4. Probability and the Binomial Distributions

4.1 Probability

The basic definition of probability is a ratio of things you can count (a ratio of their frequencies) :

(4.1)   \begin{equation*}  P(E) = \frac{n(E)}{n(S)} \end{equation*}

where

P(E) is the probability that event E happens,
n(E) is the number of ways E can happen and
n(S) is the total number of outcomes (all possibilities).

Example 4.1 : What is the probability of drawing a queen from a deck of cards :

    \[P(E) = \frac{4}{52} = 0.077 \mbox{\ \ \ (7.7\% if we were to express the result in percentages)}\]

To use P(E) mathematically we set

    \[0 \leq P(E) \leq 1 \]

Where, probability-wise:

0 means E definitely will not occur, and
1 means E definitely will occur.

This is a method we can use instead of using percent. To compute probabilities, we first need to know how to count.

Fundamental Counting Rule 

Say you have n events in order, and for event i there are k_{i} ways for it to happen. Then the number of ways for the n events to play out is :

    \[ k_1 \cdot k_2 \cdot k_3 \hdots k_n = \prod_{i=1}^{n} k_i \]

(The giant pi symbolizes a multiplication convention in the same way that a giant sigma symbolizes a summation convention as described in Section 1.3.)

Example 4.2 How many combinations are there on a lock with 3 numbers?

Lay out the events as : k_{1}=10, k_{2}=10, and k_{3}=10. Note that each number can be anything from 0 to 9 giving 10 possibilities (k_{i} = 10) for each event. So the number of possible lock combinations is

    \[ k_1 k_2 k_3 = 10 \cdot 10 \cdot 10 = 10^3 = 1000 \]

Note that you could have guessed this because the combination range from 000 to 999 — counting in base 10.

Example 4.3 Suppose that a hardware store can produce paints with the following qualities :

Colour : red, blue, white, black, green, brown, yellow (7 colours)

Type : latex, oil (2 types)

Texture : flat, semigloss, high-gloss (3 textures)

Use : indoor, outdoor (2 uses)

How many ways are there to combine these qualities to produce a can of paint?

Answer : From the above list k_{1}=7, k_{2}=2, k_{3}=3, k_{4}=2 and the number of possible paint kinds is:

    \[ 7 \cdot 2 \cdot 3 \cdot 2 = 84 \]

Applications of the Fundamental Counting Rule

We are interested in applying the fundamental counting rule to two special, important cases :

  1. Permutations.
  2. Combinations.

Let’s define each one.

  1. Permutations.

The number of ways, or permutations, of selecting r objects from a collection or n objects, while keeping track of the order of selection is [1]

    \[ {}_{n}P_{r} = \frac{n!}{(n-r)!} \]

This formula follows from the fundamental counting rule. With n objects there are k_{1} = n ways to select the first object. After selecting the first object there are n-1 ways to choose the second object so k_{2} = n-1, etc. up to k_{r} = n - r + 1 :

    \[ {}_nP_r = (n)(n-1)(n-2) \hdots (n-r+1)\]

    \[ = \frac{(n)(n-1) \hdots (2)(1)}{(n-r)(n-r-1) \hdots (2)(1)} \]

Example 4.4 : How many ways are there to choose 5 numbered balls from a bucket of 25 to make a lottery number?

Answer : 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 = 6,375,600 possibilities.

      2. Combinations.

The number of ways of selecting x objects from a collection of n objects without caring about the order is :

    \[ {}_nC_x = \frac{n!}{(n-x)!x!} = \frac{{}_nP_x}{x!} = \left( \begin{array}{c} n \\ x \end{array}\right) \]

That last symbol \left( \begin{array}{c} n \\ x \end{array}\right) is colloquially called “n choose x”. The second last expression demonstrates the application of the fundamental counting principal, it says

    \[ \left( \begin{array}{c} n \\ x \end{array}\right) = \frac{(n)(n-1) \hdots (n-x+1)}{x!} \]

where x! is just the number of ways of arranging x objects while caring about the order, x! = {}_{x}P_{x}.

As a practical matter, never try to compute n! It will usually be unimaginably big. Use the formula that directly shows the fundamental counting rule as shown in the following example.

Example 4.5 : How many ways are there to select 10 balls from a bucket of 100?

Answer :

    \[ \left( \begin{array}{c} 100 \\ 10 \end{array} \right) = \frac{100 \cdot 99 \cdot 98 \cdot 97 \cdot 96 \cdot 95 \cdot 94 \cdot 93 \cdot 92 \cdot 91} {10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\ = \frac{6.2815651 \times 10^{19}}{3,628,800} = \underline{17.3 \times 10^{12}} \]

The symbol \left( \begin{array}{c} n \\ x \end{array}\right) is also known as the binomial coefficient because it shows up in algebra when you expand expressions of the form (x+y)^{n}. For example[2]

    \[ (x+y)^n = x^2 + 2xy + y^2 \]

    \begin{eqnarray*} (x+y)^3 &=& \left( \begin{array}{c} 3\\0 \end{array} \right)x^3 + \left( \begin{array}{c} 3\\1 \end{array} \right)x^2y + \left( \begin{array}{c} 3\\2 \end{array} \right) xy^2 + \left( \begin{array}{c} 3\\3 \end{array} \right) y^3 \\ &=& x^3 + 3x^2y + 3xy^2 + y^3 \end{eqnarray*}

The binomial coefficients can be quickly computed using Pascal’s triangle :

    \[ \begin{array}{ccccccccccccccc} &&&&&&&&&&&&&& n = \\ &&&&&& 1 &&&&&&&& 0 \\ &&&&& 1 && 1 &&&&&&& 1 \\ &&&& 1 && 2 && 1 &&&&&& 2 \\ &&& 1 && 3 && 3 && 1 &&&&& 3 \\ && 1 && 4 && 6 && 4 && 1 &&&& 4 \\ & 1 && 5 && 10 && 10 && 5 && 1 &&& 5 \\ 1 && 6 && 15 && 20 && 15 && 6 && 1 && 6 \\ &&&&&& \mbox{etc.} \end{array} \]

Referring to Pascal’s triangle we can quickly write

    \[ (x+y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6 \]

for example.


  1. Recall that the definition of factorial follows 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 etc.
  2. You don't need this algebra for this statistics course. It's just interesting.

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