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December 28th, 2018, 06:06 AM  #1 
Member Joined: Mar 2017 From: Israel Posts: 85 Thanks: 2  Probability
Hello Can you help me please about the next exercise: In a certain manufacturing process, the percentage of defective products is 8%. In testing the quality of the product, there is a probability of 0.9 to disqualify the product when it is defective, but also a probability of 0.2 to disqualify the product when it is normal (not defective). 15 products were randomly selected, What is the probability that at most 2 products were disqualified in the test? I tried to solve it in many ways, and I still didn't succeed, I know that there are (15*0.08*0.9)+(15*0.92*0.2)=3.84 products which were disqualified (if I'm not wrong), and from there I lost myself. The answer is 0.22, I still didn't reach this answer. Thanks a lot! 
December 28th, 2018, 06:34 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,467 Thanks: 1342 
$P[\text{disqualified}] = P[\text{disqualified because defective}] + P[\text{disqualified in error}] = (0.08)(0.9) + (0.92)(0.2) = 0.256$ The number of samples that are disqualified, $N$, has a binomial distribution with parameters $n=15,~p = 0.256$ note: we are defining being disqualified as success here. $P[N\leq 2] = \\ \sum \limits_{k=0}^2~\dbinom{15}{k}(0.256)^k (0.744)^{15k} = \\ 0.0118466 + 0.0611438 + 0.147271 = \\0.220261$ Last edited by romsek; December 28th, 2018 at 06:38 AM. 
December 28th, 2018, 06:49 AM  #3  
Member Joined: Mar 2017 From: Israel Posts: 85 Thanks: 2  Quote:
And I'm sorry for bothering you, but how I can solve it in normal way? I mean like: (0.256/(0.256*15)) + ((2*0.256)/(15*0.256)), but I don't know why it's false because I got 0.2 (and not 0.22), it's totally like (1/15) + (2/15), I think? Edit: Ups, I understood why I did mistake, I will keep thinking how to solve it with basic math. Last edited by IlanSherer; December 28th, 2018 at 06:52 AM.  

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