Imagine that there are two boxes A, B. Box A contains 10 blue balls and
box B contains 10 red balls. In every step, we pick one ball randomly
from box A and one ball randomly from box B at the same time, and
then swap them, i.e., we put the chosen ball from box A into box B, and
vice versa.
(a) What is the expected value of the number of blue balls in box A
after the second step?
(b) What is the expected value of the number of blue balls in box A
after the third step?
(a)
The probability of drawing a blue ball from box A and drawing a red ball from box B is: (9/10)(9/10) = 81/100
Red ball from box A, Red ball from box B is (1/10)(9/10) = 9/100
A blue ball from box A, and a blue ball from box B is (9/10)(1/10) = 9/100
A Red ball from box A, a blue ball from box B is (1/10)(1/10) = 1/100
E(x) = 8(81/100)+10*(1/100)+9*(9/100)+9*(9/100) = 8.2
(b) Following a similar method:
E(x) = 7.2*(67.24/100)+9.2*(3.24/100)+8.2*(14.76/100)+8.2*(14.76/100) = 7.56
Is my understanding of this question correct?
Your answers are correct, but since the questions are only about the blue balls in box A, if you think a bit,
in one transaction, the expected number of blue balls in box $1$ change from $B$ to $0.9B + 0.1(10-B) = 0.8B +1$
$\begin{align}\text{Thus}\;B_0&&&=10,\\ B_1&= 8+1&&=9, \\ B_2 &= 7.2 + 1 &&= 8.2,\\ B_3 &= 6.56 + 1 &&= 7.56\end{align}$