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Algebra

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 Joined: 25 Mar 2012 Post: 28
4 Apr 2012 09:25:33 IST
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probability
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

There are four machines and it is known that exactly two of them are faulty.They are tested one by one in a random order till both faulty machines are identified.Then the probability that only two tests are needed,is

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Blazing goIITian

Joined: 2 Mar 2012
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4 Apr 2012 13:06:38 IST
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1/11

Blazing goIITian

Joined: 2 Mar 2012
Posts: 308
4 Apr 2012 13:08:42 IST
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1/11

Blazing goIITian

Joined: 21 Mar 2012
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4 Apr 2012 16:19:53 IST
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1/12

Blazing goIITian

Joined: 21 Mar 2012
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4 Apr 2012 16:26:22 IST
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NO I MIS PRINT IT IT IS 1/6

New kid on the Block

Joined: 25 Mar 2012
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4 Apr 2012 16:59:44 IST
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how?

Blazing goIITian

Joined: 21 Mar 2012
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4 Apr 2012 17:13:51 IST
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JUST IF IT IS GIVEN THAT 2 MACHINES ARE FAULTY AND WE HAVE TO TEST THOSE 2 MACHINES FROM 4 MACHINES SO........ SAMPLE SPACE=4c2..AND IF ONLY ONE TEST IS NEEDED IT MEANS THEY GOT DETAECT IN FIRST AND SECOND POSITION SO NO. OF WAYS ARE 2c2 SO PROBABILITY=1/6{ON SOVING}

Blazing goIITian

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5 Apr 2012 07:42:10 IST
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2C2/4C2=1/6

Scorching goIITian

Joined: 24 Aug 2011
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5 Apr 2012 08:35:54 IST
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1/6

Hot goIITian

Joined: 11 Mar 2012
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5 Apr 2012 09:49:33 IST
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Blazing goIITian

Joined: 21 Mar 2012
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5 Apr 2012 10:38:42 IST
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no rajat answer is 1/6 baby

Hot goIITian

Joined: 11 Mar 2012
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5 Apr 2012 10:47:17 IST
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No pet its 1/12......

Want Explanation???

Hot goIITian

Joined: 4 Apr 2012
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5 Apr 2012 10:52:20 IST
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Yes......

Hot goIITian

Joined: 11 Mar 2012
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5 Apr 2012 10:56:39 IST
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Let W and T denote Working machine.

Let F and G denote Faulty machine.

Alright,

Sample space={W,T,G,F}

P(F) at first go=1/4 .......

Agreed?

Now ,

P(G)=1/3

Total Probability=(1/4)*(1/3)=1/12

Agreed?

Hit like if u find it useful and right

Hot goIITian

Joined: 4 Apr 2012
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5 Apr 2012 11:01:53 IST
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nope................

You have two faulty machines.

so     n(A) = 2

now you have 2 machines out of four machines so you can select them in 4C2 ways

so n(S) = 12

so P(A) = 2/12

= 1/6

Hot goIITian

Joined: 11 Mar 2012
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5 Apr 2012 11:05:09 IST
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If you consider both machines to be identically faulty the you are correct.....

But if the are not identical mecanically faulty the i am correct

Hot goIITian

Joined: 4 Apr 2012
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5 Apr 2012 11:07:19 IST
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Yes man you are right i didn't considered such cases...wat if such questions come in IIT?

Blazing goIITian

Joined: 21 Mar 2012
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5 Apr 2012 11:07:44 IST
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yes the method is totally fair and hence they need to be identical and also it is given in only one test and u had considerd two tests

Hot goIITian

Joined: 11 Mar 2012
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5 Apr 2012 11:11:26 IST
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Such simple wont come in IIT-JEE

Hot goIITian

Joined: 4 Apr 2012
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5 Apr 2012 11:13:38 IST
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he he he he he

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