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4 numbers: 1/74 = 0.000 061 035

7 numbers: 1/77 = 0.000 001 214

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Q: What is the probability that 4 and 7 randomly chosen numbers in a row are all divisible by 7?
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What is the probability that a randomly chosen number is not divisible by 2 3 or 6?

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".


How can you find the probability that a randomly chosen point in a figure lies in the shaded region?

The probability is the ratio of the area of the shaded area to the area of the whole figure.


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.


In a school, 22% of the students are sixth graders. What is the probability that a randomly chosen student will not be a sixth grader?

50 percent


How do you find the probability of a random point being chosen in a shaded region in a circle?

The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.The probability of a single point being chosen is 0.

Related questions

What is the probability of finding five numbers of six digits chosen randomly from 100000 to 999999?

There is 100% chance.


What is the probability that a randomly chosen letter is a vowel?

5/24, or five out of twenty four, or with numbers, five out of infinity


What is the probability that a randomly chosen number is not divisible by 2 3 or 6?

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".


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 of 2 randomly chosen angles of a square being congruent?

As all the angles in a square measure 90°, the probability of 2 randomly chosen angles being congruent is 1.


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


What is the probability that a number chosen is divisible by 2?

50 50 odd or even same probability


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 randomly chosen between 1-50 what is the probability of selecting multiples of 36?

It is 0.02


How can you find the probability that a randomly chosen point in a figure lies in the shaded region?

The probability is the ratio of the area of the shaded area to the area of the whole figure.


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.


In a school, 22% of the students are sixth graders. What is the probability that a randomly chosen student will not be a sixth grader?

50 percent