If 10 different balls are to be placed in 4 boxes at random , then the probability that two of these boxes contain exactly 2 and 3 balls


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Question Number 181939 by greougoury555 last updated on 02/Dec/22

$I f 10 d i f f e r e n t b a l l s a r e t o b e p l a c e d$ $i n 4 b o x e s a t r a n d o m, t h e n t h e p r o b a b i l i t y$ $t h a t t w o o f t h e s e b o x e s c o n t a i n$ $e x a c t l y 2 a n d 3 b a l l s$
	Answered by Acem last updated on 02/Dec/22

	$a ∙ I f y o u m e a n d i f f e r e n t b o x e s, h e r e i t i s t h e s o l u .$ $P (2 B, 2, 3 b .) = \frac{C_{2}^{4} C_{2}^{10} C_{3}^{8} 2! 2^{5}}{4^{10}}$ $= \frac{3 \times 2 \times 45 \times 2^{4} \times 7 \times 2^{5}}{2^{20}} = \frac{945}{2^{10}}$ $b ∙ i f s i m i l a r b o x e s, l e t m e k n o w, b u t t h e o r i g i n a l i s t h e y a r e$ $d i f f . b e c a u s e w h a t i t c o n t a i n i s d i f f .$
	Answered by mr W last updated on 03/Dec/22

	$t o p l a c e 10 d i f f e r e n t b a l l s i n t o 4$ $d i f f e r e n t b o x e s t h e r e a r e t o t a l l y$ $4^{10} w a y s .$ $s u c h t h a t 2 b o x e s c o n t a i n e x a c t l y 2$ $b a l l s i n e a c h a n d 2 b o x e s c o n t a i n$ $e x a c t l y 3 b a l l s i n e a c h, t h e r e a r e$ $\frac{10! 4!}{{(2!)}^{3} {(3!)}^{3}} w a y s .$ $p = \frac{10! 4!}{{(2!)}^{3} {(3!)}^{3} 4^{10}} = \frac{1575}{32 768} \approx 4 . 8 %$
	Commented by Acem last updated on 03/Dec/22

	$H i S i r!$ $A c c o r d i n g t o t h e q u e s t i o n, i t a s k s a b o u t o n l y b o x$ $c o n t a i n s 2 b a l l s, a n d t h e o t h e r c o n t a i n s 3 b a l l s .$ $r i g h t ?$
	Commented by mr W last updated on 03/Dec/22

	$t h e l a n g u a g e o f t h e q u e s t i o n i s n o t$ $v e r y c l e a r . i f i t s a y s ‘ ‘ o n e b o x c o n t a i n s$ $e x a c t l y 2 b a l l s a n d a n o t h e r b o x$ $c o n t a i n s e x a c t l y 3 {b a l l s}^{″}, t h e n {i t}^{'} s$ $c l e a r .$
	Commented by Acem last updated on 03/Dec/22

	$N o p r o b l e m, b u t w e g o t a n e w q u e s t i o n!$ $m y s o l u t i o n i s d i f f r e n t o f y o u r s$ $p l e a s e, {l e t}^{'} s h e l p e a c h o t h e r$ $∙ W h i c h o f t h e 2 b o x e s w i l l c o n t a i n 2 b a l l s e a c h ?$ $C_{2}^{4}$ $∙ S e l e c t i n g t h e 2 f i r s t d i f f . b a l l s C_{2}^{10}$ $n o w t h e r e a r e t w o m e t h o d s t o p u t t h e m {i n t o}_{B_{1}, B_{2}}$ $2!_{I t h i n k t h a t y o u {w o n}^{'} t a g r e e w i t h m e i n t h i s}$ $∙ S e l e c t i n g 2 o t h e r p a i r s o f b a l l s C_{2}^{8}$ $==$ $∙ S e l e c t i n g t h e r e s t C_{3}^{6} C_{3}^{3} 2!$ $N u m w a y s = C_{2}^{4} C_{2}^{10} C_{2}^{8} 2! C_{3}^{6} C_{3}^{3} 2! = \frac{315}{2^{6}}$ $P (A) = \frac{315}{2^{14}} = 1.92 %$
	Commented by mr W last updated on 03/Dec/22

	$10 d i f f e r e n t b a l l s i n t o 4 i d e n t i c a l$ $g r o u p s w i t h 2 b a l l s, 2 b a l l s, 3 b a l l s,$ $3 b a l l s r e s p e c t i v e l y :$ $\frac{10!}{{(2!)}^{2} {(3!)}^{2} 2! 3!} w a y s$ $s i n c e t h e b o x e s a r e d i f f e r e n t,$ $\frac{10! 4!}{{(2!)}^{2} {(3!)}^{2} 2! 3!} = 50 400 w a y s$ $t o t a l l y 4^{10} = 1 048 576 w a y s$ $\Rightarrow p = \frac{50 400}{1 048 576} \approx 4 . 8 %$
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