64-points-are-in-a-plane-x-y-x-0-1-2-7-y-0-1-2-7-4-points-are-chosen-at-random-What-is-the-proabability-the-lines-connecting-them-do-not-form-a-square-or-rectangle-

Question Number 2771 by prakash jain last updated on 26/Nov/15

64 points are in a plane :

(x, y), x \in {0, 1, 2, \dots, 7}, y \in {0, 1, 2, \dots, 7}

4 points are chosen at random .

What is the proabability the lines connecting

them do not form a square or rectangle ?

Answered by Rasheed Soomro last updated on 01/Dec/15

\boldsymbol Please \boldsymbol Guide \boldsymbol me \boldsymbol if \boldsymbol I \boldsymbol am \boldsymbol wrong .

\boldsymbol I \boldsymbol {don}^{'} \boldsymbol t \boldsymbol know \boldsymbol much \boldsymbol about PROBABILITY .

\boldsymbol A s s u m i n g \boldsymbol t h a t \boldsymbol s i d e s \boldsymbol o f \boldsymbol r e c t a n g l e \boldsymbol a r e

\boldsymbol h o r i z a n t a l \boldsymbol a n d \boldsymbol v e r t i c a l .

\boldsymbol A s s u m i n g \boldsymbol t h a t \boldsymbol a l l \boldsymbol v e r t i c e s \boldsymbol a r e \boldsymbol d i f f e r e n t \boldsymbol p o i n t s .

B e i n g \overset{―}{P_{1} P_{2}} h o r i z a n t a l, P_{1} a n d P_{2} h a v e s a m e y - c o o r d i n a t e .

F o r s a m e r e a s o n P_{3} a n d P_{4} h a v e s a m e y - c o o r d i n a t e .

B e i n g \overset{―}{P_{1} P_{4}} a n d \overset{―}{P_{2} P_{3}} v e r t i c a l, P_{1} a n d P_{4} h a v e s a m e

x - c o o r d i n a t e, a n d P_{2} a n d P_{3} h a v e s a m e x - c o o r d i n a t e .

L e t

P_{1} = (x_{1}, y_{1}), P_{2} = (x_{2}, y_{1}), P_{3} (x_{2}, y_{2}) a n d P_{4} = (x_{1}, y_{2})

w h e r e x_{i} \in {0, 1, 2, \dots, 7}, y_{i} \in {0, 1, 2, \dots, 7}

\underline{W e f i r s t c o n s i d e r p o s s i b i l i t y o f b e i n g r e c t a n g l e / s q u a r e .}

\boldsymbol P o s s i b i l i t y \boldsymbol o f \boldsymbol c b o o s i n g \boldsymbol P_{1}, \boldsymbol P_{2}, \boldsymbol P_{3} \boldsymbol a n d \boldsymbol P_{4}

\boldsymbol i s \boldsymbol P o s s i b i l i t y \boldsymbol o f \boldsymbol c b o o s i n g \boldsymbol t h e i r \boldsymbol c o o r d i n a t e s

\boldsymbol x_{1}, \boldsymbol y_{1}, \boldsymbol x_{2} \boldsymbol a n d \boldsymbol y_{2}

C h o o s i n g x_{1} h a s \frac{1}{8} p o s s i b i l i t y a n d y_{1} h a s \frac{1}{8} p o s s i b i l i t y .

H e n c e c h o o s i n g P_{1} h a s \frac{1}{64} p o s s i b i l i t y .

C h o o s i n g x_{2} [\neq x_{1}] a f t e r c h o o s i n g x_{1} h a s \frac{1}{7} p o s s i b i l i t y .

C h o o s i n g x_{1}, y_{1} a n d x_{2} h a s \frac{1}{8} \times \frac{1}{8} \times \frac{1}{7} = \frac{1}{448} p o s s i b i l i t y .

C h o o s i n g y_{2} a f t e r c h o o s i n g y_{1} h a s \frac{1}{7} p o s s i b i l i t y .

C h o o s i n g a l l t h e c o o r d i n a t e s x_{1}, y_{1}, x_{2} a n d y_{2} i n s u c c e s i o n

h a s \frac{1}{8} \times \frac{1}{8} \times \frac{1}{7} \times \frac{1}{7} = \frac{1}{3136} p o s s i b i l i t y o f \underline{b e i n g} r e c t a n g l e .

P o s s i b i l i t y o f \underline{n o t b e i n g} r e c t a n g l e = 1 - \frac{1}{3136} = \frac{3135}{3136}

Commented by prakash jain last updated on 04/Dec/15

May the question was not clear but the coordiantes

of the points are (x, y) are (0, 0) . . (0, 7)

(1, 0) to (1, 7)

so you know when two lines are ⊥^{r} .

Commented by Rasheed Soomro last updated on 30/Nov/15

∙ I u n d e r s t o o d f r o m y o u r q u e s t i o n :^{'} t h e s q u a r e a r r a y

o f 64 p o i n t s o f o r d e r 8 \times 8 f r o m (0, 0) t o {(7, 7)}^{'} . A m I w r o n g ?

∙ A r e t h e s i d e s o f r e c t a n g l e h o r i z a n t a l a n d v e r t i c a l o n l y ?

Commented by Rasheed Soomro last updated on 03/Dec/15

W h a t i s t h e m e a n i n g o f^{'} t w o \boldsymbol p o i n t s a r e ⊥^{r}^{'}

Commented by prakash jain last updated on 04/Dec/15

⊥^{r} = perpendicular . A ctually I meant lines .

I was travelling for last few days returned

only today .

Commented by Rasheed Soomro last updated on 04/Dec/15

T h A n k S!

T h A n k S!