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Two dice are thrown. What is the probability that the sumof the faces equals or exceeds $10 ?$
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The correct answer is:
$1 / 6$
Let $\mathrm{E}$ be the sum of the faces equals or exceeds
Then, $\mathrm{E}=\{(5,5),(4,6),(6,4),(5,6),(6,5),(6,6)\}$
$\therefore \quad n(E)=6$
Hence, $P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$
Then, $\mathrm{E}=\{(5,5),(4,6),(6,4),(5,6),(6,5),(6,6)\}$
$\therefore \quad n(E)=6$
Hence, $P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$
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