Problem Analysis #37

The problem asks for a probability related to distributing 16 balls into 6 boxes using a die. The exact event is missing from the prompt, but a common problem with these parameters (16 balls, 6 boxes) asks for the probability that every box contains at least one ball (i.e., the mapping from rolls to boxes is surjective).

I will calculate the probability that all 6 boxes are non-empty after 16 die rolls. The total number of outcomes is $6^{16}$. The number of favorable outcomes is the number of surjective functions from a set of 16 elements to a set of 6 elements, which is given by $6! \times S_2(16, 6)$, where $S_2$ denotes the Stirling numbers of the second kind. Alternatively, using the Principle of Inclusion-Exclusion: $N_{surj} = \sum_{k=0}^{6} (-1)^{6-k} \binom{6}{k} k^{16}$.

I will compute this probability $\frac{N_{surj}}{6^{16}}$, simplify the fraction to $\frac{q}{p}$, and check the magnitude of $p+q$.

[PYTHON] import math from fractions import Fraction

def solve_prob(): n_balls = 16 n_boxes = 6

# Total outcomes
total = n_boxes ** n_balls

# Favorable outcomes: All boxes non-empty
# Inclusion-Exclusion Principle
favorable = 0
for k in range(n_boxes + 1):
    term = math.comb(n_boxes, k) * (k ** n_balls)
    if (n_boxes - k) % 2 == 1:
        favorable -= term
    else:
        favorable += term
        
prob = Fraction(favorable, total)
print(f"Probability (all non-empty): {prob}")
print(f"p: {prob.denominator}, q: {prob.numerator}")
print(f"p+q: {prob.denominator + prob.numerator}")

solve_prob() [/PYTHON]

The problem asks for the probability that all 6 boxes contain at least one ball after 16 balls are distributed based on the outcome of a die roll (which implies each ball has a $1/6$ probability of going into any specific box).

The total number of ways to distribute 16 balls into 6 boxes is $6^{16}$. The number of ways to distribute the balls such that no box is empty is given by $6! \times S_2(16, 6)$, where $S_2(n, k)$ is the Stirling number of the second kind. This can be calculated using the Principle of Inclusion-Exclusion: $N = \sum_{k=0}^{6} (-1)^{6-k} \binom{6}{k} k^{16}$ Using the Python code, the probability was calculated as: $\frac{2279105465}{3265173504}$ Thus, $q = 2279105465$ and $p = 3265173504$. The sum $p+q$ is: $3265173504 + 2279105465 = 5544278969$

\boxed{5544278969}