One urn contains three red balls, two white balls, and one blue ball. A second urn contains one red ball, two white balls, and three blue balls.

a) One ball is selected at random from each urn:

i) Describe sample space for this experiment.

ii) Find the probability that both balls will be of the same color.

b) The balls in the two urns are mixed together in a single urn, and then a sample of three is drawn. Find the probability that the three colors are represented, when

i) sampling with replacement

ii) without replacement.


closed as off-topic by JonMark Perry, Claude Leibovici, Namaste, user85798, Strants Mar 15 '18 at 15:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JonMark Perry, Claude Leibovici, Namaste, user85798, Strants
If this question can be reworded to fit the rules in the help center, please edit the question.


I will give hints and leave the rest to you:

a) i) The sample space is the set of possibilities. What are all possible outcomes of this experiment?

ii) To calculate probability of both balls being red, we calculate each probability separately and multiply them as they are independent. Probability of first ball being red is $\frac{3}{6}$ and second ball being red is $\frac{1}{6}$. Hence, $\frac{3}{36}$. The rest are easy to compute.

b) When sampling with replacement, the probability is $\frac{4}{12} * \frac{4}{12} * \frac{4}{12} * 3!$. Without replacement, it's $\frac{4}{12} * \frac{4}{11} * \frac{4}{10} * 3!$

  • $\begingroup$ I was working it out by myself, and I have a question, wouldn't the probability of both balls being red be equal to 3/36 divided by the combinatoric of 12 by 2? Since those are all the possible ways to select a ball from the two urns? $\endgroup$ – egroj97 Mar 15 '18 at 7:14
  • $\begingroup$ @JorgeRodríguez You are making two separate selections. There are six balls in the first urn, of which three are red. Hence, the probability of selecting a red ball from the first urn is $\frac{3}{6}$. By similar reasoning, the probability of selecting a red ball from the second urn is $\frac{1}{6}$. Since these choices are independent, we multiply the probabilities to find the probability that a red ball is selected from each urn. $\endgroup$ – N. F. Taussig Mar 15 '18 at 10:55

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