All probabilities lie in a range from impossible to certain. "Impossible" is represented by the value 0, and "certain" is represented by the value 1, with other degrees of certainty represented by values in between. Probabilities can be given as fractions, or decimals or percentages:
Exercise 1
For each of the following situations, decide which of the following values is closest to the correct probability for that situation:
0
0.2
0.5
0.8
1
The sun will rise tomorrow.
A randomly-chosen whole number in the range 0 to 100 turns out to be even.
A toss of a fair coin will come up tails.
My birthday in August will be a sunny day.
A 6-chamber revolver has a bullet in only one chamber. When fired, it shoots the bullet.
A fair die is tossed and comes up with the value 7.
There are three different types of probability:
Theoretical probability can be worked out using nothing more than theory and maths. For instance, we can say that if you toss a fair die (singular of "dice") then the probability of getting a "four" is 1/6 or 16.66%. We know this because there are six numbers, all equally likely, and "four" is just one of them, so 1 in 6. We know this using theory. We don't need to toss a die many times to see if we are right.
Experimental probability can be worked out by examining a large number of data points to see which ones match with what we want and which ones don't. For example, if you wanted to know how many people in a certain town were left-handed, you would speak to as many people as you could in that town and find out how many were left-handed. If you found, for example, that out of 2,700 people, and found that 243 were left-handed, then the probability of choosing a person at random in the town and getting a left-hander would be 243 / 2,700, which cancels down to 9 / 100 = 0.09 = 9%.
Intuitive probability. If you go to the doctor's with the symptoms of flu, he will tell you that there is a very high chance that you have flu. He doesn't need to examine the statistics or work out mathematical equations; he just knows from his vast experience. Similarly, astronomers tell us that there is a very high chance that there is intelligent life somewhere apart from Earth in the galaxy. This is based on their experience as well. Such "guesses" at probabilities are based on people's intuition rather than on anything more solid.
Experimental probability is always found by counting samples and forming a fraction. For instance, if you counted the number of M&Ms in a packet and found that out of the 35 M&Ms, five of them were yellow, then the probability that someone who picked an M&M at random from the packet would end up with a yellow one would be 5 out of 35, which cancels down to 1 / 7. Here is the rule for calculating the probability:
Probability of getting outcome X =
Number of things sampled that include X Total number of things sampled
Exercise 2
The following table shows details of the children in a particular class:
Boys
Girls
Children who wear glasses
6
6
Children who don't wear glasses
12
8
A child in the class is chosen at random. Calculate the following probabilities, giving your answers as decimals:
The child is a boy:
The child doesn't wear glasses:
The child is a girl with glasses:
A box contains party balloons ready to be inflated. Here is the distribution of colours of the balloons:
Colour
Red
Green
Yellow
Blue
Number
58
70
42
24
A balloon is chosen at random. Calculate the following probabilities, giving your answers as decimals:
The balloon is green:
The balloon is not blue:
The balloon is either red or yellow:
A balloon is chosen at random which isn't red.
Calculate the probability that the balloon is yellow, giving your answer to 2 decimal places:
I deliver two different types of newspaper on my paper round, the Arrow and the Mercury. 24 houses get the Arrow only, 38 houses get just the Mercury, and 18 houses get both. One morning, I choose a house randomly to knock over their garden gnomes. Calculate the probability that:
The house gets both newspapers:
The house doesn't get the Arrow:
You should be familiar with the following notation:
p(A)
The probability that outcome A happens.
p(not A)or p(A)
The probability that outcome Adoesn't happen.
p(A and B) or p(A ∩ B)
The probability that bothA and B happen.
p(A or B) or p(A ∪ B)
The probability that eitherA or B happens (or both).
p(A | B)
The probability that A happens assuming that (or "given that") B has already happened.
Exercise 3
There are 10,000 people in my local town, of whom 27 are police officers. At this precise moment, there are 20 people sitting in police cars, 14 of which are police officers. I choose a person in the town at random. Let M be The person is a police officer, and N be The person is sitting in a police car. Calculate the following, giving your answers as percentages (to 2 decimal places, if necessary):
p(M ∩ N) :
%
p(N | M) :
%
p(M | N) :
%
A survey was carried out of households that had no more than 2 children. The results are recorded in this table:
No children
172
One boy only
82
One girl only
43
One boy with a younger brother
57
One boy with a younger sister
59
One girl with a younger brother
66
One girl with a younger sister
21
A household is chosen at random. P = The household has only one child. Q = The household has two children, the younger of which is a girl. R = The household has two children, the older of which is a boy.
Calculate the following probabilities as exact percentages:
p(Q ∪ P) :
%
p(R | Q) :
%
The three laws of probability
Law 1
You get the probability of bothA and B happening by multiplying the probability of A by the probability that B happens assuming that A has already happened:
p(A and B) = p(A) . p(B given A)
p(A ∩ B) = p(A) . p(B | A)
Since there's no reason why A and B should be treated differently, you can write this equation with the letters reversed:
p(A and B) = p(B and A) = p(B) . p(A given B)
p(A ∩ B) = p(B ∩ A) = p(B) . p(A | B)
Two outcomes are said to be independent if one of them happening does not affect the probability of the other happening in any way. For example, if I toss a 10 pence piece and a 50 pence piece, then the outcome of one coin toss has no effect on the outcome of the other. If A and B are independent, then you get the probability of them both happening simply by multiplying the probabilities of the individual outcomes:
p(A and B) = p(A) . p(B) for independent outcomes only.
Law 2
You get the probability of eitherAorB (or both) happening, by adding the probabilities of the individual outcomes, then subtracting the probability that they both happen:
p(A or B) = p(A) + p(B) - p(both A and B)
p(A ∪ B) = p(A) + p(B) - p(A ∩ B)
Just as with the first law, there's no reason why you can't write the letters the other way round:
p(A or B) = p(B or A) = p(B) + p(A) - p(both B and A)
p(A ∪ B) = p(B ∪ A) = p(B) + p(A) - p(B ∩ A)
Two outcomes are said to be mutually exclusive if one outcome happening prevents the other one happening. For example, in the previous exercise, there was a question about the number of children in households. When a household was picked at random, two possible outcomes were that the household had no children, or that the household had one child. These two outcomes are clearly mutually exclusive, as they both can't take place at the same time.
For mutually exclusive outcomes, the probability of both outcomes drops to zero, so the formula can be simplified:
p(A and B) = 0 for mutually exclusive A and B. p(A or B) = p(A) + p(B) for mutually exclusive A and B.
Law 3
To get the probability of something (an outcome or combination of outcomes) not happening, subtract the probability that it does happen from 1 (100%):
p(not A) = 1 - p(A)
Exercise 4
A roulette wheel has 37 slots on it, numbered 0 to 36. The even numbers from 2 upwards are coloured "red", the odd numbers are coloured "black", with the number 0 coloured "white". The ball is rolled on the wheel and comes to rest on a slot randomly. Calculate the following probabilities as fractions in their lowest terms:
The slot will be a red one which is a mutiple of 5:
The slot won't be a black one or less than 20:
The slot won't be coloured black given that it is not coloured white:
I have to go to work today either by car or by bike. If it's raining, there is an 80% chance that I'll go by car. If it's not raining, there's a 35% chance I'll go by car. The weather forecast tells me that there is a 40% chance of rain. Calculate the following probabilities as percentages:
It will be raining and I go by bike:
%
It won't be raining and I go by bike:
%
I go by bike regardless of the weather:
%
I go by car regardless of the weather:
%
There is a flu epidemic going on in our local town, and a large number of people are turning up at the doctor's surgery with flu. The chance of any particular patient having flu is 1/4. Calculate the following probabilities as fractions in their lowest terms:
The next two patients the doctor sees will both have flu:
Only one of the next two patients the doctor sees will have flu:
In the last question, what are you assuming about the flu status of each of the next two patients?
Tree diagrams
Let's consider that question about going to work by car or bike, but put the information in the form of a tree diagram. There are two events, with the outcome of the second one being dependent on the outcome of the first.
I have to go to work today either by car or by bike. If it's raining, there is an 80% chance that I'll go by car. If it's not raining, there's a 35% chance I'll go by car. The weather forecast tells me that there is a 40% chance of rain.
There are four "leaves" at the bottom of this tree, each representing a possible joint outcome: I go by car in the rain, I cycle in the rain, I go by car in the dry, or I cycle in the dry. Each branch of the tree is marked with a probability. You'll notice that each branching has probabilities that add to 1 (=100%). This is because, as you reach each point in the tree, one of the branches that you can take must happen, and that word "must" translates into a probability of 1.
We can then calculate the probabilities of those joint events (the "leaf probabilities") by multiplying down the branches from the root of the tree at the top. For instance, to get the probability that I drive in the rain, multiply the probability that it rains (40%) by the probability that I'll drive given that it's raining (80%). You'll recognise this as applying the first law of probability (the "both" rule above).
Then, once we have the leaf probabilities, we can add them. If we want the probability that I drive by car, whether it rains or not, add the probability that I drive in the rain to the probability that I drive in the dry. You will recognise this as an application of the second law (the "or" law). The two outcomes are mutually exclusive (as I can't drive in both the dry and the wet at the same time):
p(Car) = p(Car | rain) + p(Car | no rain) = 0.53
Exercise 5
A bag contains five coloured beads: two red, one yellow and two blue. A bead is drawn out at random and examined. Then, without putting the bead back into the bag, second bead is drawn out. The following tree diagram shows the different possible outcomes, with R, Y and B representing the different colours of the beads, and the lower case orange letters representing the different possibilities at each stage.
The letters a to k represent the individual probabilities for each draw. Please fill in the correct figures, giving your answers as fractions. Don't cancel the fractions down to their lowest terms, as you will find it helpful not to later:
a =
b =
c =
d =
e =
f =
g =
h =
i =
j =
k =
The letters p to w represent the joint probabilities (e.g. p represents the probability of getting two red beads). What are the values for these letters? Again, you are advised not to cancel down the fractions.
p =
q =
r =
s =
t =
u =
v =
w =
Calculate the probability of getting two beads the same colour. This time, you should cancel the fraction, if appropriate:
Answer =
Calculate the probability that the second bead is yellow:
Answer =
In what way would the structure of that tree be different if the first bead pulled out were replaced before the second bead were pulled out?
Why does it make sense not to cancel down the fractions represented by the letters?
The Russian rocket XJ-6 has two firing mechanisms, a primary and a back-up (in case the primary fails). The primary mechanism has a chance of 75% of firing (i.e. working correctly). The back-up has a chance of 60% of firing. The following (incomplete) tree shows the possible outcomes of any firing:
Primary fires
Primary fails
Back-up fires
Back-up fails
Please fill in the final "leaf" probabilities.
What is the probability that at least one of the firing mechanisms actually works?
The bottom part of the tree has an extra layer when compared to the top part. Why is this?
