Home > Maths > Handling data > Probability
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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