# Arranging Couples

4 married couples entering a restaurant, there is only one table available, therefore the waiter put 4 people randomly near table and the 4 others near the bar, what is the probability that:

A. there are no married couples near the table
B. there are only married couples near the table
C. one couple is at the bar
D. John is not with his wife
The event $E_i$ is the event that there are $i$ married couples $(i=0,1,2)$

So I know that $|\Omega|={8\choose 4}$

A= there are 4 people and we what to make sure that we don't have any couple so we will choose a husband or a wife so $|A|=2^4$ and $P(A)=0.22$

B= we have 4 people, and we want 2 couples so we have to choose between a 2 couples out of 4 or $|B|={4\choose 2}$ and $P(B)=0.08$

C= out of all the 4 seats at the bar 2 are occupied be the couple, now we have left with 6 people and 2 seats, so we have $|C|={6\choose 2}$ and $P(C)=0.21$

Am I right with those?

• Please rephrase your question. Do you mean the waiter puts $4$ people out of the $8$ near the table and the other $4$ people near the bar? Do we know that John is one of the $8$ people? And I think you mean that $|\Omega|={8 \choose 4}$. There are also other sets we could use: for example, taking order into account. – Rory Daulton Jul 6 '15 at 12:55
• Are you sure $|\Omega|=\binom 4 8$? Because some order may be counted for placing people into the seats of both tables. – Salomo Jul 6 '15 at 12:55
• sorry, I meant $8\choose 4$. Yes the waiter put 4 near the table and 4 near the bar. The order is not taking into account, just the fact if there are married couples – gbox Jul 6 '15 at 13:04
• Consider what each of the $\binom84$ elements of your probability space looks like. For part A, which of these $\binom84$ elements have no married couples near the table? How many of those elements are there? And similarly for B, C, and D. If there is a specific problem counting the number of ways one of those events can occur, it might be better to post a question describing the specific difficulty with that specific event. – David K Jul 6 '15 at 13:37
• In question C, do you mean exactly one couple at the bar? In D, are we sure that John is one of the $8$ people? In the definition of events $E_i$ do you mean $i$ married couples near the table? Why do you give us that definition, since $E_i$ is not used in any question? – Rory Daulton Jul 6 '15 at 14:08

For all questions, let use the sample space of possible sets (unordered groups) of $4$ people sitting at (near) the table. That does make $|\Omega|={8 \choose 4}$.

Question A: For no married couples near the table, we must have $1$ out of each couple near the table. That makes $|E_0|={2 \choose 1}^4$. So

$$P(E_0)=\frac {{2 \choose 1}^4}{{8 \choose 4}}=\frac {16}{70}=\frac 8{35}$$

Question B: For only married couples near the table, two complete couples are near the table. So we choose $2$ couples out of the available $4$ couples to be at the bar. That makes $|E_1|={4 \choose 2}$. So

$$P(E_2)=\frac {{4 \choose 2}}{{8 \choose 4}}=\frac {6}{70}=\frac 3{35}$$

Question C: I will assume that you mean exactly one couple at the bar. That probability is the same as exactly one couple near the table, so let's look at that. For that to happen, one complete couple is near the table, as well as two more people, one of each out of the three remaining couples. So we choose $1$ couple to be near the table (${4 \choose 1}$ possibilities), we choose $2$ couples out of the $3$ available (${3 \choose 2}$ possibilities), out of each of those couples we choose $1$ to be near the table (${2 \choose 1}$ possibilities each). That makes ${E_1}={4 \choose 1}\cdot {3 \choose 2} \cdot {2 \choose 1}\cdot {2 \choose 1}$. So

$$P(E_1)=\frac{{4 \choose 1}\cdot {3 \choose 2} \cdot {2 \choose 1}\cdot {2 \choose 1}}{{8 \choose 4}}=\frac{4\cdot 3\cdot 2\cdot 2}{70}=\frac{24}{35}$$

As a check, we see if those three probabilities add to $1$, as they should, and they do.

Question D: I will assume that we know for sure that John is one of the $8$ people in this problem. [I gave an incorrect analysis here. Sorry!] Your answer in your comment to this answer is correct.

• in D: If John is in the bar, we have left with 3 seats so we need to choose 3 people and not his wife $6\choose 3$ and the same goes if he is in the table so it is $2*{6\choose 3}$? and overall it is $40\over 70$? – gbox Jul 6 '15 at 14:36