概率问题求解：画家完成两幅画作的概率计算及补做机会分析

Alright, let's break down this probability problem step by step to figure out which strategy gives the painter a better shot at getting both paintings done eventually. Since the original question cuts off, we'll make a reasonable assumption: if any painting isn't completed on the first attempt, the painter gets one additional chance to finish the unfinished pieces, using the same strategy as the initial try. Our goal is to calculate the probability of successfully completing both paintings for each strategy.

策略1：专注攻坚单幅画作 #

In this strategy, when focusing all effort on one painting, its on-time completion probability is 0.8, while the other painting only has a 0.4 chance of being finished on time. Let's define:

• P(X) = 0.8: Probability the focused painting is completed
• P(Y) = 0.4: Probability the non-focused painting is completed

We calculate the total success probability by breaking down all possible scenarios:

各场景概率计算 #

• Both paintings done on the first try:
  P(X ∩ Y) = 0.8 × 0.4 = 0.32
• Focused painting done, non-focused fails first but succeeds on the extra chance:
  Probability: P(X ∩ ¬Y) × P(Y) = (0.8 × 0.6) × 0.4 = 0.192
• Non-focused painting done, focused fails first but succeeds on the extra chance:
  Probability: P(¬X ∩ Y) × P(X) = (0.2 × 0.4) × 0.8 = 0.064
• Both fail first try, both succeed on the extra chance:
  Probability: P(¬X ∩ ¬Y) × P(X ∩ Y) = (0.2 × 0.6) × 0.32 = 0.0384

总成功概率 #

Add up all these scenario probabilities:
0.32 + 0.192 + 0.064 + 0.0384 = 0.6144
That's a 61.44% chance of finishing both paintings with this strategy.

策略2：同时推进两幅画作 #

Here, the painter splits effort between both pieces, so each painting has an independent 0.6 probability of being completed on time per attempt. Let P(A) = P(B) = 0.6 for the two paintings.

各场景概率计算 #

• Both paintings done on the first try:
  P(A ∩ B) = 0.6 × 0.6 = 0.36
• Painting A done, B fails first but succeeds on the extra chance:
  Probability: P(A ∩ ¬B) × P(B) = (0.6 × 0.4) × 0.6 = 0.144
• Painting B done, A fails first but succeeds on the extra chance:
  Probability: P(¬A ∩ B) × P(A) = (0.4 × 0.6) × 0.6 = 0.144
• Both fail first try, both succeed on the extra chance:
  Probability: P(¬A ∩ ¬B) × P(A ∩ B) = (0.4 × 0.4) × 0.36 = 0.0576

总成功概率 #

Add up all these scenario probabilities:
0.36 + 0.144 + 0.144 + 0.0576 = 0.7056
That's a 70.56% chance of finishing both paintings with this strategy.

结论 #

Comparing the two results:

• Strategy 1 (focus on one painting): 61.44% success probability
• Strategy 2 (work on both simultaneously): 70.56% success probability

Clearly, the painter has a higher chance of getting both paintings done on time (even with one extra attempt) if he works on both pieces simultaneously rather than focusing on a single one first.

内容的提问来源于stack exchange，提问作者vic12