Let S={1, 2, 3, 4, 5, 6, 7} If we multiply atleast 2 numbers from this set with each other, what are the chances of the product to turn out to be divisible by 3?


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Question Number 182199 by depressiveshrek last updated on 05/Dec/22

$L e t S = {1, 2, 3, 4, 5, 6, 7}$ $I f w e m u l t i p l y a t l e a s t 2 n u m b e r s$ $f r o m t h i s s e t w i t h e a c h o t h e r, w h a t$ $a r e t h e c h a n c e s o f t h e p r o d u c t t o$ $t u r n o u t t o b e d i v i s i b l e b y 3 ?$
	Answered by Acem last updated on 06/Dec/22
	$This solution is for product only 2 numbers I will solve the question later P(product_(div. by 3) )= ((11)/(21)) a×b, b×a ≡ one way, a×a is refused according to the question The method: S_1 = {1, 2, 4, 5, 7} , S_2 = {3, 6} P(product_(div. by 3) )= ((C_1 ^( 5) C_( 1) ^( 2) + C_( 2) ^( 2) _(S_2 ) )/((n(n−1))/2)) ; n= 7$
	$T h i s s o l u t i o n i s f o r p r o d u c t o n l y 2 n u m b e r s$ $I w i l l s o l v e t h e q u e s t i o n l a t e r$ $P ({p r o d u c t}_{d i v . b y 3}) = \frac{11}{21}$ $a \times b, b \times a \equiv o n e w a y,$ $a \times a i s r e f u s e d a c c o r d i n g t o t h e q u e s t i o n$ $T h e m e t h o d :$ $S_{1} = {1, 2, 4, 5, 7}, S_{2} = {3, 6}$ $P ({p r o d u c t}_{d i v . b y 3}) = \frac{C_{1}^{5} C_{1}^{2} + \underset{S_{2}}{\underset{⏟}{C_{2}^{2}}}}{\frac{n (n - 1)}{2}}; n = 7$
	Commented by mr W last updated on 06/Dec/22

	$t h e q u e s t i o n i s ‘ ‘ t a k e a t l e a s t t w o$ ${n u m b e r s}^{″}, n o t ‘ ‘ t a k e o n l y t w o$ ${n u m b e r s}^{″} a s y o u a s s u m e d .$
	Commented by Acem last updated on 06/Dec/22

	$Y e s, i {d i d n}^{'} t n o t i c e t h a t w e l l, t h a n k y o u S i r$ $f o r t h e c o r r e c t i o n$
	Answered by mr W last updated on 06/Dec/22

	$1) t a k e 7 n u m b e r s$ $t o t a l : 1 w a y$ $d i v i s i b l e b y 3 : 1 w a y$ $2) t a k e 6 n u m b e r s$ $t o t a l : 7 w a y s$ $d i v i s i b l e b y 3 : 7 w a y s$ $3) t a k e 5 n u m b e r s$ $t o t a l : C_{5}^{7} = 21 w a y s$ $d i v i s i b l e b y 3 : 1 \times C_{3}^{5} + 2 \times C_{4}^{5} = 20$ $4) t a k e 4 n u m b e r s$ $t o t a l : C_{4}^{7} = 35 w a y s$ $d i v i s i b l e b y 3 : 1 \times C_{2}^{5} + 2 \times C_{3}^{5} = 30$ $5) t a k e 3 n u m b e r s$ $t o t a l : C_{3}^{7} = 35 w a y s$ $d i v i s i b l e b y 3 : 1 \times C_{1}^{5} + 2 \times C_{2}^{5} = 25$ $6) t a k e 2 n u m b e r s$ $t o t a l : C_{2}^{7} = 21 w a y s$ $d i v i s i b l e b y 3 : 1 + 2 \times C_{1}^{5} = 11$ $t a k e a t l e a s t 2 n u m b e r s :$ $t o t a l : 1 + 7 + 21 + 35 + 35 + 21 = 120 w a y s$ $d i v i s i b l e b y 3 : 1 + 7 + 20 + 30 + 25 + 11 = 94 w a y s$ $p = \frac{94}{120} \approx 78 . 3 %$
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