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A box contain 10 screws out of which 3 are defective.If two screws are drawn at random WITH REPLACEMENT.What is the probability that no screw is defective?

my question is ,as its given With Replacement. So why not it is (7/10)*(7/10)

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@uddipto.. i guess in your ques. , the point is that we are taking 2 screws simultaneously not one by one.if it has been one by one then the one which we had taken earlier can be put into bag again but when you are taking 2 screws simultaneously then with replacement won't come into picture. it is like when you are taking only 1 object from bucket, in that case it does not matter that it is with or without replacement.not quite sure if ques is answered using this logic or not!
i was wrong! it should be 7/10 * 7/10. you are correct @uddipto
No problem! even i was confused by the answer! testbook massacre

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answered by Veteran (56.9k points) 36 189 499
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why does this formula fail (n+r-1)C2 i.e 8C2/11C2 ? since with replacement

See here the key thing is :

a) The population which is a box of  10 screws and hence finite .

b) Screws are drawn with replacement.

So the assumption of binomial distribution will hold..

So P ( no defective ) = 2C0 (7/10)2

= 0.49

answered by Veteran (88.1k points) 15 58 293
habib i did the same.. But testbook made it incorrect
Dont worry..Ur concept is correct..

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