Let p0, p1, …, pn denote the probability mass function of the binomial distribution with paramet

Let p0, p1, …, pn denote the probability mass function of...

Key Concepts

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) accumulates probability values up to a certain value, representing the probability that the random variable takes a value less than or equal to that point. For the binomial distribution, the CDF is computed by summing the individual probabilities (obtained either directly or via a recurrence relation) for each count of successes up to the specified threshold.

Computational Efficiency

In practical applications, especially when parameters are large, directly computing binomial probabilities using factorials and binomial coefficients can be computationally intensive. The recursive method reduces the number of operations by building upon previously computed values, thereby improving numerical stability and efficiency. This approach allows for quick and accurate calculation of cumulative probabilities without resorting to direct combinatorial evaluation.

Binomial Distribution

This concept refers to a discrete probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success. It is defined by its parameters n (the number of trials) and p (the probability of success on each trial), with q = 1 - p representing the probability of failure. The probability mass function is given by the formula involving binomial coefficients, which accounts for the different ways successes can be arranged.

Recurrence Relation

The recurrence relation is a method of computing successive probabilities in a distribution using a previous value rather than recalculating from the definition every time. In the context of the binomial distribution, the relation derives p_k in terms of p_(k-1) by leveraging properties of the binomial coefficients. This relation simplifies the computation and is especially useful when evaluating probabilities for large values of n.

Transcript

00:01 So we have a situation in which n is 9 ,000 and the p value is very small, .0005, and then this will be q, will be equal to that .9995.

00:14 And we are finding the probability that x is less than or equal to 4 for this value.

00:22 And we know that that means we want to find the probability that x is equal to 0, plus the probability that x is equal to 1, all the way up to the probability that x is equal to 4.

00:36 And we want to show that that can be done recursively.

00:39 So we also know that p sub 0, well let me write down the rule, our recursive rule is p sub 0 is equal to q to the n, which for us means that that would be the .9995 to the power of 9 ,000.

01:01 And we would want to find that p sub k is equal to the n minus k, which would be the 9 ,000 minus k plus 1 times p over k times q.

01:19 And so we would start with p sub 0, and let me move over here.

01:25 So we would have p sub 0, well our first probability is to find the probability of p sub 0, so that's going to be that .9995 to the 9 ,000 power, plus, now we define p, the probability of one success.

01:47 So this is the probability of having no successes, one success, two successes, and so on.

01:52 So let's put these values in.

01:54 So we have to begin with the 9 ,000 minus 1 plus 1 times p, and i'm just going to leave it as p, but that p is equal to .0005.

02:09 And then we're going to have the k, which is 1 times q, which is that .995.

02:15 And then we're going to multiply that by that value of .9995 to the 9 ,000.

02:24 So now look what happens.

02:26 This q, this is q, is going to knock down to 8 ,999, and we're going to have this value plus, we're going to have 9 ,000 times, and then we're going to have p to the first power, which is that .0005, and then times, actually why don't i write it that way.

02:52 Trying to erase .0005 to the first power times .9995 to the 8 ,999.

03:09 Now that's the same thing as we would find using the regular formula.

03:13 This is that combination of 9 ,000...

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