Each of $3$ urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black. One ball is drawn from the random urn with return in the same urn. The ball's color is white. What is the probability that the second ball drawn from the same urn is white?

I think, this is $\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$ by Bayes'theorem and Law of total probability, but can't be sure.

Thanks for any help.

  • $\begingroup$ Remember to use LaTeX! $\endgroup$ – KingLogic Oct 29 '20 at 18:17
  • $\begingroup$ Please see math.stackexchange.com/help/notation for using mathJax. $\endgroup$ – user2661923 Oct 29 '20 at 18:30
  • $\begingroup$ Has this query been garbled? All 20 of the balls are white? How can any ball drawn not be white? Does this query have a typo? $\endgroup$ – user2661923 Oct 29 '20 at 18:31
  • $\begingroup$ Thanks for remarks. Added - balls can be black and white. Ratio of white balls to all - 10 to 20, 6 to 20 and 2 to 20. $\endgroup$ – Sergio Oct 29 '20 at 18:39

I am unsure of where your math came from, and I am unfamiliar with the "Law of total probability". The following is how I would compute the probability.

Let $p(k)$ represent the probability that the 1st ball came from urn $k : k \in \{1,2,3\}.$

Then, the chance that the new drawing will also be a white ball, since the sampling is done with replacement is

$$\left[p(1) \times \frac{10}{20}\right] ~+~ \left[p(2) \times \frac{6}{20}\right] ~+~ \left[p(3) \times \frac{2}{20}\right].$$

Let $D$ (i.e. denominator) = $$\left[\frac{1}{3} \times \frac{10}{20}\right] ~+~ \left[\frac{1}{3} \times \frac{6}{20}\right] ~+~ \left[\frac{1}{3} \times \frac{2}{20}\right].$$


$$p(1) = \frac{\frac{1}{3} \times \frac{10}{20}}{D}.$$

$$p(2) = \frac{\frac{1}{3} \times \frac{6}{20}}{D}.$$

$$p(3) = \frac{\frac{1}{3} \times \frac{2}{20}}{D}.$$


Your answer is correct, it does need Bayes' Theorem and total probability, but you haven't shown how you arrived at it and I don't know what form of Bayes' Theorem you have learnt.
To keep it very rudimentary, the probability that the first white ball was found in a particular box chosen at random will equal the ratio of white balls it contains to total white balls.
Then, applying the law of total probability, we get the formula
$\Sigma$ [P(1st ball white from box $k$)$\times$ P(2nd ball white from box $k$)] = $\frac{10}{18}\cdot\frac{10}{20} + \frac6{18}\cdot\frac6{20} + \frac2{18}\cdot\frac2{20}$


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