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December 18th, 2018, 04:21 AM  #1 
Newbie Joined: Dec 2018 From: India Posts: 1 Thanks: 0  Calculating theoretical probability
Hello All, I have the following question in one of my tutorials. I need some help in resolving this. Background: A manufacturing company developed 40000 new drugs and they need to be tested. Question The QA checks on the previous batches of drugs found that — it is 4 times more likely that a drug is able to produce a better result than not. If we take a sample of 10 drugs, we need to find the theoretical probability that at most 3 drugs are not able to do a satisfactory job. a.) Propose the type of probability distribution that would accurately portray the above scenario, and list out the three conditions that this distribution follows. b.) Calculate the required probability. Thanks, Vicky Last edited by skipjack; December 18th, 2018 at 04:24 AM. 
December 18th, 2018, 09:32 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,370 Thanks: 1274 
The number of drugs out of the sample of 10 that produce a better result has a binomial distribution. $P[\text{a given drug produces a better result}] = \dfrac 4 5 = 0.8$ $P[\text{at most 3 drugs out of sample of 10 do not produce better result}] = \\ P[\text{at least 7 drugs out of sample do produce a better result} = \\ \sum \limits_{k=7}^{10}~\dbinom{10}{k}\left(\dfrac 4 5\right)^k \left(\dfrac 1 5\right)^{10k} = \dfrac{8585216}{9765625} \approx 0.88$ We could look at it another way where we define success as the drug not producing a better result. In that case we'd have $P[\text{at most 3 drugs don't produce better result}] = \\ \sum \limits_{k=0}^3~\dbinom{10}{k}\left(\dfrac 1 5\right)^k \left(\dfrac 4 5\right)^{10k} \\ \text{and this of course produces the same result}$ 

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calculating, cumulative distribution, probability, theoretical, theoretical probability 
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