Throw

We throw 2 times with 2 dices. What is the probability that the first roll will fall more than sum of 9 and the second throw have sum 3 or does not have the sum 4?

Correct result:

p =  55.5556 %

Solution:

$p=100\cdot (12-4)\mathrm{/}12\cdot (12-2)\mathrm{/}12=55.5556\mathrm{\%}$

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