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There are 8 integers under 50 divisible by 6 and 6 divisible by 8. 24 and 48 are common multiples so there are 12 qualifying integers. Probability is therefore 12 out of 50 or 24%.

(6,8,12,16,18,24,30,32,36,40,42,48.)

Q: An integer is chosen at random from the first 50 digitswhat is the probability that it is divisible by 6 or 8?

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50 50 odd or even same probability

The answer is 9*9!/9*109 = 0.0003629 approx.

The answer depends on how many are chosen at a time.

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

1001 is the lowest four-digit palindrome, which is 0 modulus 7. Adding one to the first and last digits adds 0 modulus 7. Adding one to the second and third digits adds 5 modulus 7, since 110 is also 5 modulus 7. So only 770 plus multiples of 1001 are divisible by 7. Out of 90 possibilities, there are only 9 that are divisible by 7. So the probability is 9/90, or 1/10.

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50 50 odd or even same probability

1/3

The answer is 9*9!/9*109 = 0.0003629 approx.

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.

They will certainly share the factor 1. Other than that, the probability is 1: it is nearly a certainty that that they will not share another common factor..

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 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%.

I will assume that you mean a five card poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408

Probability that a girl is chosen = 23/45 = .511 So, the probability that a boy is chosen = 1 - .511 = .489

The answer depends on how many are chosen at a time.

50%

the answer is 1 out of 26