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A warehouse uses three machines ($m_1$, $m_2$ and $m_3$) and their failure rate is 0.02, 0.03 and 0.04.

a) Find the probability that two or more machines fail.

b) The warehouse falls behind schedule if at least one of the following happens: (1) $m_1$ fails; (2) $m_2$ and $m_3$ both fail. What is the probability that the warehouse falls behind schedule?

c) Following the scenario from b), given that $m_3$ fails what is the probability that the warehouse falls behind schedule?

What I have so far:

(a) $P($two or more machines fail $) = P(m_1$ and $m_2$ fail, $m_3$ doesn't$) + P(m_1$ and $m_3 $ fail, $m_2$ doesn't$)+ P(m_2$ and $m_3$ fail, $m_1$ doesn't$) + P(m_1,m_2$ and $m_3$ fail$)$ $=(0.02)(0.03)(1-0.04)+(0.02)(1-0.03)(0.04)+(1-0.02)(0.03)(0.04)+(0.02)(0.03)(0.04)=0.002552$

(b) $P(m_1\ fails)=(0.02)(1-0.03)(1-0.04)+(0.02)(0.03)(1-0.04)+(0.02)(1-0.03)(0.04)=0.019976$ $P(m_2\ and\ m_3\ fail)=(1-0.02)(0.03)(0.04)=0.001176$ $P(m_1,m_2\ and\ m_3\ fail)=(0.02)(0.03)(0.04)=0.000024$ $P(warehouse\ falls\ behind\ schedule) = 0.019976 + 0.001176 + 0.000024 = 0.021176$

I'm confident that my answer for (a) is correct but I'm not sure about (b). I don't know how to solve (c) at all

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    $\begingroup$ Your logic for $a$ appears to be incorrect. the event "$m_1$ and $m_2$ fail " (for one) is contained in the event "all three fail" so you are badly overcounting the latter scenario. $\endgroup$
    – lulu
    Feb 8, 2018 at 20:03
  • $\begingroup$ So would it just be: $P($two or more machines fail $) = P(m_1$ and $m_2$ fail$) + P(m_1$ and $m_3 $ fail$)+ P(m_2$ and $m_3$ fail$) + P(m_1,m_2$ and $m_3$ fail$)$ $=(0.02)(0.03)+(0.02)(0.04)+(0.03)(0.04)+(0.02)(0.03)(0.04)$? $\endgroup$
    – SomebodyOnEarth
    Feb 8, 2018 at 20:06
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    $\begingroup$ That's the same as what you have above and it is incorrect for the reason I gave before. $\endgroup$
    – lulu
    Feb 8, 2018 at 20:13
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    $\begingroup$ If you want disjoint events (not a bad way to go) then take: "$m_1, m_2$ fail, $m_3$ doesn't" , "$m_1, m_3$ fail, $m_2$ doesn't", "$m_2, m_3$ fail, $m_2$ doesn't", "All three fail". $\endgroup$
    – lulu
    Feb 8, 2018 at 20:17
  • $\begingroup$ I thought, it was implied that the third machine doesn't fail if I didn't write it. I'll change that. $\endgroup$
    – SomebodyOnEarth
    Feb 8, 2018 at 20:20

1 Answer 1

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A warehouse uses three machines ($m_1, m_2, m_3$) and their respective failure rates are $0.02$, $0.03$, and $0.04$. What is the probability that two or more machines fail?

Assuming independence, your answer to this question is correct.

The warehouse falls behind schedule if at least one of the following happens: (1) $m_1$ fails or (2) $m_2$ and $m_3$ both fail. What is the probability that the warehouse falls behind schedule.

Let $F_k$ be the event that machine $k$ fails. Then the probability that the warehouse falls behind schedule is $$P(F_1 \cup (F_2 \cap F_3)) = P(F_1) + P(F_2 \cap F_3) - P(F_1 \cap F_2 \cap F_3)$$ Assuming independence, we obtain \begin{align*} P(F_1 \cup (F_2 \cap F_3)) & = 0.02 + (0.03)(0.04) - (0.02)(0.03)(0.04)\\ & = 0.02 + 0.0012 - 0.000024\\ & = 0.021176 \end{align*}

Given that $m_3$ fails, what is the probability that the warehouse falls behind schedule.

If $m_3$ fails, the warehouse falls behind schedule if at least one of the other machines fails.

Assuming independence, the desired probability is \begin{align*} P(F_1 \cup F_2) & = P(F_1) + P(F_2) - P(F_1 \cap F_2)\\ & = 0.02 + 0.03 - (0.02)(0.03)\\ & = 0.02 + 0.03 - 0.0006\\ & = 0.0494 \end{align*} More formally, the probability that the warehouse falls behind schedule is \begin{align*} P((F_1 \cap F_3) \cup (F_2 \cap F_3) \mid F_3) & = \frac{P((F_1 \cap F_3) \cup (F_2 \cap F_3))}{P(F_3)}\\ & = \frac{P(F_1 \cap F_3) + P(F_2 \cap F_3) - P(F_1 \cap F_2 \cap F_3)}{P(F_3)} \end{align*} If we assume independence, we obtain \begin{align*} P((F_1 \cap F_3) \cup (F_2 \cap F_3) \mid F_3) & = \frac{(0.02)(0.04) + (0.03)(0.04) - (0.02)(0.03)(0.04)}{0.04}\\ & = 0.02 + 0.03 - (0.02)(0.03)\\ & = 0.0494 \end{align*}

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