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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 $C$?

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    $\begingroup$ 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. $\endgroup$
    – user2661923
    Commented May 24, 2022 at 9:28

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"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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  • $\begingroup$ ...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?... $\endgroup$
    – user992622
    Commented May 24, 2022 at 4:38
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    $\begingroup$ 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. $\endgroup$
    – Graham Kemp
    Commented May 24, 2022 at 4:50
  • $\begingroup$ 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? $\endgroup$
    – user992622
    Commented May 24, 2022 at 4:58
  • $\begingroup$ 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 $\bf 3!/13^3$. $\endgroup$
    – Graham Kemp
    Commented May 24, 2022 at 5:09
  • $\begingroup$ ... 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. $\endgroup$
    – Graham Kemp
    Commented May 24, 2022 at 5:09

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