Q: "A ball is chosen at random from an urn containing 2 red and 3 green balls and is then added to a second urn which initially contains 2 green and 3 white balls. The balls in the second urn are then mixed and one is chosen at random. If this ball is green, what is the probability that the ball taken from the first urn was also green?"

At first glance this seems like a conditional probability question, am I on the right track or is this completely wrong? Can someone tell me how to approach this?

  • $\begingroup$ Do you agree with the answer? Please let me know $\endgroup$ – triomphe Oct 16 '13 at 22:59

Let $G_1$ and $R_1$ be events that first one is green and red respectively and let $G_2$ be the event that second one is green. $P(G_1)=\cfrac{3}{5}$


$P(G_2)=P(G_1)P(G_2\mid G_1)+P(R_1)P(G_2 \mid R_1)=P(G_1)\cfrac{3}{6}+P(R_1)\cfrac{2}{6}$

$P(G_2\mid G_1)=\cfrac{3}{6}$

$P(G_1 \cup G_2) = P(G_1)P(G_2\mid G_1)$

Then $\cfrac{P(G_1\cap G_2)}{P(G_2)} = P(G_1\mid G_2) $

It is long time since I did a question like this. I hope above is correct, if wrong kindly let me know I will correct it, if I know, or delete this.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.