Probability is the

likelyhood that an

event will occur. Much of the

statistical world (

inferrential statisics) relies heavily on the

notion of probability.

The most logical way of determining the probability of a given event is by finding the number of ways that that event can occur and then dividing that number by the total number of possible outcomes.

Probability of an event is equal to or greater than zero, while being less than or equal to one:

0----------------------------------------------0.5----------------------------------------------1

Impossible______________50/50chance________________Certain

The probabilities of all of the outcomes in any given sample space always adds up to 1. This is because one of the events must occur, so the probalitity of one of them occurring is 1.

Many probability problems involve finding the number of possible outcomes and then using this to determine a probability.

**Example:** Number plates consist of a sequence of three letters and three numbers. The total number of combinations is the number of possibilities for each position on the number plate multiplied by one another: That is,

26*26*26*10*10*10 = 17,576,000

Notice, this is a selection with replacement, because all letters and numbers are available. That is, once a letter/number is used, this does not restrict it from being used again.

Now say that you are asked to guess the number plate of a person chosen at random. You are given 100 guesses.

Assuming that you guess a different number plate each time, the probability is clearly going to be 100/17,576,000. Which is one chance in 175,760.

Probability is normally given as a decimal or percentage. In this case, the probability is 0.00000569 or 0.000569%.

The other common way that probability is determined is through using given probabilities. It must be remembered that the total probability of any number of non-dependent events is equal to the product of the probabilities of each event.

The notation "pr(A)" is used to denote "the probability of event A occurring"

**Example:** It is known that the probability of a head or tail on a single coin toss is 0.5 or 1/2. The probability of any single number coming up on a die throw is 0.167 or 1/6.

Our task is to determine the probability that one throw of a coin and a die will result in:

(i) A head and a 5

(ii) A tail and a 1 or 6

(iii) A head and an even number

(i) The probability of a head is 0.5 and the chance of a 5 is one in six. Therefore, the probability of both of these events occurring is their product: 1/12 or 0.0833

(ii) Pr(tail) is also 0.5. One and six are two possibilities out of six and so their chance of falling is 1/6 + 1/6 = 1/3. Therefore, pr(tail, 1 or 6) is 1/2*1/3 = 1/6. Not surprisingly, twice the probability of the previous event.

(iii) We can conclude that a head and an even number will therefore be three times as likely as the first result, indicating a probability of 1/4. That is, 1/2*1/2 = 1/4.

**FACTORIALS**

Assume that there are five different positions to be filled by five different people. Our task is to determine the number of ways that these five positions can be filled. The first position can be filled by any one of the five people. The second position by the four people remaining and the third position by the three people remaining etc. This shows a pattern for the number of people that can fill each position:

5-4-3-2-1

Hence the number of ways that this can be done is 5*4*3*2*1. This is known as "five factorial" and is denoted by 5!. n! is said to be defined as:

n! = (n)*(n-1)*(n-2)*...*(1), where n is a positive integer and 0! is defined as 1.

So, back to the five positions. The result is 5!, which by simple multiplication is 120 ways.

Factorials are used in situations where there are an equal number of objects for positions and replacement is not allowed. By finding the number of ways that something can be arranged, calculating the probability of one or more of these arangements occurring, becomes easy.

**PERMUTATIONS**

A permutation is an ordered arrangement of objects. Permutations are used to calculate the number of ways that objects can be picked or arranged from a certain number of possibilities and in a certain order. That is to say, if we wish to select r objects from n possiblities, then the number of ways we can do this (remembering order of selection is important) is:

No. of ways = nPr = n!/{(n - r)!}

For example, the number of ways we can arrange 4 books in 6 distinct positions is:

6P4 = 6!/{(6-4)!}

= 720/2

= 360 ways

**COMBINATIONS**

Combinations are similar to permutations, except that order is not important. That is, combinations can be used to determine the number of ways that r objects can be chosen from n objects, where order of selection is irrelevant. This is denoted by nCr.

Combinations are calculated in the following way:

No. of ways = nCr = n!/{r!(n - r)!}

If we take the example of a basketball team being selected. There are five players needed, but eight players are available. The number of ways that the five players can be chosen is 8C5. (Noting, of course, that the order in which the players are selected is unimportant).

8C5 = 8!/{5!(8 - 5)!}

= 40320/{120*6}

= 56 ways

We could also say, for example, that the chance of the best possible team being chosen (if it is to be chosen at random) is one in 56.

Note that there is always a larger number of permutations than combinations for any values of n and r because order is important with the former.

**Final Example**

Seven contracts are available for ten companies. There is no limit on how many contracts each company can get. Find the probability that company A receives two contracts and company B, three contracts.

First, let's work out the number of ways that seven contracts can be distributed. Each contract can be awarded to ten companies, so the no. of ways

=10*10*10*10*10*10*10

=10^7

Now, we must find the number of ways that two particular companies can receive 2 and 3 contracts respectively.

First, company A. The number of ways two contracts can be be awarded is the number of ways to select two from seven or 7C2 (remembering order won't be important). The number of ways that company B can receive the five remaining contracts is 5C3. There are now two contracts left and eight companies to receive them. The number of ways this can happen is 8*8 because any of the eight companies could receive either contract.

So, the total number of ways is:

7C2*5C3*8*8

=13440

Therefore, the probability of this happening is:

13440/10^7

Pr = 0.001344

A very small chance.