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Question
Two persons A and B take turns in throwing a pair of dice. The first person to throw 9 from both dice will be awarded the prize. If A throws first, then the probability that Bwins the game is
Options
9/17
8/17
8/9
1/9
Solution
\[ \frac{8}{17}\]
\[9 \text{ can be obtained from throw of two dice in only 4 cases as given below } :\]
\[\left[ \left( 3, 6 \right),\left( 4, 5 \right),\left( 5, 4 \right),\left( 6, 3 \right) \right]\]
\[⇒ P\left( \text{ getting } 9 \right)=\frac{4}{36}=\frac{1}{9}\]
\[ P\left(\text{ not getting } 9 \right)=\frac{32}{36}=\frac{8}{9}\]
\[\text{ Now } ,\]
\[P\left( \text{ B is winning } \right)= P\left( { \text{ getting 9 in } 2}_{nd} \text{ throw } \right)+P\left( {\text{ getting 9 in } 4}_{th} \text{ throw } \right) + P\left( {\text{ getting 9 in } 6}_{th} \text{ throw } \right) + . . . \]
\[ = \frac{8}{9} \times \frac{1}{9} + \frac{8}{9} \times \frac{8}{9} \times \frac{8}{9} \times \frac{1}{9} + . . . \]
\[ = \frac{8}{81}\left[ 1 + \frac{64}{81} + \left( \frac{64}{81} \right)^2 + . . . \right]\]
\[ = \frac{8}{81} \times \frac{1}{1 - \frac{64}{81}}\]
\[ = \frac{8}{81} \times \frac{81}{17}\]
\[ = \frac{8}{17}\]
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