Probability
Introduction
The probability of an event occurring is important in many areas of life, and many examples are in the press much of the time. We may have concerns over having the combined measles-mumps-rubella vaccine administered to our children, we may want to know the likelihood of genetically modified crops affecting natural wildlife, we may be in fear of crime and many more.
We may not use mathematics to help us make decisions, we may prefer to go on a 'gut feeling' as to the right decision. We may be influenced by the arguments that one side or another is making, based on moral or ethical arguments that are only loosely based on statistics and probabilities.
The aim of the Mathagony Aunt statistics pages is to enable you to become more confident in analysing arguments that are put forward by others, so you can make informed choices. You will not be alone, credit agencies, banks, insurance companies all use statistics and probability to make decisions about us, now is our turn.
Probability: the basics
The aim of this section is to lead you in to the idea that we can make informed choices about the outcome of an event - i.e. how probable is such an outcome. Investigating your extra-sensory perception (ESP) using our online experiment will help you understand the following discussion on probability. If you haven't yet seen the experiment here is a brief description: The computer chooses one card out of five possible cards, and displays this card on the screen. You cannot see which card it chose because it is covered, but you have to decide which card was chosen. When you have made your choice the computer reveals which card it originally picked. The computer also shows the probability of obtaining your score..
Reporting probability
You may see probabilities given as percentages. However, probabilities should be given as a number between zero and one, inclusive. If an event is definitely going to occur, the probability is given as 1. If it definitely not going to occur, its probability is given as zero. Most events, however, fall in between these two extremes.
Question:
What is the probability of choosing the same card as the computer from
the five possible cards?
Answer:
There are five possible outcomes: because there are only five cards
to choose from. Only one choice is right - or correct - (the other four
choices will be wrong). Write the probability of a particular result as
a fraction of that result over all possible outcomes:
Not surprisingly, the probability of getting a wrong card is higher as there are four times as many wrong cards as right cards:
It all adds up
After
a choice has been made, such as that above, the sum of the probability of
choosing the right answer (i.e. matching card) and the probability of choosing
the wrong answer (i.e. non-matching card) will always be ONE: 
+ 
= 
= 1.
Tree diagrams
One
way to represent choices for probability is the tree diagram. We could show
the above selection of the cards as
We
can see that there is a one in five chance that any card could be chosen.
This is shown as a probability of
for each card. Now suppose that the top card - the star - is the card that
the computer had chosen. In this case the star card is the matching card
or right card. We could display all the outcomes for all the cards, as
above, or we could group together all the cards that were not chosen as
This
grouping of the four lower, wrong, cards makes it easier to see the probability
as the fraction .
The next step is to remove the pictures, because it doesn't matter what
the right or wrong cards looked like, just that they were right or wrong.
In any case, the right card can change from one round to the next and having
to draw all the cards each time would soon become very tedious. The next
diagram shows the same round, but without any cards (the right card branch
is black/red and the wrong card branch is black/green):
If we choose the right card in the first round, in the second round we look at the branch that has the correct card (the top branch, which is coloured black/red):
Again, , with the same probabilities (in each round) as in the first round.
The alternative picture is if we had chosen the wrong card in the first round, then in the second round we follow the lower branch. Again, in the second round we either choose the right card or the wrong card (with the same probabilities as in the first round).
These two situations are put together in a single tree diagram:
Working out the probability
Now we have the complete picture, we can use the probabilities of each round to work out the probability of getting a particular result. It may help, first, to imagine a tree. You see a tree, with a large trunk topped by a canopy of branches, twigs (which are just small branches) and green leaves. At first, you climb the trunk into the canopy. There the trunk splits into a small number of large branches. You continue to climb, and follow one of these branches. This branch again splits into smaller branches and these, too, split into ever-smaller branches. Eventually, after climbing through a maze of branches and smaller branches you get to the leaves. With this picture in mind, look of the tree diagram. It's like the canopy of a tree turned on its side. The single point at the very left of the diagram is like the point at which the trunk splits into two or more branches. In this case the trunk splits into two branches. As we move, in our diagram, to the right the canopy widens, just as the canopy widens as we move up the tree. When we get to the end of all the branches of the imaginary tree we get to the leaves; in our tree diagram we get to the probability of taking the path we took.
In imagining the tree, you will see that the number of branches multiply as you go down a particular path. Say that each time a branch splits it splits in two. After splitting three times there will be 2 × 2 × 2 branches. We can see that the number of branches multiplies as we move along a particular path. Similarly, in our tree diagram the probability we get a particular result (i.e. go along a particular path), we multiply the probabilities in each round.
When we want to find the number of leaves on a tree we count them up. Similarly, if you want to find the probability of two or more branches, you add them up.
The following diagram gives the probability of getting the possible results:
The tree diagram tells us the different combinations of events that we could have after two goes with the Zenner cards, with the theoretical probability value of the events. The probabilities of these events can also be written in shorthand as P[event] = value. Here the capital P means the probability that the event shown in brackets occurs; the value is the probability of the event. For two rounds the possible events are
P[right
card followed by right card] = 
P[right
card followed by wrong card] = 
P[wrong
card followed by right card] =
P[wrong
card followed by wrong card] = 
Using
these values we can work out the probability of choosing the same card as
the computer, twice, only once or not at all. To get both cards right we
have to choose the right card in the first round and the right card in the
second round. This gives a probability of .
To get no matching cards we have to get the wrong card followed by the wrong
card, which gives a probability of .
Choosing the same card as the computer only once is a little bit more complex.
Getting one matching card
There are two ways to get only one matching card correct; these are highlighted in red in the following tree diagram, which is just a simplified version of the tree diagram given above.
In this tree diagram the routes you could take to get just one right card over two rounds are shown. The probabilities for these routes are
P[right
card followed by wrong card] =
P[wrong
card followed by right card] =
We need to know the probability of going down either branch; i.e. how many branches are there?. So we count up the branches and add their probabilities together:
P[only one right card] = P[right card followed by wrong card] + P[wrong card followed by right card]
P[only one right card] =
+
= 
Algebraically
Another way to look at the probabilities of these events is to use algebra. If you are asking why we would want to use algebra, there are some very good reasons.
Why use algebra?
There are two good reasons to use algebra. The first is because it cuts down on the amount of writing you need to do. The second is that it will enable to follow more advanced probability. Just think how complex a tree diagram would get after just five or six rounds, and Zenner card experiments are usually over 25 rounds!
When using algebra we first have to define our variables; i.e. what stands for what. This lets us remeber what is what and stops us getting confused, even though we only have two variables. In this case
P[right card] = p
P[wrong card] = q
In
this case p =
and q = ,
and these values are constant from one round to the next.
| Number correct |
Formula |
Probability |
| 2 |
p × p = p2 |
|
| 1 |
(p × q) + ( p × q) = 2(p × q) |
|
| 0 |
q × q = q2 |
|
=
In the above table the wrong (no cards match) path has only q's and the right path has only p's. In both cases they are raised to a power equal to the number of rounds.
In the intermediate case (one matching card and one non-matching card) the formula is in terms of p and q and each variable is raised to a power equal to the number of times they match or don't match. This may not be clear as, for example, p1 is usually written (as it is here) simply as p. There are, as we have seen from the tree diagram, two paths that we can follow to get only one card right. As before, we add the probabilies from each of the two routes together to get the total probability of getting only one card right.
Following on from this example, consider the probability of choosing four right cards over ten rounds. To find all the probabilities we will need to multiply a variety of p's and q's together. But in case there would be a string of ten values multiplied together, each one representing the outcome of a single round. These would range from
P[10 cards right] = p × p × p × p × p × p × p × p × p × p = p10
to
P[10 cards wrong] = q × q × q × q × q × q × q × q × q × q = q10
with a whole range of mixed values (such as the P[only one card right] in the two-round example above), including
P[4 cards right and then 6 cards wrong] = p × p × p × p × q × q × q × q × q × q = p4q6
However, this shows getting four right and then six wrong. There are other ways to get four right and six wrong, just as there are two ways to get only one card correct in two rounds. For example, we could have got one wrong, then four right and then the remaining cards wrong (note the first q, marked in bold):
P[1 card wrong, 4 cards right and then 5 cards wrong] = q × p × p × p × p × q × q × q × q × q = p4q6
The value of getting four right in this way is also p4q6, but it is just another way of getting four cards right. There are many ways to get four right and six wrong. Each way also has a probability of p4q6. To work all the possible branches by drawing a tree diagram is possible, but it would take a very long time. Therefore we need a more efficient method: the binomial theorem.
The binomial theorem has two parts, the first part tells us how many ways there are of getting a particular result, such as how many routes can you go down on a tree diagram to get four cards right and the other six wrong. The second part tells us the probability of going down any of these routes, in this case each route has a probability of p4q6. We have shown here how to work out the probability of each branch. To work out how many branches there are we need to work out the number of combinations of choosing four cards right over ten rounds. This is written in shorthand as nCr (in this case as 10C4). The value of 10C4 is 210. So there are 210 ways of getting four cards correct over ten rounds, each one with a probability of p4q6. Thus the probability of getting four cards right over ten rounds is
P[4 cards right and 6 cards wrong] =10C4 p4q6
= 210 × p4q6
= 210 ×()4
× ()6
= 0.088
To see how to calculate the probabilities of other scores are calculated have a look at the binomial theorem.