, 2021

Label the balls in order 1 through 5. Assume without loss of generality that the first transposition is $1 \leftrightarrow 2$, resulting in the order 21345. The following table shows the results of the 5 equally likely second transpositions.

OR

Let $X_i$ be the random variable that has the value 1 if ball $i$ occupies its original position after the two successive transpositions and the value 0 if it does not. The problem asks for the expected value of the sum $X_1 + X_2 + X_3 + X_4 + X_5$, which, by linearity of expectation, is the sum of the expected values. The probability that $X_i = 1$ equals $\frac{2}{5} \cdot \frac{1}{5} + \frac{3}{5} \cdot \frac{3}{5} = \frac{11}{25}$, because this will happen if ball $i$ is involved in the first transposition and the second transposition involves the same two balls, or ball $i$ is involved in neither transposition. Therefore the expected value of each $X_i$ is $\frac{11}{25}$, and the expected value of their sum is $5 \cdot \frac{11}{25} = \frac{11}{5} = 2.2$.