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What is the probability that a randomly chosen number is not divisible by 2 3 or 6?
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∙ 14y ago
The 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".
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∙ 14y ago
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What is the probability that the number 23 will be randomly picked out of 100?
Assuming then that there are 100 numbers, 1-100, the probability of the number 23 being randomly picked out of 100 is: 1/100 or 0.01.
What is the probability of prime numbers being chosen randomly from the number 1-49?
15/49ExplanationThere are 15 prime numbers between 1 and 49 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47). If you randomly choose one natural number from the 49 numbers between 1 and 49 inclusive, there is a 15/49 probability that it will be prime.
What the probability of rolling a 6 sided die and getting a 4 or a number divisible by 3?
If the die is fair then for a single roll, the probability is 1/2.
What is the probability that Lindy will roll a number divisible by 3 on a number cube?
2/6 or 1/3
What is the probability of getting tails and selecting a 3 if you toss a coin and randomly select a number?
There are infinitely many numbers and so the probability of the second event is 0. As a result the overall probability is 0.
What is the probability that a number chosen is divisible by 2?
50 50 odd or even same probability
A number is randomly chosen between 1-50 what is the probability of selecting multiples of 36?
It is 0.02
A nursing class has 13 women and 18 men If a student is chosen randomly to be the team leader what is the probability the student is a woman?
The probability that a randomly chosen student is a woman can be calculated by dividing the number of women by the total number of students in the class. In this case, there are 13 women and 31 total students, so the probability is 13/31, which simplifies to approximately 0.419 or 41.9%.
A number is chosen at random from the first twelve whole numbers. What is theThe probability that it is isExactly divisible by 3?
1/3
A number is chosen at random from the first 20 positive whole numbers What is the probability that it is not a prime number?
There are 12 composite (and 8 primes) in the first twenty whole numbers. So the probability of randomly choosing a non-prime is 12/20 or 60%.
A number is chosen at random from the first 10 whole number What is the probability that it is not exactly divisible by 3?
the first 10 whole numbers are numbers 1 to 10 and in those numbers only 3 numbers are divisible by 3 in which 3, 6 and 9 therefore the probability of from those figures that the numbers won't be divisible by 3 is 7/10 or 70%.
What is the probability that the number 23 will be randomly picked out of 100?
Assuming then that there are 100 numbers, 1-100, the probability of the number 23 being randomly picked out of 100 is: 1/100 or 0.01.
What is the Probability that if a number is chosen the number is even?
50%
What is the probability of choosing a number greater than 21 if a number is randomly chosen between 1 and 50?
Assuming the uniform continuous distribution, the answer is 29/49. With the uniform discrete distribution, the answer is 29/50.
What is the probability that a number between 1 and 200 is divisible by neither 2 3?
The probability is 67/200.
A health class has 23 women and 18 men If a student is chosen randomly to be the team leader what is the probability the student is a woman?
Total number of woman over total number of people So for this problem it's 23/41
The digits 1 through 9 are randomly arranged to make a number What is the probability that the resulting number is divisible by 18?
The answer is 4/9 or 44.444... % Any permutation of the digits 1 to 9 will be divisible by 9. So divisibility by 18 depends only on whether or not the last digit is even. If the last digit is 2, 4, 6 or 8 the number is divisible by 18 and if it is 1, 3, 5, 7 or 9 it is not. So 4 favourable outcomes out of 9 ie probability = 4/9.
