Question

In: Statistics and Probability

A batch of 500 disks contains 5 defectiv disks. Two are selected randonly without replacement, from...

A batch of 500 disks contains 5 defectiv disks. Two are selected randonly without replacement, from the batch. Let A and B denote the events that the first and the second selected disks are defective, respectively.

a) Are A and B independents events?

b) If the samplins was done with replacement, would A and B be independent?

Solutions

Expert Solution

Let events .

A = the first disk is defective. and B = the second disk is defective.

For Independent events P( A and B ) = P(A)*P(B)

Part a)

There are 5 defective disks and 495 non defective disks, total 500 disks.

We are selecting 2 diske "without replacement"

P(A) = = 5 / 500

P(A) = 0.01

P(B) = P( both disks are defective ) + P( 1st disk is non defective and 2nd disk is defective )

=

= (0.01*0.008016)+(0.99*0.01002)

P(B) = 0.01

Therefore P(A)*P(B) = 0.01*0.01

P(A)*P(B) = 0.0001

Now we have to find P(A and B) , it means both disks are defective .

P( A and B ) =

P( A and B ) = 0.00008

As P( A and B )   P(A)*P(B) , Events A and B are not independent.

Part b) We are selecting 2 disks with replacement method.

P(A) = 5/500 = 0.01

P(B) = P( both disks are defective ) + P( 1st disk is non defective and 2nd disk is defective )

=

= (0.01*0.01) + (0.99*0.01)

P(B) = 0.01

Therefore P(A)*P(B) = 0.01*0.01 = 0.0001

Now we have to find P( A and B), it means both disks are defective .

P( A and B ) =

P( A and B ) = 0.0001

As P(A and B) = P(A)*P(B) , Events A and B are independent.


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