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…
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
$\left(\frac{35}{36}\right)^{24}$
$\left(\frac{17}{18}\right)^{24}$
$\left(\frac{11}{12}\right)^{24}$
$\left(\frac{5}{6}\right)^{24}$
Solution
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 $=\frac{(35)^{24}}{(36)^{24}}$