<s>[INST] 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 $C$?
 [/INST] "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.$${\mathsf P(A,B,\overline C)+\mathsf P(A,\overline B,C)+\mathsf P(\overline A, B, C)\\=\dfrac{4\cdot 5\cdot 4}{7\cdot 9\cdot 8}+\dfrac{4\cdot 4\cdot 4}{7\cdot 9\cdot 8}+\dfrac{3\cdot 5\cdot 4}{7\cdot 9\cdot 8}\\=\dfrac {17}{42}}$$
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