Probability of Male Picking Wife as Dance Partner

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SUMMARY

The probability problem discussed involves 6 married couples at a party, where each male must select a female partner for dancing. The objective is to determine the probability that at least one male selects his own wife as his partner. The solution involves combinatorial analysis and the principle of complementary counting, leading to a definitive probability calculation.

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There are 6 married couples (12 people) in a party. If every male has to pick a female as his dancing partner, find the probability that at least one male pick his own wife as his dancing partner.
 
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Hello, anemone!

There are 6 married couples (12 people) in a party.
If every male has to pick a female as his dancing partner,
find the probability that at least one male picks his own wife
as his dancing partner.
There are $6! = 720$ possible outcomes.

If no man picks his own wife,
$\quad$ this is a derangement of 6 objects.
$d(6) \,=\,265$

$P(\text{no married couples}) \:=\:\dfrac{265}{720} \:=\:\dfrac{53}{144}$

Therefore: $\:P(\text{some married couples}) \;=\;1 - \dfrac{53}{144} \;=\;\dfrac{91}{144}$
 
soroban said:
Hello, anemone!


There are $6! = 720$ possible outcomes.

If no man picks his own wife,
$\quad$ this is a derangement of 6 objects.
$d(6) \,=\,265$

$P(\text{no married couples}) \:=\:\dfrac{265}{720} \:=\:\dfrac{53}{144}$

Therefore: $\:P(\text{some married couples}) \;=\;1 - \dfrac{53}{144} \;=\;\dfrac{91}{144}$

Well done, soroban! I solved this problem the very hard way, thus I'm happy to see this other approach to say the least.:o
 

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