probability question

Question

Bag A contains 7 blue and 3 red marbles. Bag B contains 6 blue and 4 red marbles.Firstly, a marble is chosen at random from Bag A and and placed in Bag B. Secondly, a marble is chosen at random from Bag B and placed in Bad A. Find the probability that a marble now chosen at random from Bag A is blue

Answer 1

I'm sure that there is a more nuanced way to do this, but my first take is to enumerate the probabilities of each trade, figure out the probability of a blue in A for each trade and add up the products:

Trade: Prob of Trade P of B in A after trade Product Columns 2 and 3 = P(BinA|trade)

BB 7/10 *7/11 7/10

BR 7/10*4/11 6/10

RB 3/10*6/11 8/10

RR 3/10*5/11 7/10

Then add up the last column

Answer 2

Bag A: 7b 3r

Bag B: 6b 4r

choose a random marble in Bag A and move it to Bag B

then choose a random marble from the new contents of Bag B and move it to Bag A

then choose a random marble from the new contents of Bag A. What is the probability of it will be blue.

probability the first marble from A to B was blue = 7/10

the new Bag B then has 6.7 blue and 4.3 red marbles, on average

the 2nd marble moved from B to A will be 6.7/11

Bag A will then contain 7-.7 +6.7/11 blue and 3 -.3 + 4.3/11 red

or 6.3 + 6.7/11 blue and 2.7 +4.3/11 red

or about 6.909 blue and 3.091 red which sum to the original 10 marbles

pick one at random and it's probability of being blue = about 69.09 or 69.1%

originally the probability of blue from Bag A would have been 70%

but you mixed it with a marble that was slightly less blue, so the probability decreases slightly, so the answer 69.09 or 69.1% looks about right