4.1: Probability

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

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

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 :

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

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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.

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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:

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

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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.

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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

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{3,628,800} = \underline{17.3 \times 10^{12}} \]

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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.

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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. ↵