two dice are thrown together . find the probability of that the product of the numbers on the top of the dice is (a)6 (b)12 (c( 7

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Solution

$(a) G i v e n, t w o d i c e a r e t h r o w n t o g e t h e r . S o, t o t a l n u m b e r o f o u t c o m e s = 6 \times 6 = 36 F a v o u r a b l e c a s e s o f g e t t i n g t h e p r o d u c t o f t h e n u m b e r o n t h e t o p o f t h e d i c e 6 = (2, 3), (3, 2), (6, 1), (1, 6) T o t a l n u m b e r o f f a v o u r a b l e c a s e s = 4 P r o b a b i l i t y o f g e t t i n g t h e p r o d u c t o f t h e n u m b e r o n t h e t o p o f t h e d i c e 6 = \frac{T o t a l n u m b e r o f f a v o u r a b l e c a s e s}{T o t a l n u m b e r o f o u t c o m e s} = \frac{4}{36} = \frac{1}{9} (b) F a v o u r a b l e c a s e s o f g e t t i n g t h e p r o d u c t o f t h e n u m b e r o n t h e t o p o f t h e d i c e 12 = (4, 3), (3, 4), (6, 2), (2, 6) T o t a l n u m b e r o f f a v o u r a b l e c a s e s = 4 P r o b a b i l i t y o f g e t t i n g t h e p r o d u c t o f t h e n u m b e r o n t h e t o p o f t h e d i c e 12 = \frac{T o t a l n u m b e r o f f a v o u r a b l e c a s e s}{T o t a l n u m b e r o f o u t c o m e s} = \frac{4}{36} = \frac{1}{9} (c) F a v o u r a b l e c a s e s o f g e t t i n g t h e p r o d u c t o f t h e n u m b e r o n t h e t o p o f t h e d i c e 7 = 0 T o t a l n u m b e r o f f a v o u r a b l e c a s e s = 0 P r o b a b i l i t y o f g e t t i n g t h e p r o d u c t o f t h e n u m b e r o n t h e t o p o f t h e d i c e 7 = \frac{T o t a l n u m b e r o f f a v o u r a b l e c a s e s}{T o t a l n u m b e r o f o u t c o m e s} = \frac{0}{36} = 0$

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