Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are 5 buckets, and I have 3 balls to place into these buckets. I cannot place more than one ball in any bucket.

After placing the balls in the buckets, 3 buckets are removed at random.

  1. What is the probability of there being at least 1 ball in the remaining buckets?
  2. What is the probability of not being able to retrieve any balls from the remaining buckets?
share|cite|improve this question
up vote 1 down vote accepted

Let's start with question $2$. No matter how you distribute the balls into the buckets, exactly $3$ buckets will contain a ball and exactly $2$ buckets will contain no ball. Now suppose we want to remove exactly $3$ buckets such that the remaining $2$ buckets will contain no balls. Then the number of ways this can happen is: $$ \binom{3}{3} \binom{2}{0} = 1 $$ With no restrictions, the number of ways to choose $3$ buckets from $5$ is: $$ \binom{5}{3}=\dfrac{5 \cdot 4}{2} = 10 $$ so we obtain the probability of $\boxed{\dfrac{1}{10}}$.

For question $1$, this is simply the complement, so we obtain the probability of $1-\dfrac{1}{10}=\boxed{\dfrac{9}{10}}$

share|cite|improve this answer

There are 3 buckets with one ball and 2 buckets with no balls.

1) Subtract from one the probability of picking all 3 buckets with a ball.

$1 - (\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}) = \frac{9}{10}$

2) There are only 2 buckets with no balls, and you must pick 3 buckets, so the probability of retrieving no balls is zero.


The second part of the question was clarified to mean "in the buckets that are left". In that case, the second part is the complement of the first part.

The probability of not being able to retrieve any balls from the buckets that are left means that you chose all 3 buckets with balls. So this is just like the first part, except that you don't subtract from one:

$(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}) = \frac{1}{10}$

share|cite|improve this answer
With 5 buckets and 3 balls, if I remove the 3 buckets that each contain one ball, doesn't that make the probability of retrieving no balls from the remaining buckets non-zero? – Hector Castro Jun 13 '13 at 23:21
When you say "not being able to retrieve any balls", do you mean from the removed buckets or from the buckets that are left? I had assumed that you meant in the removed buckets. – rgettman Jun 13 '13 at 23:23
Sorry, I meant the buckets that are left. – Hector Castro Jun 13 '13 at 23:33
I've updated my answer according to your clarification. – rgettman Jun 13 '13 at 23:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.