# Two urns with different number of balls, from each urn a ball is drawn at random. [closed]

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, StrantsMar 15 '18 at 15:19

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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 $\frac{4}{12} * \frac{4}{11} * \frac{4}{10} * 3!$
• @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. – N. F. Taussig Mar 15 '18 at 10:55