Three dice are thrown simultaneously, the probability of obtaining a total of 18 is

\frac{1}{6^{3}}

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\frac{1}{6^{2}}

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\frac{1}{6}

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\frac{1}{5}

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Solution

The correct option is A $\frac{1}{6^{3}}$
$T h r e e d i c e a r e t h r o w n s i m u l t a n e o u s l y, t h e p r o b a b i l i t y o f o b t a i n i n g a t o t a l o f 18 S u m o f n u m b e r s o f d i c e c a n b e 18 i s p o s s i b l e i n o n l y o n e w a y i . e 6 + 6 + 6 = 18 T h e r e f o r e p r o b a b i l i t y i s \frac{1}{6} * \frac{1}{6} * \frac{1}{6} = \frac{1}{6^{3}}$

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