We all know that the trials to the first success is a Geometric distribution. It can take one trial, two trials, three trials, etc., to see the first success. These trials are assumed a random variable X = {1, 2, 3, … }; they have a probability, i.e., P(X = 1), P(X = 2), P(X = 3), and so on.

However, Mr. Gardner needs more success. He has already sold his first “time machine” (aka bone density scanner). He’d have to sell more. He is looking for his second success, third success and so forth to get through.

The number of trials it takes to see the second success is Negative Binomial distribution.

The number of trials it takes to see the third success is Negative Binomial distribution.

The number of trials it takes to see the ‘r’th success is Negative Binomial distribution.

Are you paying attention to the pattern here? Negative Binomial distribution is Geometric distribution if r is 1 (trials to first success).

The Geometric distribution has one control parameter, p, the probability of success.

Since we are interested in more than the first success, r is another parameter in the Negative Binomial distribution. Together, p and r determine how the distribution looks.

Let’s take the example of Mr. Gardner. “I’d have to sell one more.” Each of his visit to a doctor’s office is a trial. They will buy the bone density scanner or show him the door. So Mr. Gardner is working with a probability of success, p. Let’s say that p is 0.25, i.e., there is a 25% chance that he will succeed in selling it.

Let’s assume that he sold one machine. He is looking for his second sale. r = 2.

His second success can occur in the second trial, the third trial, the fourth trial, etc. When r = 2, X, the random variable of the number of trials will be X = {2, 3, 4, …}.

When r = 3, X = {3, 4, 5, …}.

You correctly guessed my next line. When r = 1, i.e., for Geometric distribution, X = {1, 2, 3, …}.

There is a pattern. We are learning a distribution which is an extension of Geometric distribution.

Now let us compute the probability that X can take any integer value.

P(X = 2) is the probability that he makes his second sale (second success) on the second trial.

Remember the trials are independent. The second doctor’s decision is not dependent on what happened before. He buys or not with a 0.25 probability.

So, P(X = 2) is 0.25*0.25 = 0.0625. The first trial is a success and the second trial is a success.

P(X = 3) is the probability that he makes his second sale on the third trial. It means he must have made his first sale in either the first trial or the second trial, and then he makes his second sale on the third trial.

The probability of succeeding second time on the third trial is the probability of succeeding once in two trials, and the third trial is a success.

P(1 success in 2 trials) * P(3rd is a success)

The probability of the one success in two trials is computed using the Binomial distribution.

2C1*p^1*(1-p)^2-1

This probability is multiplied by p, the probability of success in the third trial.

P(X = 3) = 2C1*p^1*(1-p)^2-1*p

If this is your expression now, 😕 let’s take another case to clear up the concept.

Suppose we want P(X = 6), the probability of making the second sale on the sixth trial.

This will happen in the following way.

Mr. Gardner has to sell one machine in five trials. P(1 in 5), one success in five trials → Binomial.

5C1*p^1(1-p)^(5-1)

Then, the sixth trial is a sell. So we multiply the above binomial probability with p, the hit in the sixth.

You should have noticed the origin of the name → Negative Binomial.

To generalize,

I computed these probabilities for X ranging from 2 to 50. Here is the probability distribution. Remember r = 2 and p = 0.25.

Now I change the value for r to 3 and 5 to see how the Negative Binomial distribution looks. r = 3 means Mr. Gardner will sell his third machine.

r = 3

r = 5

Notice how the probability distribution shifts with increasing values of r.

The control parameters are r and p.

You can try different values of p and see what happens. For a fixed value of p and changing values of r, the tail is getting bigger and bigger. What does it mean regarding the number trials and their probability for Mr. Gardner?

Think about this. If he sets a target of selling three machines in the day, what is the probability that he can achieve his goal within 20 doctor visits?

What if he needs to sell five machines within this 20 visits to make his ends meet. Can he make it?

If you try out the probability distribution plots for p = 0.5, you will see that his chance of selling the fifth machine within 20 visits goes up tremendously. So perhaps he should learn the six principles of influence and persuasion to increase p, the probability of saying yes by the doctors.

People mostly prefer to say yes to the request of someone they know and like. I want our blog to have the 9000th user within nine months. See, I am requesting in the language of Negative Binomial distribution.

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