A) \[\frac{1}{4}\]
B) \[\frac{1}{6}\]
C) \[\frac{7}{12}\]
D) \[\frac{5}{36}\]
Correct Answer: D
Solution :
(d): In a simultaneous throw of two dice, \[n\left( S \right)=(6\times 6)=36.\] Let n(E) = event of getting a total of 10 or \[11=\left\{ \left( 4,6 \right),\left( 6,4 \right),\left( 5,5 \right),\left( 5,6 \right),\left( 6,5 \right). \right\}\] \[\therefore p(E)=\frac{n(E)}{n(S)}=\frac{5}{36}.\]
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