On a typical 6-sided number cube, there are three odd numbers (1, 3, 5) and three even numbers (2, 4, 6). Therefore, there are three outcomes that are desired out of a total of six possible outcomes. 3/6 = 1/2, so your probability is 1 in 2 or 50%.
1/3
Total numbers on the cube = 6 Numbers less than 5 = 4 Probability = 4/6 = 2/3 = 662/3 %
1/6 or (16 and 2/3)%
A single roll of a fair number cube: 1/2 A single toss of a fair coin: 1/2 Both events: 1/4
The experimental probability of a number cube that lands on 5 four times in a twenty toss trial is Pexp(5) = 4/20 = 1/5 = 0.20 = 20%
2
1/3
A half.A half.A half.A half.
2/6 it has to be simplified so it is now 1/3
Total numbers on the cube = 6 Numbers less than 5 = 4 Probability = 4/6 = 2/3 = 662/3 %
The probability of flipping tails on a perfect coin in a perfect toss is 0.5. The probability of rolling 1 on a die is 1 in 6. Likewise, the probability of rolling 6 on a die is 1 in 6. So the probability of rolling either 1 or 6 is 2 in 6 (which is 1 in 3).
The probability is zero. That means that it can't happen, because there's no '7' anywhere on the cube.
1/12
1/3
2:3
If the cube is honest, then every number 1 - 6 has the same probability ... 1/6 = 16 2/3 %
The question does not say which event the probability is required for!