Since a, b, c, d, and e are all variations of a similar question, I think it is fair just to do problems a and d and leave the rest.

Expected value is the average outcome of running a long-term experiment, which is calculated with the formula \(E(X) = \sum_{i = 1}^n x_iP(x_i)\) where E(X) is the expected value, x_{i} is a specific outcome, and P(x_{i}) is the probability of x_{i} occurring. Let's apply this to problem a.

Upon drawing a slip of paper from the bag, 2 possible outcomes can occur: 3 or 8 with their respective probabilities. With the unmodified bag, 3 occurs with a probability of 8/10 because there are eight 3's and ten slips of paper. Likewise, 8 occurs with a probability of 2/10 because there are two 8's and ten slips of paper. Substitute this information into the formula and the answer follows naturally.

\(\begin{align} E(X) &= 3 * \frac{8}{10} + 8 * \frac{2}{10} \\ &= \frac{24}{10} + \frac{16}{10} \\ &= 4 \end{align}\)

Therefore, E(X) = 4 for problem a.

Next, let's consider problem d. Here, the expected value is given, so E(X) = 6. Adding 8's to the original contents of the bag increases both the number of 8's and the total number of 8's in the bag. Let a represent the number of 8's to add to the bag such that E(X) = 6. This would result in the following equation.

\(\begin{align*} E(X) &= \sum_{i = 1}^n x_iP(x_i) \\ 6 &= 3 * \frac{8}{10 + a} + 8 * \frac{2 + a}{10 + a} \\ 60 + 6a &= 24 + 16 + 8a \\ 20 &= 2a \\ a &= 10 \end{align*}\)

Therefore, add ten 8's to the bag. To be a proper mathematics steward, note that 10 + a appears in the denominator. This means a ≠ -10 under any circumstance. However, this is a non-issue regardless because the variable a represents the number of 8's added, so a negative value for a is nonsensical anyway.