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