2. Introduction into the Mathematical Methods

The binomial distribution (2/2)

Rules 3 and 4 may appear to be identical, and they are for situation involving the toss of a coin or throwing a dice, but they are not necessarily identical with many real world situations. Consider the question, "If I go to the same location every day for two weeks to catch fish, what is the probability that I will catch fish every day?" You may consider that Property Four is met in this experiment, but you may also decide that Property Three is not met, since the probability of catching a fish on any individual day will depend on the tide, your level of concentration and various other factors that may mean that the probability of success on any one day is different to the probability of success on any other day.

Now if you conduct n trials and you want to know how many combinations of x results can you get from the n trials, then the answer is

(\begin{matrix} n \\ x \end{matrix})

where:

(\begin{matrix} n \\ x \end{matrix}) = \frac{n!}{(n - x)! \cdot x!} = \frac{n \cdot (n - 1) \cdot (n - 2) \cdot \dots}{[(n - x) \cdot (n - x - 1) \cdot \dots] [x \cdot (x - 1) \cdot \dots]}

The factorail ! symbol ↓

This equation for the combinations of x events from n trials can be readily established. If you have three trials, how many combinations of 2 events can you get? You can have trials (1 and 2), trials (1 and 3), and trials (2 and 3), so the answer is 3. You can easily show that

(\begin{matrix} 3 \\ 2 \end{matrix}) = 3

If you now consider 4 trials, from which you want the number of combinations of two trials, then these can be trials (1 and 2), (1 and 3), (1 and 4), (2 and 3), (2 and 4), and (3 and 4), or six in total and again you can readily show that:

(\begin{matrix} 4 \\ 2 \end{matrix}) = 6

So, what this tells us is that if there are n trials then there are

(\begin{matrix} n \\ x \end{matrix})

combinations of these n trials that will contain x successes. But if we are to determine the probability of x successes in n trials, then we need to also know the probability of success in each individual trial. Let us say that the probability of success in one trial is p. The trials are independent but achieving the goal of x successes depends on the prior results, so the probabilities can be multiplied together. Thus with three trials, where you want two successes, then the probability is p·p·(1-p), which needs then to be multiplied by the number of combinations that can have this number of successes, or

(\begin{matrix} 3 \\ 2 \end{matrix})

The general form of the Binomial equation can now be stated:

f (x) = (\begin{matrix} n \\ x \end{matrix}) p^{x} {(1 - p)}^{(n - x)}

The mean and variance of a Binomial Distribution can also be derived from the general equations for the mean and variance to give:

\begin{array}{l} E (x) = μ = n p \\ V a r (x) = σ^{2} = n p (1 - p) \end{array}

Questions:

What is the theoretical PDF for selecting a card at random of a particular number or rank (such as an Ace, 2, 3, …, Jack, etc) from a pack of 52 cards?
Shuffle a pack of 52 cards and then select a single card. Repeat this process 30 times so as to derive a sample PDF to address the question.
If you shuffle the pack once and then select 30 cards at random from the pack, how does this PDF vary from the PDF derived in Question 1?

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Classification Algorithms and Methods

2. Introduction into the Mathematical Methods

The binomial distribution (2/2)

Questions: