We align $N$ persons including person A and B in a row, what is the probability that A and B sit next to each other?
I can't find the right equation for this one, so if anyone could point me in the right direction it would help me a lot.
Thank you
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$\begingroup$ HINT : Tie A and B together and consider them as a singe person to count the cases where they sit together. $\endgroup$– najayazOct 11, 2015 at 17:21
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4 Answers
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It is enough to consider just $A$ and $B$.
$N$ seats have $N-1$ combos for for 2 adjacent seats, against $\binom{N}{2}$ total combos
thus $Pr = \dfrac{(N-1)}{\dbinom{N}{2}} = \dfrac2N$
answered Oct 11, 2015 at 18:14
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Case 1(A is at the end): A at the and probability is $\frac{2}{n}$. To choose B, there is one available slot, next to A is $\frac{1}{n-1}$.
Case 2(A is at the middle): A at the middle probability is $\frac{n-2}{n}$. To choose B, there is two available slots, next to A is $\frac{2}{n-1}$
Final: $\frac{2}{n}\frac{1}{n-1} + \frac{n-2}{n}\frac{2}{n-1} = \frac{2+2n-4}{n(n-1)} = \frac{2n-2}{n(n-1)} = \frac{2}{n}$
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Hint:
Seat person $A$ first.
Break into cases: Either person $A$ is sitting at an end, or person $A$ isn't sitting at an end.
Apply multiplication principle.
answered Oct 11, 2015 at 17:16
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$\begingroup$ @trueblueanil if my hint was not clear enough, I'm saying you should first find the probability that $A$ is sitting on an end seat, then find the probability that $B$ is sitting in the only available adjacent seat. Add this to the probability that $A$ is sitting in a center seat, then find the probability that $B$ is sitting in either of the seats next to $A$. Yes "for adjacent seats." $\endgroup$ Oct 11, 2015 at 18:30
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Hint: If A and B are sitting together, we can consider them as a single entity. Instead of n people, there will be n-1 people.
answered Oct 11, 2015 at 18:50
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