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# A company has 27 machines and 27 pumps. Each machine is

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Intern
Joined: 20 Apr 2007
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A company has 27 machines and 27 pumps. Each machine is [#permalink]  25 Oct 2007, 17:48
A company has 27 machines and 27 pumps. Each machine is hooked up to a pump. At any point of time 13 out of 27 machines are not running and 6 out of 27 pumps are not working. What is the probability that at any given time atleast one bad pump is hooked up to a machine that is not running?
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[#permalink]  25 Oct 2007, 18:01
First solve for all working:

P(All machines working and pump working) = P(M) * P(P)
P(M) = 14/27
P(P) = 21/27

P(All) = (14/27)*(21/27) = 294/729

P(at least one bad pump & bad machine) = 1-P(all) = 435/729
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Joined: 08 Jun 2005
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[#permalink]  26 Oct 2007, 01:00
yuefei wrote:
First solve for all working:

P(All machines working and pump working) = P(M) * P(P)
P(M) = 14/27
P(P) = 21/27

P(All) = (14/27)*(21/27) = 294/729

P(at least one bad pump & bad machine) = 1-P(all) = 435/729

Perfect !

[#permalink] 26 Oct 2007, 01:00
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# A company has 27 machines and 27 pumps. Each machine is

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