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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. $P[N\leq 2] = \\ \sum \limits_{k=0}^2~\dbinom{15}{k}(0.256)^k (0.744)^{15-k} = \\ 0.0118466 + 0.0611438 + 0.147271 = \\0.220261$ Thanks from IlanSherer Last edited by romsek; December 28th, 2018 at 06:38 AM. December 28th, 2018, 06:49 AM   #3
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Quote:
 Originally Posted by romsek $P[\text{disqualified}] = P[\text{disqualified because defective}] + P[\text{disqualified in error}] = (0.0 (0.9) + (0.92)(0.2) = 0.256$ 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. $P[N\leq 2] = \\ \sum \limits_{k=0}^2~\dbinom{15}{k}(0.256)^k (0.744)^{15-k} = \\ 0.0118466 + 0.0611438 + 0.147271 = \\0.220261$
Thanks!

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. Tags probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Devans99 Probability and Statistics 4 June 25th, 2018 01:06 PM rivaaa Advanced Statistics 5 November 5th, 2015 01:51 PM hbonstrom Applied Math 0 November 17th, 2012 07:11 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM

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