Answer:
Explanation:
First, we prepare a sample space for the outcomes when two dice are thrown.
[tex]\begin{gathered} \mleft(1,1\mright),\mleft(1,2\mright),\mleft(1,3\mright),\mleft(1,4\mright),\mleft(1,5\mright),\mleft(1,6\mright) \\ \mleft(2,1\mright),\mleft(2,2\mright),\mleft(2,3\mright),\mleft(2,4\mright),\mleft(2,5\mright),\mleft(2,6\mright). \\ \mleft(3,1\mright),\mleft(3,2\mright),\mleft(3,3\mright),\mleft(3,4\mright),\mleft(3,5\mright),\mleft(3,6\mright). \\ \mleft(4,1\mright),\mleft(4,2\mright),\mleft(4,3\mright),\mleft(4,4\mright),\mleft(4,5\mright),\mleft(4,6\mright). \\ \mleft(5,1\mright),\mleft(5,2\mright),\mleft(5,3\mright),\mleft(5,4\mright),\mleft(5,5\mright),\mleft(5,6\mright). \\ \mleft(6,1\mright),\mleft(6,2\mright),\mleft(6,3\mright),\mleft(6,4\mright),\mleft(6,5\mright),\mleft(6,6\mright). \end{gathered}[/tex]There are a total of 36 possible outcomes.
Next, determine the number of outcomes whose sum is a 9 or more.
The number of outcomes with a sum of 9 or more = 10
Thus, we have the following probabilities:
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