swap balls in urns; find prob. that selected ball is green

Suppose two urns are setting on a table. Call them urn A and urn B. Urn A contains 4 redballs and 6 green balls and urn B contains 6 red balls and 3 green balls. You select a ball at random from A and place it, unseen, into B. NOw you select a ball from B, again unseen, and place it into A. If you now select a ball from A what is the probability that it is green?

Don't know where to began. Please help.

Go step by step, considering the different possible outcomes, and then the probability of that happening.

A) It is possible that the first ball drawn, from urn A, is red. The probability of that is 4/10= 2/5. Once you move it to urn B, there will be 3 red, 6 green balls in A and 7 red balls, 3 green balls in B. Now

A1) You draw a red ball from B. The probability of that is 7/10. Moving that from B to A gets us again to 4 red, 6 green balls. The probability of drawing a gree ball from A is now 6/10= 3/5. The probability of all of these things happening is (2/5)(7/10)(3/5)

A2) You draw a green ball from B. The probability of that is 3/10. Moving that from B to A gets us to 3 red, 7 green balls. The probability of drawing a green ball is now 7/10. The probability of all those things happening is (2/5)(3/10)(7/10)

**************************************************************************************

B) It is possible that the first ball drawn, from urn A, is green. The probability of that is 6/10= 3/5. Once you move it to urn B, there will be 4 red, 5 green balls in A and 6 red balls 4 green balls in B. Now

You do "B1" and "B2".

leyva2389 said:

Suppose two urns are setting on a table. Call them urn A and urn B. Urn A contains 4 redballs and 6 green balls and urn B contains 6 red balls and 3 green balls. You select a ball at random from A and place it, unseen, into B. NOw you select a ball from B, again unseen, and place it into A. If you now select a ball from A what is the probability that it is green?

Click to expand...

The string \(\displaystyle R_1G_2G_3 \) stands for the first ball is red, the second is green, and the third is green.

Using conditioning we get \(\displaystyle \mathcal{P}(R_1G_2G_3)=\mathcal{P}(G_3|G_2R_1) \mathcal{P}(G_2|R_1)\mathcal{P}(R_1) \)

Pay very close attention to the order in last string.
For we know \(\displaystyle \mathcal{P}(R_1) \) therefore we know \(\displaystyle \mathcal{P}(G_2|R_1) \), etc.

Now there are three additional conditionings :
\(\displaystyle G_1G_2G_3,~R_1R_2G_3,~\&~G_1R_2G_3 \)

You have to do that for each string.