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A machine needs at least 2 of its 3 parts to work correctly. The probability of part 1 failing is 1/2, the probability of part 2 failing is p and the probability of part 3 failing is also p. We are asked to find the probability that the machine fails. We are also told that each part failing is independent of any other part failing.

In sloppy notation I let $i$ be the event that the $ith$ part fails. The probability that the machine fails is the probability that any two parts fail or all three fail. I have this written in the first line below but perhaps my interpretation is not correct. My idea is to use the compliment which converts union to intersection, then use the assumption of independence to compute each probability which I write as

\begin{array}{rcl} P(\text{machine fails}) &=& P\left[ (1\cap 2 ) \cup (1 \cap 3) \cup (2 \cap 3) \cup (1 \cap 2 \cap 3)\right] \\ &=& 1 - P\left[ \big( (1\cap 2 ) \cup (1 \cap 3) \cup (2 \cap 3) \cup (1 \cap 2 \cap 3) \big)^c \right] \\ &=& 1 - P\left[ \big( 1\cap 2 )^c \cap (1 \cap 3)^c \cap (2 \cap 3)^c \cap (1 \cap 2 \cap 3)^c \big) \right] \\ &=& 1 - P( (1\cap2) ^c)P((1 \cap 3)^c) P((2 \cap 3)^c) P((1 \cap 2 \cap 3)^c) \\ &=& 1 - (1-P(1\cap2))(1-P(1\cap 3)(1-P(2\cap 3))(1-P(1\cap 2\cap3)) \\ &=& 1 - (1-P(1)P(2))(1-P(1)P(3))(1-P(2)P(3))(1-P(1)P(2)P(3)) \\ &=& 1 - (1-p/2)(1-p/2)(1-p^2)(1-p^2/2) \end{array} This is a sixth degree polynomial in $p$ which is not one of the correct answers. I think I'm overcounting with my set setup or that I'm applying the independence assumption incorrectly. What is a better way to solve this problem, and where did my thinking go wrong?

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  • $\begingroup$ I think I see where I went wrong above. In line 4 I have assumed that each event $(i \cap j)^c$ is independent which probably does not have to be true. $\endgroup$
    – R R
    Oct 13, 2014 at 20:25

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Maybe this helps. The probability that the machine fails is the probability that parts 1 and 2 fail and 3 doesn't fail, or parts 1 and 3 fail and 2 doesn't fail, or parts 2 and 3 fail and 1 doesn't fail, or all three parts fail. These are all disjoint events, so you can sum the probabilities for each one. For example, in the first case, the probability that parts 1 and 2 fail and 3 doesn't fail is $p(1 - p)/2$, and so on.

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    $\begingroup$ Thank you. This reasoning sounds good and it is one of the plausible answers. For future reference to anyone reading the answer I get is $p.$ $\endgroup$
    – R R
    Oct 13, 2014 at 20:34

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