A) \[\frac{1}{6}\]
B) \[\frac{1}{8}\]
C) \[\frac{1}{9}\]
D) \[\frac{1}{12}\]
E) None of these
Correct Answer: C
Solution :
Explanation Option [c] is correct. In two throws of a die, \[n\left( S \right)=\left( 6\times 6 \right)=36\] Let E = event of getting a sum \[9=\left\{ \left( 3, 6 \right), \left( 4, 5 \right), \left( 5, 4 \right), \left( 6, 3 \right) \right\}\] \[\therefore \,\,\,P(E)\,=\frac{n(E)}{n(S)}=\frac{4}{36}=\frac{1}{9}\].
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