AP EAMCET · Maths · Probability
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
- A \(\left(\frac{35}{36}\right)^{24}\)
- B \(\left(\frac{17}{18}\right)^{24}\)
- C \(\left(\frac{11}{12}\right)^{24}\)
- D \(\left(\frac{5}{6}\right)^{24}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{35}{36}\right)^{24}\)
Step-by-step Solution
Detailed explanation
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 \(=\frac{(35)^{24}}{(36)^{24}}\)
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