Binomial Distribution examples | ExamSolutions

hi welcome to this tutorial on the binomial distribution this is the third one in the series and in this I'm going to show you how to calculate various probabilities from binomial distributions now I'm assuming that you're familiar what a binomial distribution is and also how to work out the probability of our successes if not please go back and look at my first two tutorials now for the first question we've got a manufacturer of a bag of sweets claims that there is a 90% chance that the bag contains some Toffees and if 20 bags are chosen what is the probability that one all the bags contain Toffees and impart two more than eighteen bags contain Toffees now this particular question is a binomial distribution why because there are a finite number of trials there's 20 in this particular problem each referring to a particular bag of sweets there are two outcomes in any trial success failure success being where we pick a bag that has Tuffy's in it failure being that the bag does not have Tuffy's the probability of success remains constant all the way through at naught point nine and the probabilities are independent of one another in any trial so typical binomial distribution but I'm not encouraging you to draw a tree diagram I just want you to be aware that this is what's going on in the background okay well let's just remove the tree diagram or we'll get on with the question now the first thing we need to do is to define a random variable so it can be any letter I'm going to choose X and make sure you write a capital letter for a random variable I'm going to let X be the random variable and I'll write RV for short and it's going to represent the number of bags with Toffees okay so we just write that in don't be lazy always write the description of what your random variable is going to say if it doesn't say it in the question and we have to define the distribution that it comes from so in this case where X is binomially distributed remember it's two parameters for the binomial distribution the number of trials which is in this case 20 for the sweets and next is the probability of success that is a bag contained Tuffy's and the probability that happening is not 0.9 for 90% and the first question probability that all the bags contain Tov is so for part one we're looking for the probability that x equals 20 and assuming that you're familiar then with the formula it will be 20 see and then the number of successes which in this case is 20 then we write down the probability of success which is not 0.9 and this occurs with a success rate of 20 so that's to the power 20 and then you've got naught point 1 the probability of failure and that would occur not know x next we need to work this out on the calculator and you should find that you get naught point 1 2 1 5 7 and so on which when rounded comes out at naught point 1 to 2 to 3 decimal places so that's a very simple basic question working out the probability of a particular value and the next example though we've got several values because we're being asked the probability that the number of bags that contain Toffees is more than 18 so X is more than 18 now the random variable X takes on the values zero because you could find you've got no bags containing Tuffy's or one bag could contain Toffees two bags and so on all the way up to 20 bags so if we're looking for X to be more than 18 it could be that we have 19 bags that contain Tuffy's or it could be 20 bags that contain Toffees and we'd have to add these two probabilities together and if we now use the formula for the probability x equals 19 it will be from 20 trials choose 19 successes probability of success is not 0.9 and that's repeated 19 times over probability of failure is not 0.1 and that would only occur once then we've got plus the probability of x equals 20 and we could either use the former again or we could just simply use the answer we had previously not 0.12 1 5 7 and so on and if we add these two values together what you should find is that you get naught point three nine one seven four and so on which is the same as naught point three nine two to three decimal places 3dp okay let's have a look at another question now in this question in clinical trials a certain drug has an 8 percent success rate of curing a known disease and if 15 people are known to have the disease what is the probability that at least two are cured lissa again is a typical tree diagram we've got a finite number of trials 15 this 2 outcomes in any trial that is that the drug is successful or it fails the probability of success is 8 percent and it remains constant throughout the problem and in any trial the probabilities of success are independent of the previous trial so a typical binomial distribution so first of all to do this let us define a random variable can be any letter remember so now why don't we choose Y okay let Y be the random variable in this particular problem and it will be the number of people cured so we define that and we now need to say where Y is distributed binomially two parameters number of trials first which is fifteen and the probability of success next which is eight percent or not point zero eight so in this particular question we're asked that the probability that at least two are cured so that's the probability that Y is greater than or equal to two now in this particular problem y can be no people being cured or one person being cured 2 people 3 4 etc all the way up to and including 15 so this is the same as the probability that y equals 2 plus the probability that y equals 3 y equals 4 and so on all the way up to the probability that y equals 15 now this would be a horrendous um to work out and there is a way around this problem what we should know is that if we were to work out the probability that y equals not plus the poverty y equals 1 plus the probability y equals 2 3 and so on all the way up to 15 that that would total 1 all the probabilities would total one so since I'm not using y equals 0 and y equals 1 I could take those probabilities away from 1 and it would give me this hour sir so I could do one minus the probability that y equals nought and I'd need to put this in a bracket plus the probability that y equals 1 these are the only other situations that I've not included here so all I need to do now is just work out these two rather than having to do all this slot so b1 minus so for the probability y equals not it would be 15 C not probability of success is not 0.08 and that is occurring no times probability of failure is not 0.92 and if we had no successes there must have been fifteen failures now we work out poverty y equals one so that B plus fifteen C one probability of success not 0.08 would have occurred just once and the probability of failure would have occurred fourteen times so you have naught point nine two to power 14 and if I leave it to you to work this sum out remember I must put that in brackets by the way if I leave it to you to work that some out you should find that you get one - not point six five nine seven and so on and if you take that away from one you end up with naught point three four zero to seven and so on which when rounded to say three decimal places is not point three four zero to three decimal places now that brings us to the end of this tutorial and I hope you've been able to follow these examples and you can apply them in similar questions now in my next tutorial what I want to show you though is how on some occasions we can do questions like this by using commutative binomial tribution tables and that can save us an awful lot of work so I hope you'll have a look at that tutorial you

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Keywords: Binomial distribution, Binomial, distribution, Statistics, examsolutions, exam solutions, A-level, probability, independent trials, trials, binomial distrubution, binominal distribution, binomial distribution calculator, binomial distribution in probability, binomial distribution statistics, binomial distribution examples, binomial distribution formula, binominal distrubution formula

Id: _0tr8Iw0tjA

Length: 11min 34sec (694 seconds)

Published: Mon Nov 02 2009