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  • A = pump 1 not working
  • B = pump 2 not working
  • T = pump 3 not working

  • P(A) = P(B) = 0.01

  • P(A given B) = 0.1

Pump 3 works independently of the two other pumps, and P(T) = 0.04

Find the probability that at least 2 of the 3 pumps are not working?

What I have been trying to do:

  • P(A and B) = 0.001
  • P(A and B and T) = 0.00004
  • P(A and T) = 0.0004
  • P(B and T) = 0.0004

Added up is equal to 0.00184 (Wrong)

The answer is 0.00172

I have tried other things, but this is the closest I have gotten to the answer.

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  • $\begingroup$ The events are not disjoint so addition will not give the probability of the union of these events. To repair you must e.g. find $\Pr(A\cap B\cap T^c)$ instead of $\Pr(A\cap B)$. $\endgroup$ – drhab Aug 1 '16 at 17:30
  • $\begingroup$ @drhab That's funny, I was exactly testing out this. Could you maybe make an answer where to use Pr(A∩B∩Tc), and when to use Pr(A∩B)? $\endgroup$ – David Lund Aug 1 '16 at 17:38
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You want the probability that two or more pumps are not working, which is

$$P(A\cap B)+P(A\cap T)+P(B\cap T)-2P(A\cap B\cap T).$$

We have to subtract twice the probability of the intersection of all three sets because we counted it three times in the other three intersections. (Drawing a Venn diagram helps.)

Thus the probability is

$$0.001+0.0004+0.0004-2(0.00004)=0.00172$$

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The event that at least $2$ of the $3$ pumps are not working is indeed:

$$(A\cap B)\cup(A\cap T)\cup (B\cap T)\cup(A\cap B\cap T)$$

But this is a union of events that are not disjoint.

Also we can write:$$(A\cap B\cap T^c)\cup(A\cap B^c\cap T)\cup (A^c\cap B\cap T)\cup(A\cap B\cap T)$$

In the second case we deal with a union of disjoint events so the probability can be calculated as$$\Pr(A\cap B\cap T^c)+\Pr(A\cap B^c\cap T)+\Pr(A^c\cap B\cap T)+\Pr(A\cap B\cap T)$$

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  • $\begingroup$ Thanks, this is how I ended up doing it. (A∩B)-(A∩B∩T)+(A∩T)-(A∩B∩T)+(B∩T)-(A∩B∩T)+(A∩B∩T). I just realized the last minus (A∩B∩T) was not necessary, but needed to understand. I drew up a venn diagram with 3 bubbles all connecting each other. $\endgroup$ – David Lund Aug 1 '16 at 19:06

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