6.025 Theoretical probability

We can make predictions for trials by first creating the sample space and then determining the theoretical probability of each outcome.

Flipping a coin

Let's say we flipped a coin. What outcomes are possible? That is, what can the coin land on?

It can either land on heads or tails. In this case there is an even chance of the coin landing on heads or tails, because the chance of it landing on heads is the same as landing on tails.

So what is the probability that when we flip the coin, we get a head?

Well there is a $1$1 out of $2$2 chance of that happening, so we can write the probability of getting a head:

• as a fraction: $\frac{1}{2}$12​
• as a decimal: $0.5$0.5
• or as a percentage: $50%$50%

Probability can be expressed as a fraction, decimal or percentage, so make sure you're comfortable converting between these forms.

 

Spin the spinner

Let's say we spun this spinner. What outcomes are possible? Are the chances of landing on each outcome equal?

Well we could land on the star (green), the pig (blue), the ball (red), or the apple (yellow), but the chances of each aren't equal.

So what is the probability of landing on the apple (the yellow segment)?

Well, the picture shows $8$8 equal pieces. $1$1 of those shows an apple on a yellow background, so the probability of this outcome is $\frac{1}{8}$18​.

 

 

Practice questions

Question 1

Question 2

 

Complementary events

A complement of an event are all outcomes that are NOT the event.

The following are examples of an event and the complement.

• Event is $\left\{\text{Heads}\right\}${Heads}, so the complement is $\left\{\text{not Heads}\right\}${not Heads} which is $\left\{\text{Tails}\right\}${Tails}
• Event is rolling a standard die and getting $\left\{2\right\}${2}, then the complement is $\left\{1,3,4,5,6\right\}${1,3,4,5,6}
• Event is $\left\{\text{Winter}\right\}${Winter}, the complement is $\left\{\text{Summer},\text{Autumn},\text{Spring}\right\}${Summer,Autumn,Spring}
• Event is $\left\{\text{Diamonds}\right\}${Diamonds}, the complement is $\left\{\text{Hearts},\text{Clubs},\text{Spades}\right\}${Hearts,Clubs,Spades}

 

Notation

We already know we can denote the probability of event $A$A as $P\left(A\right)$P(A).

Well, we denote the probability of its complement, "not $A$A", as $P\left(A'\right)$P(A′) (the prime next to the $A$A indicates negation)

The probabilities of complementary events always sum to $1$1.

$P\left(A\right)+P\left(A'\right)=1$P(A)+P(A′)=1

 

Using complementary events

Due to the fact that complementary events always sum to $1$1, we often have a choice about how to solve problems involving complementary events.

 

Worked example

Example 1

Consider the event of rolling two dice. What is the probability that you do not roll a double?

Method 1 - listing the sample space

Answering this question does not need to involve complementary events at all if we don't want to. We could just list the sample space, (the list of all possible outcomes from rolling two dice) and then count up how many are not doubles and calculate the probability. The sample space for this event is pretty large, but we can still list it:

$\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

 

The possible outcomes for two dice could also be drawn in a table(array):

		Die 1
		$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
Die 2	$1$1	$1,1$1,1	$1,2$1,2	$1,3$1,3	$1,4$1,4	$1,5$1,5	$1,6$1,6
	$2$2	$2,1$2,1	$2,2$2,2	$2,3$2,3	$2,4$2,4	$2,5$2,5	$2,6$2,6
	$3$3	$3,1$3,1	$3,2$3,2	$3,3$3,3	$3,4$3,4	$3,5$3,5	$3,6$3,6
	$4$4	$4,1$4,1	$4,2$4,2	$4,3$4,3	$4,4$4,4	$4,5$4,5	$4,6$4,6
	$5$5	$5,1$5,1	$5,2$5,2	$5,3$5,3	$5,4$5,4	$5,5$5,5	$5,6$5,6
	$6$6	$6,1$6,1	$6,2$6,2	$6,3$6,3	$6,4$6,4	$6,5$6,5	$6,6$6,6

 

$P\left(\text{not a double}\right)$P(not a double)	$=$=	$\frac{\text{total favourable outcomes}}{\text{total possible outcomes}}$total favourable outcomestotal possible outcomes​
	$=$=	$\frac{30}{6}$306​
	$=$=	$\frac{5}{6}$56​

 

Method 2 - using complementary events

Using complementary events we can see that

$P\left(\text{not a double}\right)=1-P\left(\text{double}\right)$P(not a double)=1−P(double).

This is easy enough to work out without having to list or draw the entire sample space, since we know there are only $6$6 possible doubles.

$P\left(\text{not a double}\right)$P(not a double)	$=$=	$1-\frac{6}{36}$1−636​
	$=$=	$\frac{5}{6}$56​

 

Practice questions

Question 3

Question 4