An outcome is the result of an experiment such as throwing a die or picking a card from a pack. A trial is each time the die is rolled or a card picked. The sample space is the set of all possible outcomes and an event is part of the sample space.
experiment | throwing a pair of dice | |
trial | one rolling of the dice | |
outcome | the result of throwing the dice e.g. (1, 2) | |
event | scoring an odd total | |
sample space |
all possible outcomes from |
Probabilities are expressed as fractions, decimals or percentages.
For equally likely outcomes, the probability of the event E occurring is given by:
Probability of event E happening =
When a die is tossed, what is the probability that a 5 or a 6 will face up? |
Total six possible outcomes = {1, 2, 3, 4, 5, 6} Required two outcomes= {5, 6} |
A card is picked from a normal pack of 52. What is the probability the card will be a heart? |
There are 52 cards in a pack. There are 13 hearts. P(Heart) = 13/52 = 0.25 |
Probabilities will always be in the range from 0 to 1.
If the probability of an event happening is 0, the event cannot occur.
e.g. P(throwing a six-sided die and scoring a 7) = 0
If the probability of an event happening is 1, the event is certain to occur.
With an experiment, such as throwing a die, the sum of the probabilities of all possible outcomes is 1.
Multiplying Probabilities
When two events happen independently and the probablility of the first event happening is P(E1) and the probability of the second event happening is P(E2) then
P (first event AND second event happening) = P(E1) × P(E2)
Example What is the probability of getting a head and an even number, when tossing a coin and rolling a die?
P(getting a head) = 0.5
P(rolling an even number) = 0.5P(head AND even number) = 0.5 × 0.5 = 0.25
Adding Probabilities
If the trial above was repeated what would be the probability of getting either (a head and an even number) or (a tail and an odd number)?
P( a head and an even number OR a tail and an odd number) = P(Head) x P(Even) + P(Tail) × P(Odd)0.5 x 0.5 + 0.5 × 0.5 0.25 + 0.25 0.5
Tree Diagrams
A tree diagram can help to solve probability problems or problems involving the number of ways that a combination of things can be carried out.
Example A rugby statistician has worked out that the probability that a goal kicker in a rugby game kicks a penalty successfully if it is windy is 0.6 and if it is not windy is 0.7.
The probability that the wind will blow when he takes the kick is 0.2.
Draw a probability tree to show this situation and find the probability that the kicker is successful with the penalty.
P(windy and kicks penalty) = 0.2 × 0.6 = 0.12
P(not windy and kicks penalty) = 0.8 × 0.7 = 0.56
P(kicks penalty) = P(windy and kicks penalty) or
P(not windy and kicks penalty)= 0.12 + 0.56
= 0.68
Expected value
The expected value of an event occurring from a certain number of trials is the number of times the event is expected to occur.
Expected value = Probability of event × number of trials
= P(E) × n
Example Answer If a die is thrown 60 times, how many times would you expect a 4 or a 5 to be thrown?