Probability im very confused so please help!

Bag A contains 1 red ball and one black ball, and bag B contains 2 red balls; all four balls are indistinguishable apart from colour. One ball is chosen at random from A and is transferred to B. One ball is then chosen at random from B and transferred to A.

Find the probability that, after both transfers, the black ball is in bag A.

can someone show me the step by step method to get this result (2/3)

Thank You!!

natacha said:

Bag A contains 1 red ball and one black ball, and bag B contains 2 red balls; all four balls are indistinguishable apart from colour. One ball is chosen at random from A and is transferred to B. One ball is then chosen at random from B and transferred to A.

Find the probability that, after both transfers, the black ball is in bag A.

can someone show me the step by step method to get this result (2/3)

Thank You!!

Click to expand...

After you transferred one ball from bag A to bag B -

How many balls does bag B has?

What are those? (2 situations)

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natacha said:

Bag A contains 1 red ball and one black ball, and bag B contains 2 red balls; all four balls are indistinguishable apart from colour. One ball is chosen at random from A and is transferred to B. One ball is then chosen at random from B and transferred to A.

Find the probability that, after both transfers, the black ball is in bag A.

can someone show me the step by step method to get this result (2/3)

Thank You!!

Click to expand...

Try using a tree diagram to spell out all scenarios. That sometimes is easier for one to grasp in problems like these.

what i've done is a probability tree that is a bit like this.

i dont know what to do from there....

natacha said:

what i've done is a probability tree that is a bit like this.

Bag A	Bag B	Bag A
		1/2 red
	3/3 red	1/2 black
1/2 red		-------
	0/3 black	0/1 red
	-----------	1/1 black
		-------
	2/3 red	2/2 red
1/2 black		0/2 black
		------
	1/3 black	1/2 red
		1/2 black

i dont know what to do from there....

Click to expand...

Good start. Now multiply the probabilities for each branch:

i) (1/2)(1) = 1/2
ii) (1/2)(0) = 0
iii) (1/2)(2/3) = 1/3
iv) (1/2)(1/3) = 1/6

Now scenarios (i) and (iv) represent scenarios where there is a black ball in resulting Bag A. So the probability the black ball is back in bag A is the sum of these probabilities.

natacha said:

Bag A contains 1 red ball and one black ball, and bag B contains 2 red balls; all four balls are indistinguishable apart from colour. One ball is chosen at random from A and is transferred to B. One ball is then chosen at random from B and transferred to A.

Find the probability that, after both transfers, the black ball is in bag A.

can someone show me the step by step method to get this result (2/3)

Thank You!!

Click to expand...

Transferring one ball from A to B

The red ball is chosen (1/2) - but there is a black ball left in the bag A - so in this case probability of having a black ball in the bag A is (1/2 * 1) = 1/2

The black ball is chosen (1/2) - there would be three balls in bag B with one back ball - so in this case probability of having a black ball in the bag A is (1/2 * 1/3) = 1/6

Probability of having a black ball in bag A after two transfers = 1/2 + 1/6 = 2/3