91Ó°ÊÓ

We want to find the average number of rolls it would take for all 6 sides of a fair die to appear at least once. We can define a discrete random variable \(X\) to represent the number of rolls it takes to achieve this.

The probabilities of rolling a specific number on a fair die is constant and equal to \(\dfrac{1}{6}\) for each number, since there are 6 possible outcomes and each outcome is equally likely. However, in our problem, we need a conditional probability - the probability of rolling a specific number given that we haven't rolled all the numbers yet.

Let's visualize the problem like this: after \(n-1\) rolls, we have seen \(n-1\) distinct numbers, and in the next roll, we want to see the \(n\)-th number for the first time. To calculate this probability, we can use the concept of conditional probability, using the complement rule: \(P(X = n) = P(X \leq n-1) - P(X \leq n-2)\) To calculate \(P(X \leq n-1)\) and \(P(X \leq n-2)\), we can use the fact that rolling a specific number in any roll is independent of the previous rolls, and that the probability of rolling any specific number is constant, equal to \(\dfrac{1}{6}\).

The expected value \(E(X)\) of a discrete random variable is given by the sum of the product of the values times their probabilities: \(E(X) = \Sigma_{n=1}^{\infty} n \cdot P(X=n)\) In our case, since there are only 6 possible outcomes for each roll, the sum will range from \(n = 6\) to \(n=\infty\). Using the conditional probability formula from Step 3, we can calculate the expected value as follows: \(E(X) = \Sigma_{n=6}^{\infty} n \cdot [P(X \leq n-1) - P(X \leq n-2)]\) \(E(X) = 6 \cdot (1 - 0) + 5 \cdot (\frac{5}{6} - 0) + 4 \cdot (\frac{4}{6} - \frac{3}{6}) + 3 \cdot (\frac{3}{6} - \frac{2}{6}) + 2 \cdot (\frac{2}{6} - \frac{1}{6}) + 1 \cdot (\frac{1}{6})\) \(E(X) \approx 6.6\) So, on average, it takes about 6.6 rolls of a fair die to see all six numbers at least once.