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Suppose I throw 3 balls and each ball is equally likely to land in one of 4 buckets. What's the probability no bucket has more than 1 ball in it?

I know the answer is 3/8 but for some reason I can't remember how I got the solution, I worked it out a while ago.

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  • $\begingroup$ The probability the second ball lands in a new bucket is $\frac{3}{4}$. Given this happened, the probability the third ball lands in a new bucket is $\frac{2}{4}$. Or one can do a counting version of this argument. Record where the balls landed. We have $4^3$ equally likely possibilities. Of these, $(4)(3)(2)$ are "favourable." $\endgroup$ – André Nicolas Mar 13 '14 at 19:40
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After throwing the first ball there are $3$ empty buckets left. So the probability that the second ball thrown lands in empty bucket is $\frac{3}{4}$. The probability that under this condition the third lands in an empty bucket as well is $\frac{2}{4}$ so this leads to a probability of $\frac{3}{4}\times\frac{2}{4}=\frac{3}{8}$ that all balls land in an empty bucket. This event is the same as the event that no bucket contains more than $1$ ball after throwing.

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Hints:

  • How many ways to throw three balls in order into four buckets in total?

  • How many ways to throw three balls into four buckets so none go into the same buckets? (How many for the first ball? How many for the second given it does not go in the same bucket as the first? How many for the third given it does not go in the same buckets as the first or the second?)

  • Combine the two results to give a probability

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Here is a similar problem and the answer:

(also, some formatting is lost here and may best be viewed on the original source. It is question six on page 4)

Suppose that you throw 4 balls and that each ball is equally likely to land in any of 5 buckets labeled B1,B2,B3,B4,B5. We say the ith and jth ball collide if they land in the same bucket. (1) Whats the probability that no balls collide? (2) Whats the probability that no balls land in B1? (3) Whats the probability that no balls land in B1 conditioned on no balls colliding?

(1) The event that no balls collide happens exactly when each ball goes into a different bucket. Since there are 5 buckets, the probability that the first ball thrown lands in an unoccupied bucket is 5/5. For the second ball, there is one less bucket, so the chance is now 4/5. This pattern continues with 3/5 and 2/5 for the third and fourth balls. We multiply these together to get (5/5)(4/5)(3/5)(2/5) = 24/125 .

(2) For no balls to land in bucket B1, each ball must land in one of the other four buckets. For each ball, this is a probability of 4/5, so we multiply 4/5 four times (once for each ball) and get (4/5)4 = 256/625

(3) To find P(no balls in B1 j no collisions), we use Bayes' rule to turn this into P(no balls in B1\no collisions)/P(no collisions). The numerator, the probability that we get no collisions and there are no balls in B1, can be calculated as (4/5)(3/5)(2/5)(1/5) = 24=625 because for the first ball, there are 4 buckets available (every bucket except B1), and one fewer bucket for each successive ball. For the denominator, we already calculated this in part (1) as 24/125, so the answer to the problem is (24/625)/(24/125) = 125/625 = 1/5. This should make sense because if we are given the information that no balls have collided, there must be one bucket that is empty. There's a 1-in-5 chance that it is B1.

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