A) \[\frac{1}{3}\]
B) \[\frac{1}{4}\]
C) \[\frac{3}{8}\]
D) \[\frac{29}{36}\]
Correct Answer: D
Solution :
(d): In a simultaneous throw of two dice, we have \[n\left( S \right)=(6\times 6)=36.\]4 Let E = event of getting two numbers whose product is composite. Then, \[E=\left( 1,4 \right),\left( 1,6 \right),\left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),(2,4)\], \[\left( 2,5 \right),\left( 2,6 \right),\left( 3,2 \right),\left( 3,4 \right),\left( 3,5 \right),\] \[\left( 3,6 \right),\left( 4,1 \right),~\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right),\] \[\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right),\] \[\left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right)\}.\] \[\therefore n\left( E \right)=29.\] \[p\left( E \right)=\frac{n(E)}{n(S)}=\frac{29}{36}.\] P(an even number) \[=\frac{67}{100}\]
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