仓库延误概率及三台设备故障概率相关计算咨询

Alright, let's work through this problem step by step. First, let's define our events clearly for clarity:

• Let A = event that $m_1$ fails, so P(A) = 0.02
• Let B = event that $m_2$ fails, so P(B) = 0.03
• Let C = event that $m_3$ fails, so P(C) = 0.04

We'll assume the devices fail independently (this is standard for such problems unless stated otherwise).

Part (a): Probability of two or more devices failing #

Two or more failures covers three scenarios: exactly two devices fail, or all three fail. We can calculate this by summing the probabilities of each valid scenario:

• Scenario 1: $m_1$ and $m_2$ fail, $m_3$ works: P(A ∩ B ∩ ¬C) = 0.02 * 0.03 * (1-0.04) = 0.000576
• Scenario 2: $m_1$ and $m_3$ fail, $m_2$ works: P(A ∩ ¬B ∩ C) = 0.02 * (1-0.03) * 0.04 = 0.000776
• Scenario 3: $m_2$ and $m_3$ fail, $m_1$ works: P(¬A ∩ B ∩ C) = (1-0.02) * 0.03 * 0.04 = 0.001176
• Scenario 4: All three devices fail: P(A ∩ B ∩ C) = 0.02 * 0.03 * 0.04 = 0.000024

Adding these up gives the total probability:
0.000576 + 0.000776 + 0.001176 + 0.000024 = 0.002552

Alternatively, you can calculate this as 1 - P(no failures) - P(exactly one failure) to verify:

• P(no failures) = (1-0.02)*(1-0.03)*(1-0.04) = 0.912576
• P(exactly one failure) = P(A∩¬B∩¬C) + P(¬A∩B∩¬C) + P(¬A∩¬B∩C) = 0.02*0.97*0.96 + 0.98*0.03*0.96 + 0.98*0.97*0.04 = 0.018624 + 0.028224 + 0.037624 = 0.084472
• So 1 - 0.912576 - 0.084472 = 0.002552 (same result, good check!)

Part (b): Probability of warehouse delay #

The warehouse delays if either A occurs, or B ∩ C occurs. Let D = event that the warehouse delays, so D = A ∪ (B ∩ C). Using the addition rule for probabilities:
P(D) = P(A) + P(B ∩ C) - P(A ∩ B ∩ C)

We calculate each term:

• P(B ∩ C) = 0.03 * 0.04 = 0.0012 (independent events)
• P(A ∩ B ∩ C) = 0.02 * 0.03 * 0.04 = 0.000024

Plugging in the numbers:
P(D) = 0.02 + 0.0012 - 0.000024 = 0.021176

Part (c): Probability of delay given $m_3$ has failed #

This is a conditional probability problem: we need P(D | C) (probability of delay given $m_3$ failed). Using the conditional probability formula:
P(D | C) = P(D ∩ C) / P(C)

First, let's find P(D ∩ C). Since D = A ∪ (B ∩ C), intersecting with C gives us (A ∩ C) ∪ (B ∩ C). Using the addition rule again:
P((A ∩ C) ∪ (B ∩ C)) = P(A ∩ C) + P(B ∩ C) - P(A ∩ B ∩ C)

Calculate each term:

• P(A ∩ C) = 0.02 * 0.04 = 0.0008
• P(B ∩ C) = 0.0012 (from part b)
• P(A ∩ B ∩ C) = 0.000024 (from part b)

So P(D ∩ C) = 0.0008 + 0.0012 - 0.000024 = 0.001976

Now divide by P(C) to get the conditional probability:
P(D | C) = 0.001976 / 0.04 = 0.0494

A simpler way to think about this: since we know $m_3$ has failed, the delay happens if either $m_1$ fails, or $m_2$ fails (because $m_3$ is already down, so $m_2$ and $m_3$ both failing is just $m_2$ failing). So P(D | C) = P(A ∪ B | C) = P(A) + P(B) - P(A ∩ B) (since independence means P(A|C)=P(A) and P(B|C)=P(B)). This gives 0.02 + 0.03 - (0.02*0.03) = 0.05 - 0.0006 = 0.0494—same result, which confirms we're right!

内容的提问来源于stack exchange，提问作者SomebodyOnEarth