An unbiased die with faces marked 1, 2, 3, 4, 5, and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and th... An unbiased die with faces marked 1, 2, 3, 4, 5, and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is? Read more

Steps to Solve

1. Understanding the range of values The values on the die range from 1 to 6. We need to find the probability that the minimum value rolled is at least 2, and the maximum value rolled is at most 5. This means the values we are interested in are from the set {2, 3, 4, 5}.

2. Counting total outcomes When rolling the die four times, the total number of possible outcomes is given by: $$ 6^4 $$ Calculating this gives: $$ 6^4 = 1296 $$

3. Counting successful outcomes Next, we consider the scenarios where the only values that can appear are 2, 3, 4, and 5. So, we can represent this as rolling a modified die with faces 2, 3, 4, and 5. This gives us four possible values.

The number of outcomes for rolling this modified die four times is: $$ 4^4 $$ Calculating this gives: $$ 4^4 = 256 $$

4. Calculating the probability To find the probability that the minimum face value is at least 2 and the maximum is at most 5, we calculate: $$ P = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{256}{1296} $$ Now, simplifying this fraction: $$ P = \frac{256 \div 16}{1296 \div 16} = \frac{16}{81} $$