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    The correction test is 0.038 and 10 randomly selected components define the distribution, the probability that:

    -none of the components fail the test?

    How do you do this? =\
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    None fail = all 10 pass.
    P(pass)=1- P(fail)
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    (Original post by Get me off the £\?%!^@ computer)
    None fail = all 10 pass.
    P(pass)=1- P(fail)
    So would that be 10*(0.038)^0 * (1-0.038)^10 = 6.788 (3dp) ?
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    (Original post by Nkhan)
    So would that be 10*(0.038)^0 * (1-0.038)^10 = 6.788 (3dp) ?
    No, the 10 should not be there. There is only one way for all 10 to pass.
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    (Original post by Get me off the £\?%!^@ computer)
    No, the 10 should not be there. There is only one way for all 10 to pass.
    So its just 0.038)^0 * (1-0.038)^10 ?
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    Yes.
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    (Original post by Get me off the £\?%!^@ computer)
    Yes.
    ahh okays thanks x
 
 
 
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