Continuing our discussion of probability distributions, now I want to look at the binomial distribution. The binomial distribution is used when all of the possible outcomes are discreet and have only one of two values, either success or failure. Then the probability of x successes in n trials is given by P of x where P of x is given by this expression. N factorial divided by x factorial, n minus x factorial, p raised to the x power, q raised to the m minus x power, as given in the extract from the reference handbook. And in this equation p is the probability of success in an individual trial. And q is the probability of failure in an individual trial. Which is of course equal to 1 minus p. So, let's illustrate that by means of an example. We have a bag that contains four red balls and three black balls. A ball is withdrawn and then replaced. This is repeated seven times. What is the probability that two out of the seven withdrawn balls will be red? Which of these possible alternatives? So in this case, the balls can only take on one of two colors. They're either red or black, no other color. So we'll define then, the probability of p, success, which we'll call a red ball, is equal to, well there are four red balls and seven total. So, the probability of drawing a red ball is 4 in 7, or 4 over 7, which we will call success. The probability of drawing a black ball, which we'll call a fail, is equal to 3 over 7 because there are three black balls out of a total of seven. So, we have seven trials, n equals 7, we're going to do this seven times. And we're looking for x equals 2 successes, or red balls, out of that. So, the probability then, is given by the binomial distribution as shown here. And in this case then, we have n is equal to 7, x is equal to 2, because we want two of the balls to be withdrawn, divided by 7 minus 2 factorial, 4 over 7 factorial, the probability of 4 over 7, probability of success, raised to the 2 or squared. And 3 over 7, the probability of failure, raised to the 7 minus 2. So computing that number, we find that it's equal to.099 or rounding it off is equal to .10 and the correct answer is C. So the probability is .1, or 10%. Now, we also input the cumulative binomial distribution which is the probability that at most, x samples fail out of a total of n samples, given that the probability of a single failure in a trial is p. And the table is given here. Now, the use of this is most easily shown by an example which I'll give in a minute. And let's do an example then that the binding on 20% of textbooks fail. The probability that, at most, two textbooks out of a random selection of six fail is this. Which of these alternatives? So this is the use of the cumulative distribution. And in this case, we have p, the probability of failing in an individual trial, is equal to 0.2 or 20%. The number of samples that we want to use, or the number of successes, x is equal to 2, two textbooks out of a random selection of n equals 6. So in this case we have, we can just look in the table and we find that the probability that x is less than or equal to two is equal to .9011. So if I back up a minute to the table here, we see that m is equal to 6, x is equal to 2, and .2 here. Therefore, the value that we're looking for is .9011, which is there. So the answer then, to round it off, is D. Now let's look and see how this arose. So, in this case, we're looking for the probability that at most, two textbooks fail. And so the keyword here is at most. Therefore, if this is going to be the cumulative probability that either zero textbooks fail, or one textbook fails, or two textbooks fail. And the generalized probability that exactly n fail is given by, again, this equation here. So we go through this in sequence then. First of all, the probability that zero textbooks fail is given by 6 factorial over 0 factorial, 6 factorial, .2 to the 0, .8 to the 6 power. And we can evaluate that because zero factorial is equal to 1 and .2 raised to the 0, any number raised to the 0 power is 1. So we can compute that out and the answer is that the probability is .2621. Next we get the probability that exactly one textbook fails. So here the corresponding equation is now 6 factorial divided by, now x is 1, and in minus x is 5, so computing that we get 0.3932 is the probability that one fails. And finally, the probability that exactly two books fail is given by this equation, where x is now 2, and n minus x is 4. So computing that, we get 0.2458. So the cumulative probability, the sum of the probability that either zero, or one, or two fails, in other words the probability at most two is the sum of that, which is equal to .9011. Which is, as you can see, exactly the answer that we get from the table. So this is why the cumulative table is useful, because to compute this for a large number of variables would be unreasonable, prohibitively expensive to do. So this is the value of the cumulative table and this is why it's called the cumulative table. So, this completes my discussion of the binomial distribution.
