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Sagot :
Consider the set of all possible outcomes when rolling two fair dice,
[tex]S={}\lbrace(1,1),(1,2),(1,3),...,(2,1),(2,2),...,(6,4),(6,5),(6,6)\rbrace\rightarrow36\text{ possible outcomes}[/tex]All of the combinations above are equally probable.
The combinations that give us a sum equal to 9 are,
[tex]\begin{gathered} 3+6=6+3=9 \\ 4+5=5+4=9 \end{gathered}[/tex]Therefore, out of the total number of combinations, only (3,6), (6,3), (4,5), and (5,4) satisfy the given condition; then,
[tex]P(X=9)=\frac{4}{36}=\frac{1}{9}[/tex]
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