The-number-1-2-3-4-7-are-randomly-divided-into-two-non-empty-subsets-The-probability-that-the-sum-of-the-numbers-in-the-two-subsets-being-equal-is-r-s-expressed-in-the-lowest-term-Find-r-s-

Question Number 116397 by john santu last updated on 03/Oct/20

T h e n u m b e r 1, 2, 3, 4, \dots, 7 a r e r a n d o m l y

d i v i d e d i n t o t w o n o n - e m p t y s u b s e t s .

T h e p r o b a b i l i t y t h a t t h e s u m o f t h e

n u m b e r s i n t h e t w o s u b s e t s b e i n g

e q u a l i s \frac{r}{s} e x p r e s s e d i n t h e l o w e s t

t e r m . F i n d r + s ?

Answered by mr W last updated on 03/Oct/20

1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

e a c h s u b s e t s h o u l d h a v e t h e s u m 14 .

s u b s e t s w i t h 1 n u m b e r & 6 n u m b e r s :

n = C_{1}^{7} = 7 p o s s i b i l i t i e s

n_{S} = 0

s u b s e t s w i t h 2 n u m b e r s & 5 n u m b e r s :

n = C_{2}^{7} = 21 p o s s i b i l i t i e s

n_{S} = 0

s u b s e t s w i t h 3 n u m b e r s & 4 n u m b e r s :

n = C_{3}^{7} = 35 p o s s i b i l i t i e s

7 + 3 + 4, 7 + 2 + 5, 7 + 1 + 6, 6 + 5 + 3 = 14

n_{S} = 4

p = \frac{0 + 0 + 4}{7 + 21 + 35} = \frac{4}{63} = \frac{r}{s}

\Rightarrow r + s = 4 + 63 = 67

w e c a n a l s o g e t t h e t o t a l n u m b e r o f

p o s s i b i l i t i e s v i a {_{2}^{7}} = 63 .

Commented by john santu last updated on 04/Oct/20

y e s . . t h a n k y o u

Answered by john santu last updated on 04/Oct/20

T h e t o t a l n u m b e r o f w a y s o f d i v i d i n g

t h e s e v e n n u m b e r s i n t o t w o n o n - e m p t y

s u b s e t s i s \frac{2^{7} - 2}{2} = 63 . N o t e t h a t s i n c e

1 + 2 + 3 + \dots + 7 = 28, t h e s u m o f t h e

n u m b e r i n e a c h o f t h e t w o g r o u p s i s 14 .

N o t e a l s o t h a t n u m b e r s 5, 6, 7 c a n n o t

b e i n t h e s a m e g r o u p s i n c e 5 + 6 + 7 = 18 > 14

c a s e (1) o n l y 6 & 7 i n t h e s a m e g r o u p a n d

5 t h e o t h e r g r o u p . {2, 3, 4, 5}, {1, 6, 7}

c a s e (2) o n l y 5 & 6 i n t h e s a m e g r o u p a n d

7 i n t h e o t h e r g r o u p . {1, 2, 5, 6}, {3, 4, 7}

{3, 5, 6}, {1, 2, 4, 7}

c a s e (3) o n l y 5 & 7 i n t h e s a m e g r o u p a n d

6 i n t h e o t h e r g r o u p . {2, 5, 7}, {1, 3, 4, 6}

H e n c e t h e r e a r e 4 s u c h p o s s i b i l i t i e s

T h u s t h e r e q u i r e d p r o b a b i l i t y i s

\frac{4}{63} y i e l d i n g t h a t p + q = 67