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In a game, a pair of dice is rolled 24 times. If a person wins the game by not getting 6 on both the dice in any one of the 24 rolls, then the probability that a person wins the game is
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The correct answer is:
$\left(\frac{35}{36}\right)^{24}$
No. of events in sample space when a pair of dice is rolled 24 times $=(6 \times 6)^{24}=(36)^{24}$.
No. of event of not getting on both of the dice $=36$
$\therefore$ The required probability is $=\frac{(35)^{24}}{(36)^{24}}$
No. of event of not getting on both of the dice $=36$
$\therefore$ The required probability is $=\frac{(35)^{24}}{(36)^{24}}$
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