如何计算独立事件聚合结果的概率分布

import json

# 示例数据
jobs_names = ['Job 1', 'Job 2', 'Job 3', 'Job 4', 'Job 5']
probabilities = [0.1, 0.1, 0.4, 0.6, 0.2]
hours = [1, 10, 43, 2, 5]

# 假设我们想知道获得超过10小时的概率
min_hours_desired = 10

# 1. 生成所有可能的场景
scenarios = []
num_jobs = len(jobs_names)
for i in range(2**num_jobs):
    # 将整数i转换为n位的二进制字符串
    # 例如，i=5 (二进制101) 对于5个项目会变成 '00101'
    scenario_binary_str = bin(i).split('b')[1].zfill(num_jobs)
    scenarios.append(scenario_binary_str)

# 2. 计算每个场景的概率和总工时
scenario_outcomes = []
for scenario in scenarios:
    scenario_hours_won = 0
    scenario_probability = 1.0 # 初始化场景概率为1

    for j, outcome_bit in enumerate(scenario):
        if outcome_bit == '0': # 项目j失败
            scenario_probability *= (1 - probabilities[j])
        else: # 项目j成功
            scenario_probability *= probabilities[j]
            scenario_hours_won += hours[j]
    scenario_outcomes.append((scenario, scenario_probability, scenario_hours_won))

# 打印部分场景结果（可选）
print("--- 部分场景结果示例 ---")
for i, outcome in enumerate(scenario_outcomes):
    if i < 5 or i > len(scenario_outcomes) - 5: # 打印开头和结尾的几个
        print(f"场景: {outcome[0]}, 概率: {outcome[1]:.6f}, 工时: {outcome[2]}")
print("...")


# 3. 计算获得超过指定工时的总概率
prob_desired_hours = sum([o[1] for o in scenario_outcomes if o[2] > min_hours_desired])
print(f"\n获得超过 {min_hours_desired} 小时的总概率: {prob_desired_hours:.6f}")

# 4. 验证所有场景概率之和是否为1
prob_check = sum([o[1] for o in scenario_outcomes])
print(f"所有场景概率之和（应为1）: {prob_check:.6f}")

# 5. 汇总相同工时的概率，构建工时-概率分布
possible_payouts = set(o[2] for o in scenario_outcomes) # 获取所有可能的总工时值
payout_probabilities = {} # 字典用于存储每个总工时对应的总概率

for payout in possible_payouts:
    # 汇总所有导致该总工时的场景的概率
    payout_probability = sum([o[1] for o in scenario_outcomes if o[2] == payout])
    payout_probabilities[payout] = payout_probability

print("\n--- 总工时与对应概率分布 ---")
# 按照工时大小排序输出，便于观察
sorted_payouts = sorted(payout_probabilities.items())
for payout, prob in sorted_payouts:
    print(f"总工时: {payout}, 概率: {prob:.6f}")

# 以JSON格式美观输出（可选）
# print(json.dumps(payout_probabilities, indent=2))