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June 18th, 2016, 09:24 AM  #1 
Newbie Joined: Jun 2016 From: Brazil Posts: 3 Thanks: 0  binomial distribution question
Hello guys, this is my first post in this forum and I'm sorry if I violated any rule... I'm trying to understand this question: In a particular chemical process, an quality engineer analyzes 13 samples. It's known that the test normally has 75% success rate. a) What is the chance of 10 samples being aproved? b) What is the chance of at least 10 samples being rejected? The letter a) I made it and just want to check if its right My Resolution: $\displaystyle P(x = 10) = (13!)/(10!*3!) * 0,75^(10) * 0,25^3$ In letter b) I followed this thought: if I have at least 10 rejected samples, I can have a maximum of 3 approved samples, right? So I did this: $\displaystyle P(x<=3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)$ Which results in 0,000126121 I will attach the complete resolution in the post bellow. Ps: I'm a brazilian, sorry for the bad english. Ty you all... Last edited by emanuelhuber; June 18th, 2016 at 09:31 AM. 
June 18th, 2016, 10:11 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
I'm sorry, but I cannot read your attachment. I do, however, agree with your result. $P(x = 0) = \dbinom{13}{0} * 0.75^0 * 0.25^{13} \approx 0.0000000149.$ $P(x = 1) = \dbinom{13}{1} * 0.75^1 * 0.25^{12} \approx 0.0000005811.$ $P(x = 2) = \dbinom{13}{2} * 0.75^2 * 0.25^{11} \approx 0.0000104606.$ $P(x = 3) = \dbinom{13}{3} * 0.75^3 * 0.25^{10} \approx 0.0001150668.$ $P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) \approx 0.0001261234.$ And your English is quite good, astronomically better than my Portugese. 
June 18th, 2016, 10:19 AM  #3 
Newbie Joined: Jun 2016 From: Brazil Posts: 3 Thanks: 0 
Awesome! Thank's man. My Regards 

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