The probability is 67/200.
On a single roll is it 2/3.
Well, numbers divisible by 2 are even numbers, and the even numbers from 1 to 6 are 2,4,and 6. Since there are 3 out of 6 even numbers on a number cube, there is a probability 1/2 to roll an even number on a number cube.
Evens or 50/50 as either 2 or 4 or 6 (all divisible by 2) or 1 or 3 or 5 (not divisible by 2)
1/2
AnswerThe probability that a randomly chosen [counting] number is not divisible by 2 is (1-1/2) or 0.5. One out of two numbers is divisible by two, so 1-1/2 are not divisible by two.The probability that a randomly chosen [counting] number is not divisible by 3 is (1-1/3) = 2/3.Similarly, the probability that a randomly chosen [counting] number is not divisible by N is (1-1/N).The probability that a random number is not divisible by any of 2, 3 or 6 can be reduced to whether it is divisible by 2 or 3 (since any number divisible by 6 can definitely be divided by both and so it is irrelevant). This probability depends on the range of numbers available. For example, if the range is all whole numbers from 0 to 10 inclusive, the probability is 3/11, because only the integers 1, 5, and 7 in this range are not divisible by 2, 3, or 6. If the range is shortened, say just from 0 to 1, the probability is 1/2.Usually questions of this sort invite you to contemplate what happens as the sampling range gets bigger and bigger. For a very large range (consisting of all integers between two values), about half the numbers are divisible by two and half are not. Of those that are not, only about one third are divisible by 3; the other two-thirds are not. That leaves 2/3 * 1/2 = 1/3 of them all. As already remarked, a number not divisible by two and not divisible by three cannot be divisible by six, so we're done: the limiting probability equals 1/3. (This argument can be made rigorous by showing that the probability differs from 1/3 by an amount that is bounded by the reciprocal of the length of the range from which you are sampling. As the length grows arbitrarily large, its reciprocal goes to zero.)This is an example of the use of the inclusion-exclusion formula, which relates the probabilities of four events A, B, (AandB), and (AorB). It goes like this:P(AorB) = P(A) + P(B) - P(AandB)In this example, A is the event "divisible by 2", and B is the event "divisible by 3".
The probability is 67/200.
On a single roll is it 2/3.
If the die is fair then for a single roll, the probability is 1/2.
Well, numbers divisible by 2 are even numbers, and the even numbers from 1 to 6 are 2,4,and 6. Since there are 3 out of 6 even numbers on a number cube, there is a probability 1/2 to roll an even number on a number cube.
2/6 or 1/3
Evens or 50/50 as either 2 or 4 or 6 (all divisible by 2) or 1 or 3 or 5 (not divisible by 2)
1/2
Even numbers are divisible by two, and half of all numbers are even, so there is a 50 percent chance that a four-digit number is divisible by two.
1-2
1-2
If the die is rolled often enough, the event is a certainty - probability = 1. For a single roll, the probability is 1/2.