Answer
If the fair die is rolled four times, the probability of not rolling a 2 or 3 = (16/81)
Explanation
The question tells us that a fair die (a fair die is the regular die with 6 sides numbered from 1 to 6 with each side equally likely to turn up when it is rolled) is rolled 4 times, we should calculate the probability that we wont have a 2 or a 3 turning up.
Like I said, a fair die has 6 sides numbered 1 to 6.
If we don't want a 2 or a 3, the allowable numbers are 1, 4, 5 and 6.
The probability of obtaining a 1, 4, 5 or 6 in a single roll of a die is (4/6)
Note that it is (4/6) because the probability of an event (rolling a 1,4,5 or 6) is given as the number of elements in that event (1,4,5 and 6 are only four numbers) divided by the number of elements in the sample space (1,2,3,4,5 and 6 are six numbers in total).
So, now, if the die is rolled once
The probability of not rolling a 2 or 3 = The probability of rolling a 1, 4, 5 or 6 = (4/6) = (2/3)
So, if this same fair die is rolled four times, the probability of not rolling a 2 or 3 will be
(2/3) × (2/3) × (2/3) × (2/3) = (16/81)
(Since each roll is independent from one another, we can just multiply the probability of not rolling a 2 or a 3 in one trial in four places)
Hope this Helps!!!