Probability of getting at least one job offer based on interview odds
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 Feb 7 at 20:02
Eevee Trainer
7,03321337
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
$endgroup$
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 Feb 7 at 20:02
Eevee Trainer
7,03321337
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
$endgroup$
$begingroup$
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
Feb 7 at 19:49
$begingroup$
Out of curiosity, how do you know your probability at each interview? Are you assuming the applicants have uniform probability?
$endgroup$
– Prince M
Feb 8 at 4:48
$begingroup$
I know the number of spots and the number of people interviewed for the spots, and I am just assuming all applicants have equal chances.
$endgroup$
– Yi Calvert
Feb 8 at 13:31
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 Feb 7 at 20:02
Eevee Trainer
7,03321337
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
$endgroup$
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 Feb 7 at 20:02
Eevee Trainer
7,03321337
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
edited Feb 7 at 20:02
Eevee Trainer
7,03321337
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
edited Feb 7 at 20:02
Eevee Trainer
7,03321337
edited Feb 7 at 20:02
Eevee Trainer
7,03321337
edited Feb 7 at 20:02
Eevee Trainer
7,03321337
7,03321337
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
asked Feb 7 at 19:47
Yi CalvertYi Calvert
311
311
$begingroup$
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
Feb 7 at 19:49
$begingroup$
Out of curiosity, how do you know your probability at each interview? Are you assuming the applicants have uniform probability?
$endgroup$
– Prince M
Feb 8 at 4:48
$begingroup$
I know the number of spots and the number of people interviewed for the spots, and I am just assuming all applicants have equal chances.
$endgroup$
– Yi Calvert
Feb 8 at 13:31
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
Feb 7 at 19:49
$begingroup$
Out of curiosity, how do you know your probability at each interview? Are you assuming the applicants have uniform probability?
$endgroup$
– Prince M
Feb 8 at 4:48
$begingroup$
I know the number of spots and the number of people interviewed for the spots, and I am just assuming all applicants have equal chances.
$endgroup$
– Yi Calvert
Feb 8 at 13:31
$begingroup$
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
Feb 7 at 19:49
$begingroup$
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
Feb 7 at 19:49
$begingroup$
Out of curiosity, how do you know your probability at each interview? Are you assuming the applicants have uniform probability?
$endgroup$
– Prince M
Feb 8 at 4:48
$begingroup$
Out of curiosity, how do you know your probability at each interview? Are you assuming the applicants have uniform probability?
$endgroup$
– Prince M
Feb 8 at 4:48
$begingroup$
I know the number of spots and the number of people interviewed for the spots, and I am just assuming all applicants have equal chances.
$endgroup$
– Yi Calvert
Feb 8 at 13:31
$begingroup$
I know the number of spots and the number of people interviewed for the spots, and I am just assuming all applicants have equal chances.
$endgroup$
– Yi Calvert
Feb 8 at 13:31
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
$$P(A) = 1 - P(textnot ; A)$$
That is to say, more relevant to your case,
$$P(textgetting at least one job offer) = 1 - P(textgetting no job offers)$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$beginalign
P(textgetting no offers) &= (1 - P(textgetting job #1)) \
× (1 - P(textgetting job #2)) \
× (1 - P(textgetting job #3)) \
&... \
× (1 - P(textgetting job #10))
endalign$$
With these two facts in mind you should find it easy to complete.
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
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 - $(frac910)^7 * frac4350 * frac4750 * frac1920$
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
answered Feb 7 at 20:00
zadachozadacho
1265
$endgroup$
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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
$$P(A) = 1 - P(textnot ; A)$$
That is to say, more relevant to your case,
$$P(textgetting at least one job offer) = 1 - P(textgetting no job offers)$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$beginalign
P(textgetting no offers) &= (1 - P(textgetting job #1)) \
× (1 - P(textgetting job #2)) \
× (1 - P(textgetting job #3)) \
&... \
× (1 - P(textgetting job #10))
endalign$$
With these two facts in mind you should find it easy to complete.
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
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
$$P(A) = 1 - P(textnot ; A)$$
That is to say, more relevant to your case,
$$P(textgetting at least one job offer) = 1 - P(textgetting no job offers)$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$beginalign
P(textgetting no offers) &= (1 - P(textgetting job #1)) \
× (1 - P(textgetting job #2)) \
× (1 - P(textgetting job #3)) \
&... \
× (1 - P(textgetting job #10))
endalign$$
With these two facts in mind you should find it easy to complete.
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
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
$$P(A) = 1 - P(textnot ; A)$$
That is to say, more relevant to your case,
$$P(textgetting at least one job offer) = 1 - P(textgetting no job offers)$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$beginalign
P(textgetting no offers) &= (1 - P(textgetting job #1)) \
× (1 - P(textgetting job #2)) \
× (1 - P(textgetting job #3)) \
&... \
× (1 - P(textgetting job #10))
endalign$$
With these two facts in mind you should find it easy to complete.
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
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
$$P(A) = 1 - P(textnot ; A)$$
That is to say, more relevant to your case,
$$P(textgetting at least one job offer) = 1 - P(textgetting no job offers)$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$beginalign
P(textgetting no offers) &= (1 - P(textgetting job #1)) \
× (1 - P(textgetting job #2)) \
× (1 - P(textgetting job #3)) \
&... \
× (1 - P(textgetting job #10))
endalign$$
With these two facts in mind you should find it easy to complete.
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
answered Feb 7 at 20:00
Eevee TrainerEevee Trainer
7,03321337
7,03321337
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 - $(frac910)^7 * frac4350 * frac4750 * frac1920$
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
answered Feb 7 at 20:00
zadachozadacho
1265
$endgroup$
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
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 - $(frac910)^7 * frac4350 * frac4750 * frac1920$
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
answered Feb 7 at 20:00
zadachozadacho
1265
$endgroup$
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
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 - $(frac910)^7 * frac4350 * frac4750 * frac1920$
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
answered Feb 7 at 20:00
zadachozadacho
1265
$endgroup$
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 - $(frac910)^7 * frac4350 * frac4750 * frac1920$
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
answered Feb 7 at 20:00
zadachozadacho
1265
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
edited Feb 7 at 20:32
Mike Earnest
23.9k12051
23.9k12051
answered Feb 7 at 20:00
zadachozadacho
1265
answered Feb 7 at 20:00
zadachozadacho
1265
answered Feb 7 at 20:00
zadachozadacho
1265
1265
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
add a comment |
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
There's an error in the first term of your product. It should read $(9/10)^7.$
$endgroup$
– Robert Shore
Feb 7 at 20:10
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
$begingroup$
thank you, corrected :)
$endgroup$
– zadacho
Feb 7 at 20:13
add a comment |
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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
Feb 7 at 19:49
$begingroup$
Out of curiosity, how do you know your probability at each interview? Are you assuming the applicants have uniform probability?
$endgroup$
– Prince M
Feb 8 at 4:48
$begingroup$
I know the number of spots and the number of people interviewed for the spots, and I am just assuming all applicants have equal chances.
$endgroup$
– Yi Calvert
Feb 8 at 13:31