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Question Number 138060 by mathocean1 last updated on 09/Apr/21

W e t h r o w 2 t i m e s a c u b i c d i c e . I t s

f a c e s a r e n u m e r o t e d 0, 0, - 1, 1, 1, 1 .

W e n o t e a t h e r e s u l t o f t h e f a c e t h r o w

a n d b t h e r e s u l t o f s e c o n d o n e .

1) C a l c u l a t e t h e p r o b a b i l i t i e s t o h a v e

a a n d b w h i c h v e r i f y :

∙ a = - 1 a n d b = 0

∙ a = 1 a n d b = 0

∙ a^{2} + b^{2} = 2

∙ a - b = 1

2) S h o w t h a t t h e p r o b a b i l i t y t o h a v e

a - b = 1 k n o w i n g t h a t a^{2} + b^{2} = 2 i s 0.75

Answered by floor(10²Eta[1]) last updated on 09/Apr/21

1)

\frac{1}{6} \times \frac{2}{6} = \frac{1}{18}

\frac{3}{6} \times \frac{2}{6} = \frac{1}{6}

a^{2} + b^{2} = 2 \Rightarrow (a, b) = (1, 1), (1, - 1), (- 1, 1), (- 1, - 1)

\frac{4}{6} \times \frac{4}{6} = \frac{4}{9}

a - b = 1 \Rightarrow (a, b) = (0, - 1), (1, 0)

\frac{2}{6} \times \frac{1}{6} + \frac{3}{6} \times \frac{2}{6} = \frac{2}{9}

2) if a^{2} + b^{2} = 2 so a and b {can}^{'} t be 0 so

a - b \neq 1 . the probability is 0