What is the Expected Number of Defective Keyboards?

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SUMMARY

The discussion focuses on calculating the expected number of defective keyboards produced by a computer company, where 10% of the keyboards are defective. The appropriate model for this scenario is the binomial probability distribution, defined by the formula P(i) = _nC_i p^i(1-p)^{n-i}. For this case, with n=3 keyboards and p=0.10, the expected value can be calculated using the formula NP, resulting in an expected number of 0.3 defective keyboards.

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Homework Statement



10% of the keyboards a computer company manufactures are defective. 3 keyboards come off the assembly line. Determine the probability distribution, where x=the number of defective keyboards. What would the expected number of defected keyboards be?


Homework Equations



I haven't done data management in over a year, this is the reason I can't do the question...I don't know what the equation is for probability distribution.


The Attempt at a Solution



After finding the probability distribution, I suppose I can just multiply it by 3?
 
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There is no one probability distribution. What you want is the "binomial probability distribution". If, on each trial, an event happens with probability p, then the probability the event will happen i times in n trials is [itex]P(i)= _nC_i p^i(1-p)^{n-i}[/itex] where [itex]_nC_i[/itex] is the "binomial coefficient". Here, n= 3 and p= 0.10.

No, you cannot just multiply the probability distribution by 3. The expected value in 3 "trials" is 0P(0)+ 1P(1)+ 2P(2)+ 3P(3). That is, for i= 0, 1, 2, 3, you multiply the probability that many are faulty by the number, then add.
 
Alternatively, you can just use the fact that for a binomial distribution with number N and probability P, the expected value is NP.
 

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