Question
Two dice are thrown. What is the probability that the sum of the faces equals or exceeds $$10\,?$$
A.
$$\frac{1}{{12}}$$
B.
$$\frac{1}{4}$$
C.
$$\frac{1}{3}$$
D.
$$\frac{1}{6}$$
Answer :
$$\frac{1}{6}$$
Solution :
Let $$E$$ be the sum of the faces equals or exceeds. Then,
$$\eqalign{
& E = \left\{ {\left( {5,\,5} \right),\,\left( {4,\,6} \right),\,\left( {6,\,4} \right),\,\left( {5,\,6} \right),\,\left( {6,\,5} \right),\,\left( {6,\,6} \right)} \right\} \cr
& \therefore \,n\left( E \right) = 6 \cr
& {\text{Hence,}}\,\,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{6}{{36}} = \frac{1}{6} \cr} $$