In a game,a pair of dice is rolled 24 times. | vedclass

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(A) When a pair of dice is rolled,the total number of possible outcomes is $6 	imes 6 = 36$. The outcome where $6$ appears on both dice is $(6, 6)$,w

Vedclass · Accepted Answer

$\left(\frac{35}{36}\right)^{24}$

Vedclass · Answer

(A) When a pair of dice is rolled,the total number of possible outcomes is $6 	imes 6 = 36$. The outcome where $6$ appears on both dice is $(6, 6)$,which is only $1$ outcome. The probability of getting $6$ on both dice in a single roll is $P(E) = \frac{1}{36}$. The probability of not getting $6$ on both dice in a single roll is $P(E') = 1 - \frac{1}{36} = \frac{35}{36}$. Since the dice are rolled $24$ times independently,the probability of not getting $6$ on both dice in any of the $24$ rolls is the product of the probabilities for each roll. Therefore,the required probability is $\left(\frac{35}{36}ight)^{24}$.

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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