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A number is chosen at random from the first twelve whole numbers. What is the probability that it is exactly divisible by 3?
Wiki User
∙ 14y ago
Number of choices = 12
Number of successes = 4
Probability of success = 4/12 = 1/3rd = [ 33 and 1/3rd] percent
Wiki User
∙ 14y ago
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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 if a number is chosen the number is even?
50%
What if a class of 25 students has 15 girls and boys if one student is chosen at random what is the probability it is a girl?
The probability is 15/25 = 3/5
A circle is inscribed in a square with a side length of 4 If a point in the square is chosen at random what is the probability that the point is in the square but not in the circle?
21.5%
How many times has number 25 come out as the bonus number in the lottery?
What lottery? What country? Please note that having it appear a certain number of times in the past will have NO INFLUENCE WHATSOEVER on the probability of this number appearing in the future. Numbers are chosen at random, without consideration to previous draws, so each number has the same probability to appear. For more details, read the Wikipedia article on "Gambler's fallacy".
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%.
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
What is the probability that 4 and 7 randomly chosen numbers in a row are all divisible by 7?
4 numbers: 1/74 = 0.000 061 035 7 numbers: 1/77 = 0.000 001 214
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".
What is the probability of choosing the 8 winning lottery numbers when the numbers are chosen at random from 0 - 9?
1/3628800
An integer is chosen at random from the first 50 digitswhat is the probability that it is divisible by 6 or 8?
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.)
Coin numbered 1-200 are placed in a hat. what is the probability that a coin chosen at random is a multiple of 6?
There are 33 numbers between 1 and 200 that are evenly divisible by 6 (6, 12, 18,... 198), so the probability of pulling on a single draw a single such coin is 33/200, or 16.5%.
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.
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 random selected poker hand contains exactly 3 aces given that it contains at least 2 aces?
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
If a number from 1 to 20 is chosen what is the probability of getting 7?
1 out of 20 this is because there are 20 numbers in total, and there is only one 7 in there. (Assuming that there is the same probability for each number to be chosen, and that 17 is excluded as an affirmative outcome)
