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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...
Attached Images 2016-06-18-141442.jpg (88.0 KB, 2 views)

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. Thanks from emanuelhuber June 18th, 2016, 10:19 AM #3 Newbie   Joined: Jun 2016 From: Brazil Posts: 3 Thanks: 0 Awesome! Thank's man. My Regards Tags binomial, distribution, question 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 Bhuvaneshnick Probability and Statistics 3 January 7th, 2015 10:10 PM thomas2608 Advanced Statistics 1 July 27th, 2014 01:47 PM r-soy Probability and Statistics 2 November 16th, 2013 04:05 AM koricha25 Advanced Statistics 3 April 1st, 2012 06:08 PM latkan Advanced Statistics 1 May 11th, 2009 05:49 AM

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