Arranging Couples

Question

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?

Answer 1

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.