91Ó°ÊÓ

First, let's understand the problem. When we roll a fair 6-sided die, there is a \(\frac{1}{6}\) probability for each side to appear. The expected value is the average number of rolls needed for a specific event to occur, given that the probability distribution is known. The formula for expected value is: \(E(x) = \sum_{i=1}^{n} P(x_i) * x_i\) where \(E(x)\) is the expected value, \(P(x_i)\) is the probability of a specific event occurring, and \(x_i\) is the value associated with that event. In this problem, we are interested in finding the expected number of rolls before all 6 sides of the die appear.

Now, let's find the expected value for each roll. We will break this problem into 6 different events corresponding to the appearance of each side of the die. Event 1: Getting the first unique side. In this case, since all sides are unique, the probability of getting the first unique side is 1, and this will happen in the first roll. Event 2: Getting the second unique side. After rolling once, there are 5 remaining unique sides. The probability of getting one of these is now \(\frac{5}{6}\). So, \(P(x_2) = \frac{5}{6}\), meaning that there is a \(\frac{5}{6}\) chance of getting the second unique side on the second roll. Event 3: Getting the third unique side. After rolling twice and getting two unique sides, there are 4 remaining unique sides. The probability of getting one of these is \(\frac{4}{6}\). So, \(P(x_3) = \frac{4}{6}\), meaning that there is a \(\frac{4}{6}\) chance of getting the third unique side on the third roll. Continue this pattern for the next events: Event 4: \(P(x_4) = \frac{3}{6}\) Event 5: \(P(x_5) = \frac{2}{6}\) Event 6: \(P(x_6) = \frac{1}{6}\)

Now that we have the probabilities for each event, we can calculate the expected number of rolls for each event: Event 1: \(E(x_1) = 1 * 1 = 1\) Event 2: \(E(x_2) = \frac{6}{5} (\frac{5}{6}) = \frac{6}{5}\) Event 3: \(E(x_3) = \frac{6}{4} (\frac{4}{6}) = \frac{6}{4}\) Event 4: \(E(x_4) = \frac{6}{3} (\frac{3}{6}) = \frac{6}{3}\) Event 5: \(E(x_5) = \frac{6}{2} (\frac{2}{6}) = \frac{6}{2}\) Event 6: \(E(x_6) = \frac{6}{1} (\frac{1}{6}) = \frac{6}{1}\) Now sum up the expected number of rolls for each event: \(E(total) = E(x_1) + E(x_2) + E(x_3) + E(x_4) + E(x_5) + E(x_6)\) \(E(total) = 1 + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2} + 6\) Approximate the sum: \(E(total) \approx 1 + 1.20 + 1.50 + 2 + 3 + 6\) \(E(total) \approx 14.7\) So, on average, you would expect to roll the die approximately 14.7 times (rounded up to 15 times) before all 6 sides have appeared at least once.