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$4$ males and $4$ females are randomly directed to $8$ seats, arranged into $2$ opposite rows (each row has $4$ seats). What is the probability that each male sits opposite to a female?

The key is $\dfrac{16}{35},$ but mine is only half of that. I'm not sure what I got wrong.

There are $8!$ ways to arrange the group. Let $A$ and $B$ be the two rows. There are $8$ ways to select a person for the first seat in row $A$ and $4$ ways to select a person from the other sex for the opposite seat in row $B$. Similarly, we have the next pairs $(6; 3), (4; 2), (2; 1)$ to fill up the rest of the seats, so the probability is: $P = \dfrac{8.4.6.3.4.2.2.1}{8!} = \dfrac{8}{35}.$

What am I missing ?

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    $\begingroup$ Your answer is correct. $\endgroup$ Commented May 18 at 8:49

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Here's another way to see that you're right.

After you fill the first seat, the probability that you're still alive after filling the seat across from it is $\frac 47$ because of the $7$ remaining people, $4$ are of the opposite sex from the first seat. Similarly, assuming you're still alive, the probability of staying alive after filling the next pair of seats is $\frac 35$ and the conditional probability that you're still alive after filling the third pair of seats is $\frac 23$.

Multiplying these three probabilities gets you $\frac{8}{35}$.

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