A round table probability question Hi guys I am writing my P exam for the second time and I remembered two question that confused me when writing the exam. I asked my prof. but it confused him as well. For simplicity I will ask one question here and post the other one after.
So if you could please help me I would really be greatful.
Question:
You have 3 Actuaries, 4 Social Workers, and 3 clients. They all sit at a ROUND table. What is the probability that NO actuaries sit beside each other.
My attempt:
I labelled 3 A, 4S, 3C. Then I labelled 1 to 10 on a circle. I placed A in one spot and I knew no other actuaries sit beside each other so we have 7 spaces left. so now how can we place those other 2 actuaries so they dont sit beside each other.
Then I got confused.Please help out.
Thank you
 A: Seat the first A. It is $2/9$ that the second one is one seat away, and $5/8$ that the third is OK. Otherwise it is $5/9$ the second is farther, then $4/8$ the third is OK. Total $30/72=5/12$
Expanded:  We don't care where the others are, only the A's.  The probability that the first A will sit next to another A is zero.  When the second A sits, with $\frac 29$ probability he sits next to the first and we will fail.  With probability $\frac 29$ he sits with one empty chair between him and the first A.  There are then five of the seats out of eight available to the third A-all but the one in between and the two on the outside of each A.  So the chance we succeed this way is $\frac 29 \cdot \frac 58=\frac 5{36}$  With probability $\frac 59$ he sits in a seat at least two chairs away from the first A.  Then four of the remaining eight are prohibited to the third A, so we succeed this way $\frac 59 \cdot \frac 48=\frac 5{18}$.  As these probabilities are disjoint, we add them to get $\frac 5{18}+\frac 5{36}=\frac 5{12}$ for a final answer.
A: The way you are asking the question it seems to me that whether it is social workers or clients has no relevance. So basically we have 10 people seated around a round table and 3 specific people must not sit next to each other.
So label the seats 1 through 10 as you suggested: You can choose 3 out of 10 in $\binom{10}{3}=120$ ways.
You can construct all ways to choose 3 non-adjacent seats by choosing $3$ out of $7$, say $i<j<k\in\{1,2,...,7\}$ and map them to $i+1,j+2,k+3$ which points to three seats among $\{2,3,...,10\}$ separated by at least one seat each. This produces $\binom{7}{3}=35$ succesful patterns.
But this does not account for patterns where the first seat chosen is $1$. These paterns can be constructed by fixing the first seat as $i=1$ and then choose $j,k\in\{2,3,...,7\}$ and mapping these to $j+1,k+2$ in $\{3,...,9\}$. This produces another $\binom{6}{2}=15$ succesful patterns.
To sum up we have $35+15=50$ succesful patterns out of $120$ possible patterns. Thus we have
$$
P=\frac{50}{120}=\frac{5}{12}
$$
probability of succes.
A: Let's think of putting all $10$ people in a line arrangement and then having the end points meet to make the table arrangement.  For starters there are $10!$ ways to do this. This will be our denominator. 
First we line up the $7$ non-actuaries. There are $7!$ ways to do this. Now we seat the $3$ actuaries. We can seat them in any of the $6$ spots between the non-actuaries plus one of the end seats (but not both since in the table arrangement they'd be next to each other). So there are $7$ choices for the first actuary, $6$ for the second and $5$ for the third.  In total we have:
$$7!\times7\times6\times5$$ 
ways to do this and likewise when allowing one of the actuaries to sit at the other endpoint. Summing these up we get twice our number above. But then in total this is double counting all of the arrangements when the actuaries are all on the inside (i.e. none are at the end seats in the line arrangement). The count of these ways is $7!\times6\times5\times4$. So in total we have the probability as:
$$\frac{2\times7!\times7\times6\times5-7!\times6\times5\times4}{10!}$$
$$=\frac{7!\times6\times5\times2(7-2)}{10!}$$
$$=\frac{300}{720}$$
$$=\frac{5}{12}$$
Good luck on your exams. 
