# There are $3$ urns $A,B$ and $C$. Urn $A$ contains $4$ red balls and $3$ black balls. Urn $B$ contains $5$ red balls and $4$ black balls. Urn $C$

There are $$3$$ urns $$A,B$$ and $$C$$. Urn $$A$$ contains $$4$$ red balls and $$3$$ black balls. Urn $$B$$ contains $$5$$ red balls and $$4$$ black balls. Urn $$C$$ contains $$4$$ red balls and $$4$$ black balls. One ball is drawn from each of these urns . What is the probability that $$3$$ balls drawn consists of $$2$$ red balls and a black ball?

My solution goes like this:

Considering events $$A,B$$ and $$C$$ as picking up a red ball from urn $$A$$, picking up a red ball from urn $$B$$,picking up a red ball from urn $$C$$. The probability of picking up two balls of red color from urn $$A$$ and $$B$$ and a black ball from urn $$C$$ is $$P(A)P(B)P(\overline{C})=\frac{4.5.4}{7.9.8}$$ . Now, this selection can be made in three different ways so the total probabiblity in this case is $$P(A)P(B)P(\overline{C})=\frac{4.5.4.3}{7.9.8}$$. Now, the probability of picking up two balls of red color from urn $$A$$ and $$C$$ and a black ball from urn $$B$$ is $$P(A)P(C)P(\overline{B})=\frac{4.4.4}{7.8.9}$$. This selection can be made in three different ways so the total probabiblity in this case is $$P(A)P(C)P(\overline{B})=\frac{4.4.4.3}{7.8.9}$$. The probability of picking up two balls of red color from urn $$B$$ and $$C$$ and a black ball from urn $$A$$ is $$P(B)P(C)P(\overline{A})=\frac{5.4.3}{9.8.7}$$.This selection can be made in three different ways so the total probabiblity in this case is $$P(B)P(C)P(\overline{A})=\frac{5.4.3.3}{9.8.7}$$. So, the total probability is $$\frac{4.5.4.3}{7.9.8}+\frac{4.4.4.3}{7.8.9}+\frac{5.4.3.3}{9.8.7}$$.

However, this is not a valid probability as you can see as the probability is greater than $$1$$. Where the problem is occuring ? Why this method is not valid? I am not getting it? Is it because the question does not support taking all arrangements in consideration ? Does the question is made for only the specific order i.e first drawing two red balls and then drawing a black ball? Does the question only stand valid for the previous case mentioned?Should the problem specify this that the order in which the balls of a particular color apre chosen follows a soecific order i.e say we must choose first from urn $$A$$ then urn $$B$$ and then urn $$C$$?

• See also this answer, which attacks a related problem. In the linked answer, I advocate using a Combinatorics approach, rather than a probability of events approach. I advocate the same thing (i.e. the Combinatorics approach) for your problem. Commented May 24, 2022 at 9:28

"Now, this selection can be made in three different ways"... No. It cannot. You can select red from A, red from B, and black from C in one way (each with a probability of success). So $$\mathsf P(A, B, \overline C)=\dfrac{4\cdot 5\cdot 4}{7\cdot 9\cdot 8}$$

The "three ways" a black ball may be drawn from an urn is accounted for in your summation.$${\mathsf P(A,B,\overline C)+\mathsf P(A,\overline B,C)+\mathsf P(\overline A, B, C)\\=\dfrac{4\cdot 5\cdot 4}{7\cdot 9\cdot 8}+\dfrac{4\cdot 4\cdot 4}{7\cdot 9\cdot 8}+\dfrac{3\cdot 5\cdot 4}{7\cdot 9\cdot 8}\\=\dfrac {17}{42}}$$

• ...Thanks a lot! Just one more thing I multiplied $3$ because of the number of ways we can do the thing say: we first choose a red ball from $A$ then a black ball from $C$ and again a red ball from $B$. This is just one out of the three ways possible. Shouldn't we take this into account?...
– user992622
Commented May 24, 2022 at 4:38
• Changing the order you choose to select balls from the urn does not change the probability for selecting a particular set of colours from those urns. Commented May 24, 2022 at 4:50
• That's true! However , well consider an example of a different problem: Three cards are drawn with replacement from a well shuffled pack of $52$ cards . Find the probability that the cards are king , queen and jack. So, the required probability here is $\frac{1.6!}{13^3}$ . We did take the order of selection into consideration right? So , in this case the same thing should have been valid?
– user992622
Commented May 24, 2022 at 4:58
• The probability for drawing a king, queen, and then jack (as first, second, and third card) is $1/13^3$. There are 5 other such arrangements of king, queen, and jack, so to get the probability of "a king, queen, and jack in any order", you add all six together.$$\mathsf P(K_1, Q_2, J_3)+\mathsf P(K_1, Q_3, J_2)+\cdots+\mathsf P(K_3,Q_2, J_1)$$ It is only because these all have the same probability that you can multiply any one probability by $3!$ to get the answer of $\bf 3!/13^3$. Commented May 24, 2022 at 5:09
• ... In your balls and urns example, the probabilities for "1 black in a this urn, and red in the others" are different for each urn, so you must add the three probabilities together instead of multiplying one probability by three. Commented May 24, 2022 at 5:09