There are 4 people being interviewed for a job, one of them being me. There are 3 jobs. What is the probability of me getting a job? I think it's 75%.
[Math] 4 people apply for 3 jobs; what’s the probability that I will get a job
probability
Related Solutions
I think I find this line of reasoning rather too speculative, I'm afraid. :-)
First, you assert that $P(I) = 1/2$. I don't see any symmetry that justifies this kind of application of the principle of indifference; it's not as though the only difference between being selected for an interview and not being selected is the opportunity to go to the interview. From the job poster's perspective, they have limited time to interview candidates, whether five people apply for the posting, or five hundred.
But suppose we put that aside for the moment. You then assert that $P(J \mid I) = 1/4$, on the assumption that once half of the possibilities have been eliminated (presumably, the $\neg I$ portion), only half of the half remain, or $1/4$. But $P(J \mid I)$ is already a conditional probability—it expresses the probability of the more specific compound event $I$ and $J$ as a fraction of the probability of the condition $I$! If you are to apply the principle of indifference again, you should treat the two possibilities equally; that is, $P(J \mid I) = P(\neg J \mid I)$. And since, by excluded middle, $P(J \mid I)+P(\neg J \mid I) = 1$ necessarily, it should be the case that
$$ P(J \mid I) = P(\neg J \mid I) = \frac{1}{2} $$
Instead, you have
$$ P(J \mid I) = \frac{1}{4} $$
which implies
$$ P(\neg J \mid I) = \frac{3}{4} $$
This has no more basis than the assertion that $P(I) = 1/2$, and violates the principle of indifference to boot. (Of course, I said that the principle of indifference shouldn't be applied here, but since you appear to want to do it...)
In other words, when you arrive at $1/4$, you are determining the probability that you get the interview and you get the job (and that's $P(I, J)$), not the probability that you get the job given that you got the interview (that's $P(J \mid I)$). If you want to analyze the problem that way, there's nothing wrong with that (modulo indifference), but then you should just leave that value alone, and not multiply, again, the probability that you get the interview. That's already been accounted for in the joint probability.
I think the most we can say about it is as follows: Suppose the probability of being selected for a given job posting is $\sigma$. Then, by independence, the probability of being selected for at least one of two job postings is $1-(1-\sigma)^2$; the probability of being selected for at least one of three job postings is $1-(1-\sigma)^3$; and the probability of being selected for at least one of $k$ job postings is $1-(1-\sigma)^k$. Even independence seems rather daring, but I can see a case for it better than I can for indifference.
First of all, I upvoted your query. I regard your query in a very positive light, because you are (in effect) stretching your intuition. I regard intuition as indispensable in problems involving probability &/or combinatorics.
In your initial calculation, I am unfamiliar with your RHS terminology, but your LHS expression is accurate.
Critiquing your alternate approach is challenging, because I have to speculatively reverse-engineer your #'s. I was not able to do this. That is, I could not come to terms with your thinking.
However, although I do not regard this alternative approach as to be preferred in this problem, I do think that it is still highly educational to explore it. So...
The first thing to consider is how you are going to express your sample space, which normally becomes the denominator (D) of a fraction. Then when you compute the numerator (N), in a manner consistent with how you computed the denominator, the answer becomes $(N/D).$
The easy way is to set $D = (99)^8,$ which requires than when computing $N$, you construe order as important. That is, assuming that the people get a number in ascending order (i.e. Person-1 before Person-2), then [Person-1 = 5, Person-2 = 20] must be counted separately from [Person-1 = 20, Person-2 = 5].
The alternative method of setting $D$ is to accept the complexity that order must not be deemed important. I am going to reject this approach out of hand, because setting $D = (99)^8$ is simple and foolproof, as long as you are extremely careful when computing $N$.
When considering the different ways that a match occurred (i.e. computing $N$), one of the challenges is to avoid over-counting. There are two ways to approach this challenge.
(1) Inclusion-Exclusion : Given sets A, B, and letting |X| denote the
number of elements in set X, you have
$|A \cup B| = |A| + |B| - |A \cap B|.$
(2) Partition the cases of matched numbers into mutually exclusive groups whose union represents all pertinent cases. The mutual exclusivity is critical for avoiding over counting.
For this problem, I would rather eat ground glass than try (1) above, so against the backdrop of $D = (99)^8,$ I am going for (2).
First of all, totally ignoring what the matching numbers are, in how many different ways can at least two of the people have matching numbers?
I would identify the various ways into cases (i.e. each case is to be a mutually exclusive group), be sure that I haven't omitted any cases, be sure that each case is mutually-exclusive from all of the others, and then count the # of elements in each case (i.e. each group). Note that the counting method used must be consistent with setting $D = (99)^8.$
Group (1):
All 8 people had the same #.
The # of ways is 99, since once the # to be matched is to be identified,
there is only one way that it can occur.
Group (2):
7 people had the same #, and there was one odd man out.
There are 8 ways of choosing the odd man out.
Here, it would be a serious mistake to assume that there are
$\binom{99}{2}$ ways of choosing the pertinent numbers,
because having 7 people match #50, and 1 person getting #49 is different
from having 7 people match #49, and 1 person getting #50.
Therefore you have to
compute the # ways of choosing the numbers to be selected as
$99 \times 98$.
This makes the # of elements in this group:
$8 \times 99 \times 98.$
If pursued to conclusion, this answer would be very long. One of the challenges would be to precisely identify each of the groups.
If you represent the first two groups as (for example)
Group 1 = (8), Group 2 = (7,1),
Then (in my opinion) the next groups would be represented by
Group 3 = (6,2), and Group 4 = (6,1,1).
Instead, as one last examination, I am going to explore the group
(4,2,2), because I think that it has educational value.
There are $\binom{8}{4} \times \binom{4}{2} = \frac{8!}{4! \times 2! \times 2!} $
ways that the 8 people can be appropriately grouped into (4,2,2).
At this point it is essential to ask yourself some hard questions.
For example, in the above computation method, is the grouping of (for example)
the people into (1,2,3,4), (5,6), (7,8) over counted.
I say yes it has been over counted.
The easiest way to see this, is that once people (1,2,3,4) are grouped together
person-5 may go with either person-6, person-7, or person-8.
Therefore, the correct enumeration is $\binom{8}{4} \times 3.$
Keeping in mind the backdrop of $D = (99)^8,$ it is clear
that there are 99 choices for the first #. Then, the question becomes
are the further choices to be computed as $98 \times 97$, or
$\frac{98 \times 97}{2}.$
Keeping in mind that the grouping of (1,2,3,4), (5,6), (7,8)
has not been over counted, and keeping in mind that you are looking
for a counting method consistent with $D = (99)^8$,
I regard $(98 \times 97)$ as correct.
The easiest way to see this is to pretend that #99 was assigned to
the group (1,2,3,4).
Then, regardless of who Person-5 is partnered with
there are 98 choices for the number that is to be assigned
to Person-5 and his partner.
Then, there are 97 choices for the number that
is to be assigned to the final 2 people.
Thus, for the group (4,2,2), the computation should be
$\binom{8}{4} \times 3 \times 99 \times 98 \times 97.$
I realize that I have (in effect) dodged your actual question. I did this only because, as I say, I could not reverse engineer your thinking. Hopefully, this answer will show you the complexities involved.
Addendum
Response to Lunatic Flowers' comment of 2020-09-29, at about 10:30am PST
First of all, good thinking. Your comment indicates that you attach importance to the idea that the cases you identify must be mutually exclusive. This leaves three challenges:
(1)
In addition to the cases being mutually exclusive, they must also be comprehensive.
That means that any eventuality that could occur (such as 5 people getting one
number and 3 people getting a different number) must also be assigned a separate case.
(2)
You must stick with a single idea for the sample space (i.e. denominator $= D$).
In this case, I somewhat arbitrarily chose $D = (99)^8$, simply because it is the
simplest way of defining the sample space.
(3)
When enumerating the # of ways that a specific case can occur, you must be
extremely careful. That is, once you chose a strategy for defining
your sample space, then for each case that you enumerate, you must ensure that
its enumeration is perfectly consistent with your strategy, re defining the
sample space.
Assume that you accept that $D = (99)^8.$
Assume that you are focusing on the specific case where exactly $x$ people got the same #, and the other $(8-x)$ people each got a different number, so that of the numbers 1 through 99, $[8 + 1 - x]$ of these numbers were distributed.
To facililitate a precise enumeration, I would pretend that each person was assigned a name-tag (i.e. Person-1, Person-2, ... Person-8). I would then assume that the specific #'s (re from 1 to 99) to be distributed would be assigned as follows:
First, a number would be assigned to the group of $x$ people.
Then, with respect to the $(8-x)$ remaining people, the person with the lowest name-tag (e.g Person-1, if he is not part of the group of $x$ people), would then be assigned a number. Following that, of the people that had (still) not yet been assigned a number, the person with the lowest name-tag would then be assigned a number, and so forth.
The idea is that if (for example) neither Person-7 nor Person-8 are part of the matching group of $x$ people, the possibility of Person-7 being assigned #51 (for example) and Person-8 being assigned #50 must be considered as distinct when compared with the possibility of Person-7 being assigned #50 and Person-8 being assigned #51.
This approach would guarantee that the numerator for this specific case would be enumerated in a manner consistent with the denominator of $D = (99)^8.$
With this in mind, there are $\binom{8}{x}$ ways of selecting $x$ people to represent the group that had matching #'s. Further, once this group is selected, then one number must be assigned to this group, and the other $(8-x)$ people must each be assigned a separate number.
Therefore, my enumeration of the numerator for this specific case is
$\displaystyle \binom{8}{x} \times \frac{99!}{\left(99 - [8 + 1 - x]\right)!}.$
As an example, when $x = 3,$ the enumeration is
$\displaystyle \binom{8}{3} \times 99 \times 98 \times 97 \times 96 \times 95 \times 94.$
Again, I am unable to grapple with how you enumerate this possibility. I hope that you are able to understand my formula for the numerator when $x=3$, and for the general value of $x$. Please leave further comments, if you have more questions/challenges/disagreements.
Best Answer
This is not really a probability question as your ability to do the job and your ability to sell yourself at the interview will take part.
For example if you call you prospective new boss a moron the probability of getting the job is going to be close to zero.
If one or more of the other applicants insults the new boss and you don't then the probability of you getting the job will be higher.
But, If we assume the interview is not taken into account and the new employees are chosen at random then yes the probability would be $\frac{3}{4}$ or 75%.