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A new Machine is bought by a factory to produce 2 parts, A

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A new Machine is bought by a factory to produce 2 parts, A [#permalink] New post 06 Jun 2004, 21:26
A new Machine is bought by a factory to produce 2 parts, A and B. A and B when assembled produce C. During the testing stages of the machine, 1 out of 10 in A is defective and 2 out of 10 in B are defective. What is the probability of C being defective.

a. 0.02
b. 0.20
c. 0.28
d. 0.32
e. 0.72
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 [#permalink] New post 07 Jun 2004, 07:33
P(non-defective)=(9/10)*(8/10)=72/100=0.72
P(defective)=1-0.72=0.28
C.
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 [#permalink] New post 07 Jun 2004, 07:35
That "Guest" was me :) .Forgot to login before replying.
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another approach.. [#permalink] New post 09 Jun 2004, 08:40
P(C) defective = P(A) defective + P(B) defective - P (A and B) defective

= .2 + .1 - .2 * .1 = .3 - .02 = .28
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 [#permalink] New post 16 Jun 2004, 01:02
Sorry for the late reply

The correct answer is indeed 0.28.

Thanks for all the explaination.
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 [#permalink] New post 23 Jun 2004, 20:13
Another great one where no calculation is required. We know that the upper limit is 30% ( if all defective A's and B's are seperated)and lower limit is 20% (if all defective A's and B's are joined together). Only one answer left

Should I be thinking like this or should I be looking to calculate
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 [#permalink] New post 24 Jun 2004, 04:14
HI Maneesh

In my opinion, it is best to calculate in any probability question. I used to strulle in Combinatins and Prob questions. But thru this forum, things looks much easier. All it needs is to put the words in to figures. Mental calculations or close approximation does not help in here.

Example. You may have answers whic are very close 1/3 or2/5 ..it is difficult. Practice using calculation. The more you do, the better you get. I bet on this.

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 [#permalink] New post 24 Jun 2004, 06:55
Hi guys, actually we have solved this question before...

Of course, C is the answer.
  [#permalink] 24 Jun 2004, 06:55
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A new Machine is bought by a factory to produce 2 parts, A

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