Question

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. \( \left(\frac{5}{6}\right)^{24} \)

Hint:

When dealing with repeated independent events, multiply the probabilities of the individual events to find the overall probability.

Answer 1

The Correct Option is A

In this game, we roll a pair of dice 24 times. To win the game, the person must not get a 6 on both dice in any one of the 24 rolls. The total number of outcomes when rolling two dice is 36 (since each die has 6 faces, so \( 6 \times 6 = 36 \) possible outcomes). The probability of not getting a 6 on both dice in a single roll is: \[ 1 - \frac{1}{36} = \frac{35}{36} \] This is the probability that the person does not get a 6 on both dice in any single roll. Now, since the dice are rolled 24 times, and we need the person to avoid rolling a 6 on both dice in all 24 rolls, the probability of winning the game is: \[ \left( \frac{35}{36} \right)^{24} \] Thus, the correct answer is \( \left( \frac{35}{36} \right)^{24} \).