1
$\begingroup$

Suppose there are 10 people apart of a club: A, B, C, D, E, F, G, H, I, and J. They decide to go to a restaurant for a club outing, but there isn't one table to seat all of them, so they decide to take one table that seats four people and two tables that seat three people. Based on this, what is the probability that Person G and Person J sit at the same table?

  • Ways for both Person G and Person J to both sit at the 4-person table: 6 = (4 choose 2)
  • Ways for both Person G and Person J to both sit at the 3-person table: 3 = (3 choose 2)
  • Ways for both Person G and Person J to both sit at the other 3-person table: 3 = (3 choose 2)

Total of $6+3+3=12$ ways for Person G and Person J to both sit at the same table.

I've determined the number of ways that Person G and Person J can sit together at each table, but I don't know how to properly count the total different ways (i.e., the denominator) to determine the probability.

$\endgroup$
  • 1
    $\begingroup$ Are we counting the two tables of $3$ as distinct tables? $\endgroup$ – scoopfaze Jan 30 at 2:08
  • 1
    $\begingroup$ so far off. think about the other tables, if they chose their spots, if reordering changes things, all these affect the answer... 6 just aint cutting it. $\endgroup$ – user645636 Jan 30 at 2:33
  • $\begingroup$ Is the seat of each person important, or only at which table they seat? $\endgroup$ – Alain Remillard Jan 30 at 2:40
  • $\begingroup$ @AlainRemillard Only the table $\endgroup$ – 324 Jan 30 at 2:44
  • $\begingroup$ so not the other people at it, not where they sit within the table seating, not anything interesting ? if so you over estimated by a factor of 2. $\endgroup$ – user645636 Jan 30 at 2:49
3
$\begingroup$

As pointed out in the comments, it depends whether the 3-persons tables are distinguishable or not.

If the 3-persons table are distinguished

There is Three cases to look at.

  • If $G$ and $J$ sit at the 4-persons table, we need two more to seat with them, from remaining eight. Then choose the three, from remaining six for first 3-persons table, the last three goes at the last table. $${8\choose2} {6\choose3} {3\choose3}=560$$

  • if they sit at the first 3-persons table, we need one more to complete the table. Then split the remaining seven. $${8\choose1} {7\choose4}{3\choose3}=280$$

  • If they sit at the second 3-persons table, it is exactly the same as above. $${8\choose1} {7\choose4}{3\choose3}=280$$

The number of possibilities that $G$ and $J$ are seated together. $$560+280+280=1120$$

For the total possibilities, we choose four, from ten, for the 4-persons table, then three from remaining six for the first 3-persons table, the last three goes on the last table. $${10\choose4}{6\choose3}{3\choose3}=4200$$

The probability that $G$ and $J$ are seated together is $$\frac{1120}{4200}=\frac{4}{15}$$

If the 3-persons tables are not distinguished

Funny thing, the probability is the same. Simply divide by two, since every possibility has been counted twice. $1120\div2=560$ ways to seat them together, and $4200\div2=2100$ total possibilities. $$\frac{560}{2100}=\frac{4}{15}$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Here is another approach. We assume that the $10$ people arrange themselves in some random order and then sit down, so there are $10!$ possible orders, all of which we assume are equally likely, with the first $4$ people at the first table, the next $3$ at the second, and the last $3$ at the third.

If G and J sit together at the first table, then their places can be chosen in $4 \times 3$ ways, and the remaining people can be placed in $8!$ ways.

If G and J sit at the second or third tables, then their places can be chosen in $3 \times 2$ ways, and then the remaining people can be placed in $8!$ ways.

So the probability that G and J sit at the same table is $$\frac{4 \times \ 3 \times 8! + 2 \times 3 \times 2 \times 8!}{10!}= \frac{4}{15}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ clever approach +1 ! $\endgroup$ – G Cab Jan 30 at 19:00
1
$\begingroup$

By conditional probability, we have

$$P(J\text{ is at $G$’s table}) = P(G\text{ at $4$-table})P(J\text{ at $G$’s table} | G\text{ at $4$-table}) + P(G\text{ at a $3$-table})P(J\text{ at $G$’s table}| G\text{ at a $3$-table})$$

$$\left(\frac4{10}\right)\left(\frac39\right)+\left(\frac6{10}\right)\left(\frac29\right) = \frac2{15} + \frac2{15} = \frac4{15}$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I am considering the tables to be identical because you haven, t clarified that in the question, so in question we have 6 seats in two tables, where each table contains exact three seats, and one table which contain exact 4 seats.

So we treat both ,G, J as together one. Case 1=when both seat on table 1 consist of 4 people So we have 8 options left to choose 2 spots because rest 2 were already occupied by them. , and similarly for next table we have only 6 options left to choose three people and left 3 people have only one choice that is last, and for cases 2)you can count the similar way. Now total umber of possible cases are we choose 4 out of 10 people from table one, and 3 out of 6 people, and left three with left chairs. So by calculating the value you will get the answer $4/15$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.