exactly one job offer, but of getting at least one job offer, right? OK, then here's a HINT: Calculate the chance of not getting any job offer at all! Do you see how to do that? And how, if you have that, you can calculate the chance of getting at least one job offer on the basis of that?
$endgroup$
– Bram28
4 hours ago
$begingroup$
So the assumption is that in the first 7 interviews, each interviewee has an equal chance to get the offer? ... not very realistic IMHO.... Anyway, it sounds like you're not looking for the chance of getting exactly one job offer, but of getting at least one job offer, right? OK, then here's a HINT: Calculate the chance of not getting any job offer at all! Do you see how to do that? And how, if you have that, you can calculate the chance of getting at least one job offer on the basis of that?
$endgroup$
– Bram28
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a problem in which the complementary approach will be the most fruitful - let's instead consider how likely you are to not get a job. We know that, for an event $A$, then
$$P(A) = 1 - P(text{not} ; A)$$
That is to say, more relevant to your case,
$$P(text{getting at least one job offer}) = 1 - P(text{getting no job offers})$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$begin{align}
P(text{getting no offers}) &= (1 - P(text{getting job #1})) \
× (1 - P(text{getting job #2})) \
× (1 - P(text{getting job #3})) \
&... \
× (1 - P(text{getting job #10}))
end{align}$$
With these two facts in mind you should find it easy to complete.
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
$endgroup$
add a comment |
$begingroup$
It depends if you're looking for the odds to get exactly one job, or at least one job.
To get at least one job you have to substract the probability to get none of the jobs from 1.
Pr(at least one job) = 1 - Pr(zero jobs) = 1 - $(frac{9}{10})^7 * frac{43}{50} * frac{47}{50} * frac{19}{20}$
edited 3 hours ago
Mike Earnest
22.5k12051
answered 4 hours ago
zadachozadacho
314
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zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
This is a problem in which the complementary approach will be the most fruitful - let's instead consider how likely you are to not get a job. We know that, for an event $A$, then
$$P(A) = 1 - P(text{not} ; A)$$
That is to say, more relevant to your case,
$$P(text{getting at least one job offer}) = 1 - P(text{getting no job offers})$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$begin{align}
P(text{getting no offers}) &= (1 - P(text{getting job #1})) \
× (1 - P(text{getting job #2})) \
× (1 - P(text{getting job #3})) \
&... \
× (1 - P(text{getting job #10}))
end{align}$$
With these two facts in mind you should find it easy to complete.
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
$endgroup$
add a comment |
$begingroup$
This is a problem in which the complementary approach will be the most fruitful - let's instead consider how likely you are to not get a job. We know that, for an event $A$, then
$$P(A) = 1 - P(text{not} ; A)$$
That is to say, more relevant to your case,
$$P(text{getting at least one job offer}) = 1 - P(text{getting no job offers})$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$begin{align}
P(text{getting no offers}) &= (1 - P(text{getting job #1})) \
× (1 - P(text{getting job #2})) \
× (1 - P(text{getting job #3})) \
&... \
× (1 - P(text{getting job #10}))
end{align}$$
With these two facts in mind you should find it easy to complete.
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
$endgroup$
add a comment |
$begingroup$
This is a problem in which the complementary approach will be the most fruitful - let's instead consider how likely you are to not get a job. We know that, for an event $A$, then
$$P(A) = 1 - P(text{not} ; A)$$
That is to say, more relevant to your case,
$$P(text{getting at least one job offer}) = 1 - P(text{getting no job offers})$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$begin{align}
P(text{getting no offers}) &= (1 - P(text{getting job #1})) \
× (1 - P(text{getting job #2})) \
× (1 - P(text{getting job #3})) \
&... \
× (1 - P(text{getting job #10}))
end{align}$$
With these two facts in mind you should find it easy to complete.
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
$endgroup$
This is a problem in which the complementary approach will be the most fruitful - let's instead consider how likely you are to not get a job. We know that, for an event $A$, then
$$P(A) = 1 - P(text{not} ; A)$$
That is to say, more relevant to your case,
$$P(text{getting at least one job offer}) = 1 - P(text{getting no job offers})$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$begin{align}
P(text{getting no offers}) &= (1 - P(text{getting job #1})) \
× (1 - P(text{getting job #2})) \
× (1 - P(text{getting job #3})) \
&... \
× (1 - P(text{getting job #10}))
end{align}$$
With these two facts in mind you should find it easy to complete.
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
answered 4 hours ago
Eevee TrainerEevee Trainer
5,9021936
5,9021936
add a comment |
add a comment |
$begingroup$
It depends if you're looking for the odds to get exactly one job, or at least one job.
To get at least one job you have to substract the probability to get none of the jobs from 1.
Pr(at least one job) = 1 - Pr(zero jobs) = 1 - $(frac{9}{10})^7 * frac{43}{50} * frac{47}{50} * frac{19}{20}$
edited 3 hours ago
Mike Earnest
22.5k12051
answered 4 hours ago
zadachozadacho
314
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
add a comment |
$begingroup$
It depends if you're looking for the odds to get exactly one job, or at least one job.
To get at least one job you have to substract the probability to get none of the jobs from 1.
Pr(at least one job) = 1 - Pr(zero jobs) = 1 - $(frac{9}{10})^7 * frac{43}{50} * frac{47}{50} * frac{19}{20}$
edited 3 hours ago
Mike Earnest
22.5k12051
answered 4 hours ago
zadachozadacho
314
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
add a comment |
$begingroup$
It depends if you're looking for the odds to get exactly one job, or at least one job.
To get at least one job you have to substract the probability to get none of the jobs from 1.
Pr(at least one job) = 1 - Pr(zero jobs) = 1 - $(frac{9}{10})^7 * frac{43}{50} * frac{47}{50} * frac{19}{20}$
edited 3 hours ago
Mike Earnest
22.5k12051
answered 4 hours ago
zadachozadacho
314
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It depends if you're looking for the odds to get exactly one job, or at least one job.
To get at least one job you have to substract the probability to get none of the jobs from 1.
Pr(at least one job) = 1 - Pr(zero jobs) = 1 - $(frac{9}{10})^7 * frac{43}{50} * frac{47}{50} * frac{19}{20}$
edited 3 hours ago
Mike Earnest
22.5k12051
answered 4 hours ago
zadachozadacho
314
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Mike Earnest
22.5k12051
edited 3 hours ago
Mike Earnest
22.5k12051
edited 3 hours ago
Mike Earnest
22.5k12051
22.5k12051
answered 4 hours ago
zadachozadacho
314
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
zadachozadacho
314
answered 4 hours ago
zadachozadacho
314
314
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
zadacho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
add a comment |
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
There's an error in the first term of your product. It should read ${(9/10)}^7.$
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
3 hours ago
add a comment |
Yi Calvert is a new contributor. Be nice, and check out our Code of Conduct.
Yi Calvert is a new contributor. Be nice, and check out our Code of Conduct.
Yi Calvert is a new contributor. Be nice, and check out our Code of Conduct.
Yi Calvert is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
So the assumption is that in the first 7 interviews, each interviewee has an equal chance to get the offer? ... not very realistic IMHO.... Anyway, it sounds like you're not looking for the chance of getting exactly one job offer, but of getting at least one job offer, right? OK, then here's a HINT: Calculate the chance of not getting any job offer at all! Do you see how to do that? And how, if you have that, you can calculate the chance of getting at least one job offer on the basis of that?
$endgroup$
– Bram28
4 hours ago