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Suppose that in a company there are three groups of employees:

  • $80$ associate consultants
  • $50$ consultants
  • $20$ senior consultants

The president of the company is calling out names of each of the $150$ employees one by one and giving them a ticket to the Universal Studios. The names of the employees are called out at random without any specific preference to any one group of employees and each employee has a different name. A group is considered "completed" if all the members of the group receive the ticket. Find the probability that the group of associate consultants gets "completed" before consultants and senior consultants.

My attempt:

I thought of finding the favorable cases by considering say after calling out the K-th name, the associate consultants group gets completed. So there should be atleast one person each of consultant and senior consultant in remaining 150-K names that should be called out. Then I summed over all possible values of K, but neither did this result in a closed form solution nor do I think that it is correct because cases are being repeated.

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  • $\begingroup$ Welcome to Math.SE ! You should show us what you have attempted to solve the question, why it failed, etc. Look at meta.math.stackexchange.com/questions/9959/… to find out how to ask a good question. $\endgroup$
    – Tom-Tom
    Commented Oct 26, 2016 at 10:16
  • $\begingroup$ I have edited the question. Thanks $\endgroup$ Commented Oct 26, 2016 at 10:20

2 Answers 2

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To avoid misunderstandings I will speak of associates, consultants and seniors (so that the word "consultant" will not be overworked).


The answer is: $$\frac{50}{150}\frac{20}{100}+\frac{20}{150}\frac{50}{130}=\frac{23}{195}\simeq0.117949$$


First term:

Here $\frac{50}{100}$ is the probability that the consultants are completed as last and $\frac{20}{100}$ is the probability that the seniors are not completed as first under the extra condition that the consultants are completed as last.

Second term:

Here $\frac{20}{150}$ is the probability that the seniors are completed as last and $\frac{50}{130}$ is the probability that the consultants are not completed as first under the extra condition that the seniors are completed as last.


So actually if $E$ denotes the event that the associates are completed as first, $C$ denotes the event that the consultants are completed as last and $S$ denotes the event that the seniors are completed as last, then we calculated:$$\Pr\left(E\right)=\Pr\left(E\cap C\right)+\Pr\left(E\cap S\right)=\Pr\left(C\right)\Pr\left(E\mid C\right)+\Pr\left(S\right)\Pr\left(E\mid S\right)$$


Observe that e.g. the event that the consultants are completed as last is the same as the event that the last name mentioned by the president is the name of a consultant. This makes it easy to find the corresponding probability. Under the condition that this indeed happens we now look at the $149$ precedents and become aware of the fact that the $49$ consultants among them are not relevant anymore. So we focus on the $100$ non-consultants (i.e. associates and seniors). We now find likewise a probability of $\frac{20}{100}$ that the last mentioned of them is a senior.

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  • $\begingroup$ Nice answer, nicely explained ! (+1) $\endgroup$ Commented Oct 28, 2016 at 9:08
  • $\begingroup$ @trueblueanil Thank you for your compliment. I like the fact that this problem has such an "easy" solution. $\endgroup$
    – drhab
    Commented Oct 28, 2016 at 9:09
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I simulated this and the probability is approximately $0.118$. Here is the code in Matlab:

n=1e4;          %number of iterations
A=1:80;         %index assignment
S=81:100;
C=101:150;
s=0;
for j=1:n
    index = randperm(150);  %random permutation
    for k=20:150    % only check after 20 calls
        if sum(ismember (index(1:k),S))==20 % all members in S are called
            break
        end
        if k>49
            if sum(ismember (index(1:k),C))==50 % all members in C are called
                break
            end
            if k>79
                if sum(ismember (index(1:k),A))==80 %all members in A are called
                    s=s+1;
                    break;
                end
            end
        end
    end
end
s/n         %probability 

To analytically solve this problem, I can give the following hints:

Assume $k$ is the number of people whose name is called. Let $\tt A$, $\tt S$, $\tt C$, be the lists of associate consultants, senior consultants and concultants, respectively. We have the following cases:

  1. $k<20$ : $\hspace{15mm}$ none of the lists can be completed.
  2. $20\le k<50$ : $\hspace{5mm}$ only $\tt S$ can be completed
  3. $50\le k<80$ : $\hspace{5mm}$ $\tt S$ or $\tt C$ can be completed
  4. $80\le k<150$ :$\hspace{3.5mm}$ $\tt S$ or $\tt C$ or $\tt A$ can be completed

We need to be in case 4 whose length is $150-80=70$. So an upperbound for the probability is $\frac{70}{150}$.

The undesired events are:

  • completion of $\tt S$ in 2, whose probability is $\frac{30}{150-20}$
  • completion of $\tt S$ in 3, whose probability is $\frac{30}{150-20}$
  • completion of $\tt C$ in 3, whose probability is $\frac{30}{150-50}$
  • completion of $\tt S$ or $\tt C$ in 4 before completion of $\tt A$.
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  • $\begingroup$ It seems that at least one of us must be wrong. Could you apply your simulation also to find the approximate probability that the group of consultants, respectively senior consultants are completed first? My outcomes are $\frac{161}{1365}\simeq0.117949$, $\frac{320}{1365}\simeq0.234432$, $\frac{884}{1365}\simeq0.647919$ so nicely add up to $1$. Yours too? $\endgroup$
    – drhab
    Commented Oct 27, 2016 at 8:26
  • $\begingroup$ The simulation does not disprove your answer, but actually confirms it @drhab. It is correct indeed. $\endgroup$
    – msm
    Commented Oct 27, 2016 at 11:48
  • $\begingroup$ Nice to hear that! Thank you for checking. $\endgroup$
    – drhab
    Commented Oct 27, 2016 at 11:49
  • $\begingroup$ @drhab You're welcome and thank you for the comment. $\endgroup$
    – msm
    Commented Oct 27, 2016 at 11:50

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