bag A contains 3 red balls,and 2 green balls...?

Question

bag A contains 3 red balls,and 2 green balls...bag B contains 2 red balls and 3 green balls.2 are withdrawn from bag A to bag B.Then 2 balls are taken out from B to be put in bag A.What is probablity that bag A now contains 3 red bals and 2 green balls.?

Answer 1

What a peculiar question!

Look at all of the possibilities.
These are :
(1) two red balls withdrawn, two red balls returned
(2) one red and one green, one red and one green
(3) two green and two green.

OK, now we can work it out:
(1) 3/5 * 2/4 * 4/7 * 3/6 (RR-RR)
(2) 3/5 * 2/4 * 3/7 * 4/6 (RG-RG)
+ 3/5 * 2/4 * 4/7 * 3/6 (RG-GR)
+ 2/5 * 3/4 * 3/7 * 4/6 (GR-RG)
+ 2/5 * 3/4 * 4/7 * 3/6 (GR-GR)
(3) 2/5*1/4 * 5/7 * 4/6 (GG-GG)

Well, this is tedious, but now just add them all up. So:
6/20*12/42
+ 6/20*12/42 *4
+ 2/20*20/42

which equals
(72+72*4+40)/840
=400/840
=10/21

Hope this helps in understanding!
.

Answer 2

there are a total of nine outcomes.the chance that bag A contains 3 red balls and 2 green balls is when the combination of the 2 balls taken from bag B to bag A is the same combination of the 2 balls taken from bag A to bag B.this can only happen 3 times because of two color combination.therefore 3 chances over 9 outcomes is 33.33% probability.

Answer 3

mmm... interesting question.
if 2 green balls r taken from bagA & again r put in it,then there will 3 red balls and 2 green balls again but if 2red balls r taken and 2 green balls r put in it then 1red and 4green will left .Therefore, the answer depends on which colour balls r taken and which r put in it.

Answer 4

This problem amounts to the probability of restoring bag A to its original composition. This can be accomplished in three ways.

1) Moving two red balls to B and two red balls back to A

P(2R) = [(3C2)/(5C2)] * [(4C2)/(7C2)] = (3/10)*(6/21) = 3/35

2) Moving one red ball and one green ball to B and one red ball and one green ball back to A

P(1R 1G) = [(3C1)(2C1)/(5C2)] * [(3C1)(4C1)/(7C2)]
P(1R 1G) = [(3*2)/10]*[(3*4)/(21)] = (6/10)*(12/21) = 12/35

3) Moving two green balls to B and two green balls back to A

P(2G) = [(2C2)/(5C2)] * [(5C2)/(7C2)] = (1/10)*(10/21) = 1/21
_________

The probabililty of restoring bag A to its original composition is:

P(2R) + P(1R 1G) + P(2G) = 3/35 + 12/35 + 1/21 = 10/21