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March 9th, 2017, 03:54 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, 04:12 AM  #2  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
 
March 9th, 2017, 04:23 AM  #3 
Newbie Joined: Mar 2017 From: South Africa Posts: 2 Thanks: 0  
March 9th, 2017, 12:36 PM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 908 Thanks: 489  Quote:
$p(k) = \binom{n}{k}p^k (1p)^{nk}$ $p(k)=\binom{6}{k}(0.15)^k(0.85)^{6k}$ 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.  

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