Zero
It is 1.
Assuming the 5 cubes are standard, unbiased dice with the numbers 1-6 on the faces: There are 4 numbers {1, 2, 3, 4} which are less than 5. Each has a probability of 1/6 of appearing on 1 die, thus each die has a probability of 4 × 1/6 = 4/6 = 2/3 of showing a number less than 5 → probability of all 5 dice showing a number less than five = (2/3)⁵ = 32/243 ≈ 0.132 = 13.2 %
If there are five dice, there are four numbers less than five, giving each die a 2/3 chance. Over five iterations that is (2/3) to the 5th power or 32/243 or a 13.16 percent chance.
That depends how many numbers it's possible to get.
Zero
It is 1.
Assuming the 5 cubes are standard, unbiased dice with the numbers 1-6 on the faces: There are 4 numbers {1, 2, 3, 4} which are less than 5. Each has a probability of 1/6 of appearing on 1 die, thus each die has a probability of 4 × 1/6 = 4/6 = 2/3 of showing a number less than 5 → probability of all 5 dice showing a number less than five = (2/3)⁵ = 32/243 ≈ 0.132 = 13.2 %
The probability is 57/216 = 19/108
With a single cube, the probability is 100%With two cubes, the probability is 15/36 = 5/12 = 412/3 %
If there are five dice, there are four numbers less than five, giving each die a 2/3 chance. Over five iterations that is (2/3) to the 5th power or 32/243 or a 13.16 percent chance.
yes
Total numbers on the cube = 6 Numbers less than 5 = 4 Probability = 4/6 = 2/3 = 662/3 %
That depends how many numbers it's possible to get.
1/6 because the other numbers don't affect the probability
There are 2 numbers less than 3, so the probability in this case is 2 in 8, or 1 in 4. There are 3 numbers greater than 5, so the probability in this case is 3 in 8. There are 5 numbers less than 3 or greater than 5, so the probability in this case is 5 in 8.
573 / 1719