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June 18th, 2016, 09:24 AM   #1
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Question 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...
Attached Images
File Type: jpg 2016-06-18-141442.jpg (88.0 KB, 2 views)

Last edited by emanuelhuber; June 18th, 2016 at 09:31 AM.
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June 18th, 2016, 10:11 AM   #2
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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.
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June 18th, 2016, 10:19 AM   #3
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Awesome!

Thank's man.

My Regards
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