+769 Suppose we have a bag with 10 slips of paper in it. Eight slips have a 3 on them and the other two have an 8 on them.
a) What is the expected value of the number shown when we draw a single slip of paper?
b) What is the expected value of the number shown if we add one additional 8 to the bag?
c) What is the expected value of the number shown if we add two additional 8's (instead of just one) to the bag?
d) How many 8's do we have to add to make the expected value equal to 6?
e) How many 8's do we have to add before the expected value is at least 7?
+189 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, xi is a specific outcome, and P(xi) is the probability of xi 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.
