Maths
Probability
Probability is the maths of chance. A probability is a number that tells you how likely (probable) something is to happen.
This Revision Bite covers:
We often make judgments 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 that the sun will come up tomorrow, or that it is impossible to find somebody who is more than 3m tall.
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.
Now try to order the probabilities in the activity below.
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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 ^{3}/_{6} (= ^{1}/_{2} or 0.5 or 50%)
The probability of an outcome = number of ways the outcome can happen ÷ total number of possible outcomes
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'?
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 2/11.
What is wrong with the following statement?
The probability of obtaining a 6 when I throw a die is ^{1}/_{6}  so if I throw the die 6 times I should get exactly 1 × 6.
In theory this statement is true, but in practise it might not be the case. Try throwing a die 6 times  you won't always get one 6.
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) How many times would you expect Kate or Josh to obtain a 6?
a) In theory, Kate should obtain a 6 on ^{1}/_{6} of her throws. Therefore, in theory she should throw a 6 on 5 of her 30 throws.
b) Josh should also throw to 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. However, it is very unlikely that Kate and Josh obtained exactly the same results, or that either of them threw a 6 on 5 occasions.
As you can see, if an experiment is repeated, the results are not necessarily the same each time.
However, it is more likely that their combined results were closer to the expected outcome (10 × 6s) than their individual results.
In other words, if you do a large number of trials you will get a more accurate result.
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 whether they prefer caravanning or camping holidays. Seven out of 10 say that they prefer caravanning, so Alex writes the dramatic headline you see above.
What is wrong with Alex's method?
Alex has based his probability on a very small survey. This is equivalent to throwing a die 10 times, throwing a 6 nine times by chance, and stating that the probability of obtaining a 6 is ^{9}/_{10}! To estimate probabilities from the results of surveys, you must gather a large number of results.
If you toss a coin, the probability of obtaining a head is ^{1}/_{2} and the probability of obtaining a tail is also ^{1}/_{2}.
P(head) + P(tail) = ^{1} / _{2} + ^{1} / _{2} = 1
If we choose a letter at random from the word 'SUMS', the probability of obtaining the letter 'S' is ^{2} / _{4}, the probability of obtaining the letter 'U' is ^{1} / _{4} and the probability of obtaining the letter 'M' is ^{1} / _{4}.
P(S) + P(U) + P(M) = ^{2} / _{4} + ^{1} / _{4}+ ^{1} / _{4} = 1
Remember that the sum of the probabilities of all possible outcomes is 1
The probability that I am late for work tomorrow is ^{2}/_{9}. What is the probability that I am not late for work?
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  ^{2}/_{9} = 7/9
You already know that the probabibility of an outcome is:
number of ways an outcome can happen ÷ 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.
Two coins are tossed, once each. What is the total number of possible outcomes when they land, with either their heads or their tails uppermost?
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 or use a table.
1. Using a list
1st coin  2nd coin 

H  H 
H  T 
T  H 
T  T 
2. Using a table
1st coin  

H  T  
2nd coin  H  (H, H)  (H, T) 
T  (T, H)  (T, T) 
From both these methods we can see that there are four possible outcomes. We can use this fact to calculate probabilities.
Eg: there is only one way of obtaining two heads, so the probability (P) of obtaining 2 heads is ^{1}/_{4} .
We can say P (two heads) = ^{1}/_{4}
There are two ways of obtaining a head and a tail, so P (head and tail) = ^{2}/_{4} = 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 (foursided) dice are thrown
Copy and complete the following table, which shows the sum of their scores:

1st die  


1 
2 
3 
4 

2nd die 
1 
2 
3 
4 

2 





3 





4 




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?

1st die  


1 
2 
3 
4 

2nd die 
1 
2 
3 
4 
5 
2 
3 
4 
5 
6 

3 
4 
5 
6 
7 

4 
5 
6 
7 
8 
a) 5 is the most likely outcome
b) The probability of getting the sum 3 = ^{2} / _{16} = 1/8
c) The probability of a sum greater than 5 = ^{6} / _{16} = 3/8
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