Bernoulli, Binomial, and Poisson

This is for my friends who want to learn a little more about statistics and probability. If you see any errors, please feel free to let me know. I will still try to update this even after I post it.

There are quite a few discrete distributions that are commonly used when talking about statistics and probability. I will give a quick overview of some of these distributions:

• Bernoulli/Binomial
• Poisson

Bernoulli/Binomial

A quick glance!

Bernoulli and Binomial are very closely tied. When someone mentions Bernoulli, I think of a Bernoulli Trial. A Bernoulli Trial deals with an event with only two outcomes, a success and a failure, labeled as 1 and 0, respectively. A success has a probability of p, and a failure has a probability of 1-p=q, where 0<p<1 and p+q=1. To demonstrate a quick a example of a Bernoulli Trial, l will use a cup game.

In this game, there are 4 cups where only 1 cup contains a ball. I will shuffle the cups randomly shuffled and have you randomly choose a cup. Thus, the probability of success(or n=1) is

P(n=1) = 1/4 = .25

The expected value of the trial is .25, which is the probability of a success.

The probability of failure(or n=1) is,

P(0) = q = 1-p = .75

Now that a quick example of a Bernoulli Trial is explained, this can be taken a step further. Instead of 1 instance of a game, let there be 5 where the outcomes were [0,0,0,0,1]. The probability of this specific sequence is,

Which is about 7.910%.

This can still be taken a step further and where Binomial Distributions come in handy. The probability of the sequence [0,0,0,0,1] is the same as [1,0,0,0,0] and other sequences where only 1 game is won. In each sequence, there is a probability of

where n is the number of games played, and k is the number of games won. In this example, n=5 and k=1. Now, I need the number of ways 5 games can have 1 win. This is where combinations comes into play. Using combinations, I can count the total number of ways sequences have 1 win out of all the possible outcomes, where order does not matter. Order will not matter since sequences that have 1 victory have the same probability outcome. The formula for combinations is

Bringing these together, I am finding the probability of success of 1 game out of 5 games total. This is the same as a Binomial Distribution. The expected value of a Binomial Distribution is how many games I expect to be won. Thus,

Poisson Distribution

To put it simply, using a Poisson Distribution and process can help find an n number of events withing a given interval. For example, if I were to work at a call center, I can calculate the number of call I expect between 9 A.M to 12 P.M. If I had a rope of 100 ft., I can find the number of imperfections this rope would have.

There are three very important conditions that must be met in a Poisson process:

1. The events are independent of one another
2. The rate at which each event can occur is constant
3. The probability of an event occurring is proportional to the interval.

The quick breakdown is

Lambda is the rate of each occurrence multiplied by the time.

To show how this works, I will again use another example. I am working in the customer service department of my favorite store. I usually have a call from 3 customers in 5 minutes. I have a lunch break in 15 minutes and I don’t want to deal with more than 10 customers. I want to calculate the probability that I will have 10 or more customers.

I will check the conditions first. Each call is independent from one another, I am not having a customer call based on another customer. I established that 3 customers call in 5 minutes. If I have 15 minutes, the number of calls I get is proportional to this time. That means I should expect about 9 calls, and so on for every five minutes I add.

Now that the conditions are met, I will continue with the probability.

My lambda will be 9=(3/5)(15). This will stay constant. Now, since I want 10 or more calls, I want P(X≥10)=1-P(X≤9). This is because I can calculate the probability for each X and sum the results for X between 0 and 9. If I did it starting from 10, the calculation would be a lot harder. Thus,

Which comes out to about 71.49%. Looks like my lunch will have to wait. :(