My Math Forum Probability Distributions

 Probability and Statistics Basic Probability and Statistics Math Forum

 March 9th, 2017, 02:54 AM #1 Newbie   Joined: Mar 2017 From: South Africa Posts: 2 Thanks: 0 Probability Distributions A wheat miller will only accept a consignment of wheat if the wheat has dried out completely. From past experience, the miller knows that approximately 15% of all wheat bags delivered is too wet to be processed immediately. If a random sample of 6 bags are drawn from each consignment he receives, find the probability that: 2.1.1. Exactly 3 bags being too wet 2.1.2. Less than 4 bags being too wet 2.1.3. At least 5 bags being too wet 2.1.4. Exactly 2 bags being dry enough Any help is greatly appreciated! i don't know how use these questions in the formula..
March 9th, 2017, 03:12 AM   #2
Member

Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

Quote:
 Originally Posted by Kyle0607 A wheat miller will only accept a consignment of wheat if the wheat has dried out completely. From past experience, the miller knows that approximately 15% of all wheat bags delivered is too wet to be processed immediately. If a random sample of 6 bags are drawn from each consignment he receives, find the probability that: 2.1.1. Exactly 3 bags being too wet 2.1.2. Less than 4 bags being too wet 2.1.3. At least 5 bags being too wet 2.1.4. Exactly 2 bags being dry enough Any help is greatly appreciated! i don't know how use these questions in the formula..
Each time you test a bag, the only possibilities are wet or dry. As each trial only ever has two possibilities, that means the distribution is binomial.

March 9th, 2017, 03:23 AM   #3
Newbie

Joined: Mar 2017
From: South Africa

Posts: 2
Thanks: 0

Quote:
 Originally Posted by Prove It Each time you test a bag, the only possibilities are wet or dry. As each trial only ever has two possibilities, that means the distribution is binomial.
So would i be correct in saying
2.2.1) 0.27648
2.2.2) 0.04096
2.2.3) 0.995904
2.2.4) 0.059535

March 9th, 2017, 11:36 AM   #4
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,102
Thanks: 1093

Quote:
 Originally Posted by Kyle0607 So would i be correct in saying 2.2.1) 0.27648 2.2.2) 0.04096 2.2.3) 0.995904 2.2.4) 0.059535
these numbers aren't what I get at all

$p(k) = \binom{n}{k}p^k (1-p)^{n-k}$

$p(k)=\binom{6}{k}(0.15)^k(0.85)^{6-k}$

1) $p(3) = \binom{6}{3}(0.15)^3(0.85)^3 \approx 0.0414534$

similar for the others. Try again or show how you came about the above results.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ConcreteJungle Advanced Statistics 0 January 6th, 2016 08:06 AM chemdude Advanced Statistics 0 November 6th, 2015 03:35 PM chapsticks Probability and Statistics 1 October 28th, 2012 07:19 PM Chee Probability and Statistics 1 May 31st, 2012 11:48 PM Natalie89 Advanced Statistics 1 February 13th, 2011 04:44 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top