Question 10, 9.2.B-13 The following box contains the 8 letters shown. The letters are drawn one-by-one without replacement, and the results are recorded in order. Find the probability of the outcome EXCESSES. P(EXCESSES) \( =\square \) (Type an integer or a simplified fraction.)

To find the probability of drawing the outcome "EXCESSES" from the letters contained in the box, we first note that the letters in "EXCESSES" consist of: E, X, C, E, S, S, E, S. This means we have three E's, two S's, one C, and one X. The total count of letters in "EXCESSES" is 8. The number of ways to arrange the letters "EXCESSES" is calculated using the formula for permutations of multiset: \[ \text{Total Arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \] In our case: - \( n = 8 \) (total letters) - \( n_1 = 3 \) (for E's) - \( n_2 = 2 \) (for S's) - \( n_3 = 1 \) (for C) - \( n_4 = 1 \) (for X) So, \[ \text{Total Arrangements} = \frac{8!}{3! \times 2! \times 1! \times 1!} = \frac{40320}{6 \times 2 \times 1 \times 1} = \frac{40320}{12} = 3360 \] The total distinct arrangements for the letters contain 8 characters from the box (assuming it's a mix of E, X, C, and S). Next, the probability of that specific arrangement occurring when all letters are selected from a bag of 8 unique letters is based on there being only one way to pick EXCESSES from a multiset as determined above out of 8 selected letters. Thus, the probability can be calculated as: \[ P(EXCESSES) = \frac{1}{Total Arrangements} = \frac{1}{3360} \] So the final answer will be \(\boxed{\frac{1}{3360}}\).