Answer:
The correct answer is 21.25%.
The simplest way to solve this question is to find the complementary event of the event described in the question, i.e. find the probability that the die will roll less than 2 ‘six’s in 5 rolls. Then we can find the probability that the die will roll at least 2 ‘six’s in 5 rolls using the following equation:
P(at least two 6’s in 5 rolls) = 1 - P(zero 6’s in 5 rolls) - P(exactly one 6 in 5 rolls).
The probability of getting a ‘six’ in a normal die is 1⁄6.
The probability of getting a ‘six’ in the loaded die is 1⁄6 *1.05 = 0.175.
Therefore, the probability of getting a number other than ‘six’ in the loaded die is 1 - 0.175 = 0.825.
The probability of not getting a ‘six’ in any of the 5 rolls equals the probability of getting a number other than ‘six’ in each of these rolls:
P(zero 6’s in 5 rolls) = 0.825 * 0.825 * 0.825 * 0.825 * 0.825 = 0.8255 ≈ 0.3822
The probability of getting exactly one ‘six’ in 5 rolls equals the probability of getting a ‘six’ in either one of the rolls and a number other than ‘six’ in all other rolls.
The probability of getting a ‘six’ in either one of the rolls equals 0.175*5. The reason for multiplying the probability of getting a ‘six’ by 5 (the number of rolls) is that we do not care in which of the 5 rolls we would get a ‘six’.
The probability of getting a number other than ‘six’ in all other rolls is 0.825 * 0.825 * 0.825 * 0.825.
Therefore, P(exactly one 6 in 5 rolls) = 0.175 * 5 * 0.8254 ≈ 0.4053
Thus, P(at least two 6’s in 5 rolls) = 1 - 0.3822 - 0.4053 = 0.2125 = 21.25%.