Describing probabilities
We often make judgements as to whether an event will take place, and use words to describe how probable that event is.
For example, we might say that it is likely to rain tomorrow, or that it is impossible to find somebody who is more than \({3}~{m}\) tall.
Other commonly used words to describe the chance of an event happening include:
- certain
- very likely
- even chance
- unlikely
- very unlikely
The probability scale
Maths uses numbers to describe probabilities.
Probabilities can be written as fractions, decimals or percentages.
You can also use a probability scale, starting at \({0}\) (impossible) and ending at \({1}\) (certain).
Here are some events placed on the probability scale.
Finding probabilities
When you throw a die (plural: dice), there are six possible different outcomes.
It can show either \(1\), \({2}\), \({3}\), \({4}\), \({5}\) or \({6}\).
But how many possible ways are there of obtaining an even number?
There are three possibilities: \({2}\), \({4}\) and \({6}\).
The probability of obtaining an even number is \(\frac{3}{6} (= \frac{1}{2}\) or \(0.5\) or \(50\%)\)
Key point
If every possible outcome has the same chance of occurring, the probability of an outcome equals the number of ways the outcome can happen divided by the total number of possible outcomes.
Questions
Q1. How many outcomes are there for the following experiments? List all the possible outcomes.
a) Tossing a coin
b) Choosing a sweet from a bag containing \(1\) red, \(1\) blue, \(1\) white and \(1\) black sweet.
c) Choosing a day of the week at random.
Q2. Sindhu writes the letters of the word 'MATHEMATICS' on separate cards and places them in a bag.
She then draws a card at random.
What is the probability that Sindhu chooses the letter 'A'?
Answer
A1.
a) There are two possible outcomes (head and tail).
b) There are four possible outcomes (red, blue, white and black).
c) There are seven possible outcomes (Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday).
A2.
There are \(11\) letters in MATHEMATICS, \(2\) of which are A.
So the probability that Sindhu chooses the letter A is \(\frac{2}{11}\).
Experimental probability
Experimental probability
What is wrong with the following statement?
The probability of obtaining a \(6\) when I throw a die is \(\frac{1}{6}\), so if I throw the die \(6\) times I should expect to get exactly one \(6\).
In theory this statement is true, but in practice it might not be the case.
Try throwing a die \(6\) times - you won't always get exactly one \(6\).
Question
Kate and Josh each throw a die \(30\) times.
a) How many times would you expect Kate to obtain a \(6\)?
b) How many times would you expect Josh to obtain a \(6\)?
c) What is the total number of sixes you would expect Kate and Josh to obtain between them?
Answer
a) In theory, Kate should obtain a \(6\) on \(\frac{1}{6}\) of her throws.
Therefore, in theory, you should expect Kate to throw a \(6\) on \(5\) of her \(30\) throws.
b) Josh should also obtain a \(6\) on \(5\) of his \(30\) throws.
c) In total, Kate and Josh have thrown the die \(60\) times.
You would expect them to obtain a \(6\) on \(10\) of those throws.
It is very unlikely that either Kate or Josh would have obtained exactly five \(6\)s, or that together they would have thrown ten \(6\)s.
However, it is more likely that their combined results were closer to the expected outcome (ten \(6\)s) than their individual results.
If an experiment is repeated, the results are not necessarily the same each time.
However, as it is more likely that the combined results will be closer to the expected outcome, we can see that if you do a large number of trials you will get a more accurate result.
Probability surveys
Probability is often based on surveys, because you get a more accurate measure of probability by basing your calculation on a large number of results.
Alex is doing some reporting for his local paper. The subject of his article is 'holidays'.
As part of his report he decides to question \(10\) of his friends about whether they prefer caravanning or camping holidays.
Seven out of \(10\) say that they prefer caravanning, so Alex writes the headline you can see above.
What is wrong with Alex's method?
Answer
Alex has based his probability on a very small survey.
This is equivalent to throwing a die \(10\) times, and getting a \(6\) nine times by chance, and then stating that the probability of obtaining a \(6\) is \(\frac{9}{10}\).
To estimate probabilities from the results of surveys, you must gather a large number of results.
Sum of probabilities
If you toss a coin, the probability of obtaining a head is \(\frac{1}{2}\) and the probability of obtaining a tail is also \(\frac{1}{2}\).
P (head) + P (tail) = \(\frac{1}{2} + \frac{1}{2} = 1\)
If we choose a letter at random from the word 'SUMS', the probability of obtaining the letter 'S' is \(\frac{2}{4}\), the probability of obtaining the letter 'U' is \(\frac{1}{4}\), and the probability of obtaining the letter 'M' is \(\frac{1}{4}\).
P(S) + P(U) + P(M) = \(\frac{2}{4} + \frac{1}{4}+ \frac{1}{4} = 1\)
Remember that the sum of the probabilities of all possible outcomes is 1.
Question
The probability that I am late for work on any morning is \(\frac{2}{9}\).
What is the probability that I am not late for work?
Answer
There are two possible outcomes - being late and not being late.
The sum of their probabilities must add up to \(1\), so the probability of not being late is:
\(1 - \frac{2}{9} = \frac{7}{9}\)
Probability of combined events
If every possible outcome has the same chance of occurring, the probability of an outcome is:
number of ways an outcome can happen \(\div\) total number of possible outcomes.
However, finding the total number of possible outcomes is not always straightforward - especially when we have more than one event.
Question
Two coins are tossed, once each. What is the total number of possible outcomes when they land?
Answer
The total number of possible outcomes is not three (two heads, a head and a tail or two tails).
To find the true number of possible outcomes, we must list the results as below:
Method 1 - Using a list
Method 2 - Using a table
From both these methods we can see that there are four possible outcomes.
We can use this fact to calculate probabilities.
For example, there is only one way of obtaining two heads - so the probability (P) of obtaining \(2\) heads is \(\frac{1}{4}\).
We can say P (two heads) = \(\frac{1}{4}\)
There are two ways of obtaining a head and a tail, so P (head and tail) = \(\frac{2}{4} = \frac{1}{2}\)
When listing possible outcomes, try to be as logical as possible. If you repeat or forget any of them, it will affect the rest of your answers.
Two tetrahedral (four-sided) dice are thrown.
Copy and complete the following table, which shows the sum of their scores:
a) What is the most likely outcome?
b) What is the probability that the sum of the scores will be \(3\)?
c) What is the probability that the sum of the scores will be greater than \(5\)?
Answer
a) \(5\) is the most likely outcome
b) The probability of getting the sum \(3 = \frac{2}{16} = \frac{1}{8}\)
c) The probability of a sum greater than \(5 = \frac{6}{16} = \frac{3}{8}\)
Test section
Question 1
How would you describe the probability that a Non-Premier League football team, wins the Premier League this year?
Certain, Impossible or Very unlikely?
Answer
A non-premier league team cannot play in the premier league therefore it is impossible for them to win.
Question 2
With one throw of a \({6}\)-sided die, what's the probability of throwing a \({5}\)?
Answer
There is one \({5}\) and six possible outcomes, so the probability of getting a \({5}\) is \(\frac{1}{6}\).
Question 3
A bag holds lego pieces which are all the same size.
There are \({6}\) red pieces, \({3}\) blue pieces, \({2}\) green pieces and \({4}\) white pieces.
What's the probability of pulling a red piece out of the bag?
Answer
\({6}\) red pieces and \({15}\) pieces altogether which gives \(\frac{6}{15}\).
You can simplify this fraction to \(\frac{2}{5}\).
Question 4
Gwen throws a conventional \({6}\)-sided die \({300}\) times.
Approximately how many times would you expect her to throw an odd number?
Answer
\({150}\)
There are \({3}\) odd numbers out of the \({6}\) possible outcomes, so you'd expect about half her outcomes to be odd numbers.
Question 5
The probability that Gary will forget his homework is \({0.2}\).
What's the probability that Gary won't forget his homework?
Answer
The total probability of all possible outcomes is \({1}\).
So to get the answer you need to calculate \({1}-{0.2}={0.8}\).
Question 6
Erin tosses a \({50}{p}\) coin twice.
What's the probability that she will get two tails?
Answer
There are four possible outcomes (HH, HT, TH and TT), and only one gives TT.
So the answer is: \(\frac{1}{4}\)
Question 7
Sally tosses a \({50}{p}\) coin and throws a conventional \({6}\)-sided die.
What's the probability that she gets heads and a \({3}\)?
Answer
To get the probability of two independent outcomes, you need to multiply the probabilities \(\frac{1}{2}\) and \(\frac{1}{6}\), to get \(\frac{1}{12}\).
Question 8
Draw a two-way table showing all possible outcomes of throwing two conventional \({6}\)-sided dice.
Which total is most likely?
Answer
\({7}\) appears six times, so \({7}\) is most likely.
Question 9
When throwing two conventional \({6}\)-sided dice, what's the probability of getting a total that's higher than \({8}\)?
Answer
\({10}\) of the \({36}\) possible outcomes give a total of \({9}\) or more.
So the probability is \(\frac{10}{36}=\frac{5}{18}\).
Question 10
Mark tosses three coins.
By making a list, find the probability that he'll get \({2}\) heads and one tail?
Answer
There are \({8}\) possible outcomes, and \({3}\) of them comprise two heads and one tail (HHT, HTH, THH).
So the answer is \({\frac{3}{8}}\).
More on Probability
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