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Re: Three tools are taken out of a tool box having ten different machine [#permalink]
Carcass, why it is not correct to solve this question in the following way?
select 1 defective tool out of 3--> 1/3
select 2nd defective tool out of 2--> 1/2
selct 3rd tool out of non defective tools--> 1/7

So total probability=1/3*1/2*1/7

How the question should be rephrased, so it is possible to solve a question based on the logic above?
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Re: Three tools are taken out of a tool box having ten different machine [#permalink]
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No even though the logic of the question is very basic you have to solve in the correct way of probability

1) No of ways of Selecting 2 tools from the three defective ones and 1 from the 7 non-defective ones

2) No of ways of Selecting 3 tools from the total 10 tools.

The first scenario is 3/2*7/1

The second is 10/3

Now, A/B=3*7/120=7/40

I hope this helps
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Re: Three tools are taken out of a tool box having ten different machine [#permalink]
Expert Reply
or alternatively

3 defective, 7 proper tools are present

Total ways: 10C3 =120

Required: 3C2 x 7C1=21

Probability is 21/120= 7/40
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Re: Three tools are taken out of a tool box having ten different machine [#permalink]
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Re: Three tools are taken out of a tool box having ten different machine [#permalink]
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