1
$\begingroup$

Question. Four people sit down for dinner at a round table. They have assigned seating. What is the probability that 1, and only 1, person sits in their assigned seat.

My try:

We fix one of the members to a seat that is not theirs. Suppose the events $A_i, i=1,2,3$ are denoted as someone sitting in their correct seat. Then $$ \begin{align*} P(A_1 \cup A_2 \cup A_3) &= P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - \dots + P(A_1 \cap A_2 \cap A_3)\\ &= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} \cdot \frac{1}{2} - \dots + \frac{1}{3} \cdot \frac{1}{2} \\ &= 1 - \frac{1}{2} + \frac{1}{6} \\ &= \frac{2}{3}. \end{align*} $$

However, the answer is $\frac{1}{3}$. Meaning we are taking the complement. So, am I wrong in assuming that the fixed person is not in their assigned chair? If we fix him to be in his assigned chair then the complement makes sense (i.e. we need none of the other 3 in the correct spot). What is the basis for fixing the assigned person to their correct spot though?

$\endgroup$
2
$\begingroup$

Assume that person 1 is in his seat. Then 2,3,4 should not be in their assigned seats and this can be done in 2 ways (derangement on three symbols). Thus the number of ways in which only one is in the assigned seat is $4 \times 2 = 8$. Thus the required probability is $\frac{8}{24} = \frac{1}{3}$

| 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.