The probability of having a sum of either 6 or 10 is [tex]\frac{2}{9}[/tex]
The sample space has the size [tex]|S|=36[/tex]
Let
[tex]A=\text{the event of having a sum of 6}\\B=\text{the event of having a sum of 10}[/tex]
then
[tex]A=\{(1,5),(2,4),(3,3),(4,2),(5,1)\}\\|A|=5[/tex]
and
[tex]B=\{(4,6),(5,5),(6,4)\}\\|B|=3[/tex]
then, the probability of having a sum of either 6 or 10 will be
[tex]P(A\cup B)=P(A)+P(B)[/tex]
since [tex]A[/tex] and [tex]B[/tex] are mutually exclusive events. Substituting
[tex]P(A\cup B)=P(A)+P(B)\\=\frac{|A|}{|S|}+\frac{|B|}{|S|}\\=\frac{2}{9}[/tex]
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