A) \[\frac{9}{17}\]
B) \[\frac{8}{17}\]
C) \[\frac{8}{9}\]
D) \[\frac{1}{9}\]
Correct Answer: B
Solution :
The probability of throwing 9 with two dice \[=\frac{4}{36}=\frac{1}{9}\] \[\therefore \]The probability of not throwing \[9\] with two dice \[=\frac{8}{9}\] If A is to win he should throw 9 in 1st or 3rd or 5th attempt If B is to win, he should throw, 9 in 2nd, 4th attempt B¢s chances\[=\left( \frac{8}{9} \right)\,.\,\frac{1}{9}+{{\left( \frac{8}{9} \right)}^{3}}.\,\frac{1}{9}+....=\frac{\frac{8}{9}\times \frac{1}{9}}{1-{{\left( \frac{8}{9} \right)}^{2}}}=\frac{8}{17}\].
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