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    A factory is considering two methods of checking the quality of production of the batches of items it produces.
    Method 1 A random sample of 10 items is taken from a large batch and the batch is accepted if there are no defectives in this sample. If there are 2 or more defectives the batch is rejected. If there is only 1 defective then another sample of 10 is taken and the batch is accepted if there are no defectives in this second sample, otherwise the whole batch is rejected.

    Method 2
    A random sample of 20 items is taken from a large batch and the batch is accepted if there is at most 1 defective in this sample, otherwise the whole batch is rejected.
    The factory knows that 1% of items produced are defective and wishes to use the method of checking the quality of production for which the probability of accepting the whole batch is largest.

    a. Decide which method the factory should use.
    b. Determine the expected number of items sampled using method 1.

    I solved a. but I cannot find a way to solve b. I even searched on solution bank, but still I havent got a clue....
    can anyone help me please
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    In method 1, either one sample will be taken (and there will be a probability associated with that) or two samples are taken (and there will be a probability for that too). You need to define a random, say Y, to represent the number of samples taken (which can only be 1 or 2). Then you need to work out:

    P(Y=1) = ....
    P(Y=2) = ....

    Just check that those to add up to 1 before proceeding further, then you can work out E(Y) using the normal method for any discreet random variable. The number of items sampled will then be 10 x E(Y) as there are ten items per sample.
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    (Original post by old_engineer)
    In method 1, either one sample will be taken (and there will be a probability associated with that) or two samples are taken (and there will be a probability for that too). You need to define a random, say Y, to represent the number of samples taken (which can only be 1 or 2). Then you need to work out:

    P(Y=1) = ....
    P(Y=2) = ....

    Just check that those to add up to 1 before proceeding further, then you can work out E(Y) using the normal method for any discreet random variable. The number of items sampled will then be 10 x E(Y) as there are ten items per sample.
    thank you but I... still havent solved it haha...
    I worked out those two probabilities
    P(Y=1)= 10C1 * 0.01 * 0.99^9 = 0.09135.....
    P(Y=2)= 10C2 * 0.01^2 * 0.99^8 = 0.00415...

    Then I got lost.... do I just simply multiply each of them by 10 and add them together?
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    It looks as though what you've worked out there is the probabilities of there being one and two faulty items in the first sample. What we're after here is something different, i.e. the probability that either one or two samples will be tested. According to the methodology described in the question:

    * If there are no defective items in the first sample, the whole batch will be accepted (i.e. only one sample will be tested, and Y=1 in my terminology);

    * If there is exactly one defective item in the first sample, a second sample will be tested (Y=2 in my terminology)

    * If there are two or more defective items in the first sample, the whole batch will be rejected (i.e. no second sample test and Y=1 in my terminology)

    Does that help?
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    your P(y=1) is wrong, is should be: P(none have defects in first sample) + P(2 or more have defects in first sample)
    While P(Y=2): P(exactly one defect in first sample)

    Then you work out the expected value by doing (P(Y=1)*1 + P(Y=2)*2)*10
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    (Original post by Darth_Narwhale)
    your P(y=1) is wrong, is should be: P(none have defects in first sample) + P(2 or more have defects in first sample)
    While P(Y=2): P(exactly one defect in first sample)

    Then you work out the expected value by doing (P(Y=1)*1 + P(Y=2)*2)*10
    Thank you. I got the answer and you two helped me really a lot thanks
    but can I just ask you where that *2 came from?
 
 
 
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