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A box contain 20 machine parts, 5 of them standard. A worker

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Avtar
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A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   10 Jun 2004, 18:08
A box contain 20 machine parts, 5 of them standard. A worker takes out 3 parts at random. Find the probababilty that at least 1 out of three turns out to be standard.



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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   10 Jun 2004, 18:19
all possible ways = 20C3
number of ways to pick all three non-standard = 15C3

P = 1 - (15C3/20C3) = 15*14*13/20*19*18 = 0.6
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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   10 Jun 2004, 20:04
:yes I concur with the reasoning but answer should be 1 - 91/228 = 137/228
Oh well, that was a very close approximation.
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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   11 Jun 2004, 05:50
Avtar wrote:
A box contain 20 machine parts, 5 of them standard. A worker takes out 3 parts at random. Find the probababilty that at least 1 out of three turns out to be standard.


Atleast one will drive three cases :

Total : 20C3 = 1140
One standard and another 2 non-standard : 5C1 * 15C2 = 525
Two standard and another 1 non-standard : 5C2 * 15C1= 150
All three standard : 5C3 = 10

So answer should be: 685 / 1140 = 137 / 228

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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   11 Jun 2004, 06:23
Avtar wrote:
A box contain 20 machine parts, 5 of them standard. A worker takes out 3 parts at random. Find the probababilty that at least 1 out of three turns out to be standard.


P{1<=x<=3}=1-P{x=0} = 1-(15C3/20C3)= 1 - (15!*3!*17!)/(12!*3!*20!) = 1- 15*14*13/(20*19*18) = 1-91/228 = 137/228
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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   12 Jun 2004, 01:54
Avtar wrote:
A box contain 20 machine parts, 5 of them standard. A worker takes out 3 parts at random. Find the probababilty that at least 1 out of three turns out to be standard.


P(at least 1 is standard) = 1 - P(all are non-standard) = 1 - C[3,15]/C[3,20] = ...
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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   17 Jun 2004, 09:41
I got the process down, but keep forgetting to do
1- probability of getting an unfavorable outcome

and you know that the ETS will conveniently place the other answer among the choices
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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink] New post   17 Jun 2004, 09:43
lastochka wrote:
I got the process down, but keep forgetting to do
1- probability of getting an unfavorable outcome

and you know that the ETS will conveniently place the other answer among the choices


Yes, you should be aware of such traps in GMAT...



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Re: A box contain 20 machine parts, 5 of them standard. A worker [#permalink]
17 Jun 2004, 09:43
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