Given: Two unbiased dice are thrown.
So, we now have to determine the probability of getting the sum of digits on dice greater than 10
Total number of possible outcomes is 62=36
n (S) = 36
Let E be the event of getting same number on all the three dice
E = {(5,6) (6,5) (6,6)}
n (E) = 3
P (E) = \( \dfrac{ n (E) }{ n (S)}\)
=\( \dfrac{3 }{ 36}\)
= \( \dfrac{ 1 }{ 12}\)
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