A fair die is rolled 4 times. What is the probability of having no 3 and no 4 among the rolls? Round your answer to three decimal places

66.667%

Step-by-step explanation:

In a fair die, there are 6 possible rolls. Out of the 6, 1 number is chosen each time. Since order does not matter in this problem, we will be applying combination, so this would be written as 6C1 x 4. In order to exclude 3 and 4, from 6, the choices must be narrowed down to 4. Written out as an equation, this would be 4C1 x 4. To get the percentage, simply divide the two equations.

(4C1 x 4)/(6C1 x 4) = 0.66667 = 66.667%

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )