Problem Analysis: We need to find the expected total number of balls in the two extreme slots (slot 1 and slot 6) when 10 balls are randomly distributed among 6 slots.

Expected Value Calculation Using Linearity

We can solve this using the linearity of expectation. Let $X$ be the total number of balls in the two extreme slots. We can define indicator random variables for each ball.

1. Define Indicator Variables: Let $B_j$ be an indicator variable for the $j$-th ball ($j = 1, 2, ..., 10$).
   • $B_j = 1$ if the $j$-th ball lands in either slot 1 or slot 6.
   • $B_j = 0$ otherwise.
2. Calculate Probability for One Ball: For any single ball, the probability of landing in any specific slot is $\frac{1}{6}$, as there are 6 slots and the distribution is random. The probability of landing in an extreme slot (slot 1 or slot 6) is: $P(B_j = 1) = P(\text{ball } j \text{ in slot 1}) + P(\text{ball } j \text{ in slot 6})$ $P(B_j = 1) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$
3. Calculate Expected Value of Indicator Variable: The expected value of an indicator variable is the probability of the event it indicates: $E[B_j] = 1 \times P(B_j = 1) + 0 \times P(B_j = 0) = P(B_j = 1) = \frac{1}{3}$
4. Apply Linearity of Expectation: The total number of balls in the extreme slots is the sum of the indicator variables for all 10 balls: $X = \sum_{j=1}^{10} B_j$. The expected total number of balls is the sum of the expected values of the indicator variables: $E[X] = E\left[\sum_{j=1}^{10} B_j\right] = \sum_{j=1}^{10} E[B_j]$
5. Final Calculation: Substituting the expected value of each indicator variable: $E[X] = \sum_{j=1}^{10} \frac{1}{3} = 10 \times \frac{1}{3} = \frac{10}{3}$

Conclusion

The expected total number of balls in the two extreme slots is $\frac{10}{3}$.