SOLUTION: A bag contains 3 red balls and 1 blue ball. A second bag contains 1 red ball and 1 blue ball. A ball is picked at random out of each bag and is then placed in the other bag. What i

Question 1088382: A bag contains 3 red balls and 1 blue ball. A second bag contains 1 red ball and 1 blue ball. A ball is picked at random out of each bag and is then placed in the other bag. What is the expected number of red balls in the first bag? (4marks
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
A bag contains 3 red balls and 1 blue ball. A second bag contains 1 red ball and 1 blue ball. A ball is picked at random out of each bag and is then placed in the other bag. What is the expected number of red balls in the first bag?

Solution:
Interpretation of question:
Given bags A (3r+1b) and B (1r+1b).
We were told:
"A ball is picked at random out of EACH bag and is then placed in the other bag." which is interpreted as follows:
A bag is randomly selected, a ball is taken out, and placed in the other bag, THEN a ball is taken out of the other bag and put back in the first bag.

We want to know what is the expected number of red balls in bag A.

We define event
r=picking a red ball, and therefore
b=picking a blue ball.
A=picking bag A first, and
B=not picking bag A first
There are two cases, each of which has two steps.
1. Bag A was selected as the first bag.
P(Arr)=picking a red from A [ therefore B has 2R+1B] and a red from B
=P(A)*P(r)*P(r)
=(1/2)(3/4)(2/3)
=1/4 [ A contains 3r]
similarly
P(Arb)=(1/2)(3/4)(1/3)=1/8 [A contains 2r]
P(Abr)=(1/2)(1/4)(1/3)=1/24 [A contains 4r]
P(Abb)=(1/2)(1/4)(2/3)=1/12 [A contains 3r]
(check: 1/4+1/8+1/24+1/12=(6+3+1+2)/24=1/2 ... ok.

2. Bag B was selected as the first bag
P(Brr)=(1/2)(1/2)(4/5)=4/20 [A contains 3r]
P(Brb)=(1/2)(1/2)(1/5)=1/20 [A contains 4r]
P(Bbr)=(1/2)(1/2)(3/5)=3/20 [A contains 2r]
P(Bbb)=(1/2)(1/2)(2/5)=2/20 [A contains 3r]
(check: 4/20+1/20+3/20+2/20=(4+1+3+2)/20=1/2....ok

The expected number of red balls in A is given by
E[n]=∑P(n)*n
=3(1/4)+2(1/8)+4(1/24)+3(1/12) + 3(4/20)+4(1/20)+2(3/20)+3(2/20)
=169/60
=2.82 approximately.

Note: if the above interpretation is different, you can proceed similarly with adjustments of the calculations.

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