n married couples are invited to a dance party. for the first dance n paires are radomly selected. what′s the probability that no woman dances with her own husband? 1) if a pair must be of different genders. 2) if a pair can also be of the same gender.


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Question Number 207787 by mr W last updated on 26/May/24

$n m a r r i e d c o u p l e s a r e i n v i t e d t o$ $a d a n c e p a r t y . f o r t h e f i r s t d a n c e$ $n p a i r e s a r e r a d o m l y s e l e c t e d .$ ${w h a t}^{'} s t h e p r o b a b i l i t y t h a t n o w o m a n$ $d a n c e s w i t h h e r o w n h u s b a n d ?$ $1) i f a p a i r m u s t b e o f d i f f e r e n t$ $g e n d e r s .$ $2) i f a p a i r c a n a l s o b e o f t h e s a m e$ $g e n d e r .$
	Answered by A5T last updated on 26/May/24

	$F o r c a s e 1)$ $T h e n u m b e r o f w a y s s u c h t h a t n o w o m a n$ $d a n c e s w i t h h e r o w n h u s b a n d i s$ $D_{n} = n! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + . . . \underline{+} \frac{1}{n!})$ $(e n d s w i t h + \frac{1}{n!} w h e n n i s e v e n, - o t h e r w i s e)$ $\Rightarrow P r o b a b i l i t y = \frac{D_{n}}{n!} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + . . . \underline{+} \frac{1}{n!}$
	Commented by mr W last updated on 26/May/24
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	Commented by A5T last updated on 27/May/24

	$D_{n} = n! [\underset{n = 0}{\sum^{n}} {(- 1)}^{n} \frac{1}{n!}]$
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