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Total people $2n$on that $n$are couples.

They are randomly divided into $n$pairs.

Here, $W_i$is the event and $ith$woman paired with her husband.

Number of ways to $i$th woman can paired with others $=2n-1$

Number of ways to $ith$woman can paired with her husband $=1$

woman $i$is ready to pair with any of the remaining $"2n-1"$people.

So the probability to pair with her husband is$P\left(W_i\right)=\frac{1}{2n鈭�1}$

$P\left(W_i\right)=\frac{1}{2n鈭�1}$

Total people $2n$on that $n$are couples.

They are randomly divided into $n$pairs.

Here,$W_i$is the event.

In this case, we will see the probability of $i$th woman to pair with her husband.

But, $i$is likely to be pair with any people.

That means,$2n鈭�3$

Number of ways $i$th woman to pair with her husband is :$1$

Hence, the probability will be$P\left(W_i鈭�W_j\right)=\frac{1}{2n鈭�3}$

$P\left(W_i鈭�W_j\right)=\frac{1}{2n鈭�3}$

Total people $2n$on that role="math" localid="1646663902940" $n$are couples.

They are randomly divided into $n$pairs.

Here, $W_i$is the event.

For large sample size $n$

It becomes Poisson distribution with mean $位$equal to the probability that no couple is paired together.

If $n$is large then the possibility to pair with their husbands will approximately be Poisson with the following mean

$位=\underset{i=1}{\overset{n}{鈭�}}鈥�P\left(w_i\right)$

$=\frac{n}{2n鈭�1}$

$位=\frac{1}{2}$

probability that no wife is paired with their husband is:

$p\left(w_i=0\right)=\frac{e^{鈭�位}{位}^0}{0!}$

$=\frac{e^{鈭�\left({\displaystyle \frac{1}{2}}\right)}{\left(\frac{1}{2}\right)}^0}{1}$

$=e^{鈭�(1/2)}$

Therefore, the probability is$e^{鈭�\left(0.5\right)}$

The probability is$e^{鈭�\left(0.5\right)}$

Total people $2n$on that $n$are couples.

They are randomly divided into $n$pairs.

Here,$W_i$ is the event.

If the pairing must consist of a men and women, then this experiment reduces to match problem.

The experiment reduces to match problem.