Two dice are thrown together, find the probability that the product of the numbers on the top of two dice is 12.

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Solution

$\begin{matrix} I f t h e t w o d i c e a r e t h o r w n t h e n t h e t o t a l o u t c o m e = 36 F o r g e t t i n g 12 F a v o u r a b l e o u t c o m e = {(2, 6) (3, 4) (4, 3) (6, 2)} P (p r o d u c t i s 12) = \frac{4}{36} = \frac{1}{9} H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{1}{9} . \end{matrix}$

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Two dice are thrown together, find the probability that the product of the numbers on the top of two dice is 12.

Ifthetwodicearethorwnthenthetotaloutcome=36Forgetting12Favourableoutcome={(2,6)(3,4)(4,3)(6,2)}P(productis12)=436=19Hence,therequiredprobabilityis19.