5.02 Review of probability fundamentals

Let's now look at how to describe and calculate probabilities. 

In day to day life, there are many places where the language of probability is used. For example:

• The chance of rain today is 75%
• A medication has an effectiveness of 90%
• 4 in 5 people agree that a particular brand of toothpaste is more effective than another
• I will probably go to the gym today

All probabilities lie on the interval $0\le P\left(A\right)\le1$0≤P(A)≤1, where a probability of zero indicates the event cannot possibly occur and a probability of one indicates the event is certain to occur. This can be visualised as follows: 

 

We can calculate the probability of an event occurring as:

$P(event)=\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$P(event)=number of favourable outcomestotal possible outcomes​

If all the outcomes can be easily listed, then the process of counting favourable and total outcomes is relatively straightforward. For large or complex sample spaces we may wish to employ some advanced counting methods such as combinatorics or use properties of probability that relate to different characteristics of the event such as the complement of the event, if the event is mutually exclusive, if it is a compound event or conditional event.

Let's look at complementary events and mutually exclusive events and the properties we can use.

Complementary events

Recall we can denote the probability of event $A$A as $P\left(A\right)$P(A) and we denote the probability of its complement, not $A$A as $P\left(A'\right)$P(A′).

Complementary events have the property that their probabilities always add to $1$1. Since it is certain either event $A$A occurred or it did not occur. That is:

In some cases it may be easier to count the possibilities where an event does not happen and use this fact to find the probability that the event does happen more efficiently. Let's look at an example:

Worked example 

Example 1

Let's consider the experiment of rolling two dice.  What is the probability that you do not roll a double?

Method 1

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 2 dice) and then count up how many are not doubles and calculate the probability.  The sample space for this event is quite large, but we can still  list it or write it out in a table: 

$S=\left\{\left(\text{Roll 1, Roll 2}\right)\right\}$S={(Roll 1, Roll 2)}

$=$=

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

 

Hence, $P\left(\text{not a double}\right)=$P(not a double)=$\frac{\text{number of favourable outcomes }}{\text{total possible outcomes }}=\frac{30}{36}$number of favourable outcomes total possible outcomes ​=3036​ $=$= $\frac{5}{6}$56​

 

Method 2

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)

We do not have to list the sample space here, there will be one double for each number $1$1 through to $6$6.

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

 

Mutually exclusive events

When calculating the probability for an event that involves either/or statements we are calculating the probability of the union of the individual events. For example: finding the probability of rolling either a $6$6 or an even number on a die roll would be equivalent to finding $P\left(\text{Rolling a 6}\cup\text{Rolling an even number}\right)$P(Rolling a 6∪Rolling an even number). To find such probabilities we should consider if the events are mutually exclusive so when we count the number of favourable outcomes we do not double count outcomes that are common to both events.

 

Worked example

Example 2

Consider an experiment of drawing one card from a deck of $52$52 cards. And define two events $A$A: 'Drawing a $7$7 card' and $B$B: 'Drawing a 10 card'. What is the probability of drawing a $7$7 or a $10$10?

Think: We are trying to find $P\left(A\cup B\right)$P(A∪B). Are the events mutually exclusive? Yes, as on a single card draw we cannot draw a card that is both a seven and a ten.

Do: Since there are no elements common to both events we can count the 'favourable outcomes' by just adding the outcomes of the individual events and then calculating the probability. Alternatively, we can use the rule to add the probability of the individual events.

There are $4$4 cards of each number in the deck of $52$52 cards, so there are $8$8 favourable outcomes.

$P\left(A\cup B\right)$P(A∪B)	$=$=	$\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$number of favourable outcomestotal possible outcomes​
	$=$=	$\frac{8}{52}$852​
	$=$=	$\frac{2}{13}$213​

Alternative, calculation:

$P\left(A\cup B\right)$P(A∪B)	$=$=	$P\left(A\right)+P\left(B\right)$P(A)+P(B)
	$=$=	$\frac{4}{52}+\frac{4}{52}$452​+452​
	$=$=	$\frac{8}{52}$852​
	$=$=	$\frac{2}{13}$213​

 

Example 3

Consider an experiment of drawing one card from a deck of $52$52 cards. And define two events $A$A: 'Drawing a Club card' and $B$B: 'Drawing a $7$7 card'. What is the probability of drawing a club or a $7$7?

Think: We are trying to find $P\left(A\cup B\right)$P(A∪B). Are the events mutually exclusive? No, the events share the outcome of the $7$7 of clubs.

If we picture the Venn diagram, we can see if we were to add the number of elements in $A$A to the number of elements in $B$B we would be double counting any elements in the intersection.

 

Do: We can count the 'favourable outcomes' by adding the outcomes of the individual events and subtract the number of elements in the intersection. Alternatively, we can use the rule and the probability of the individual events.

Favourable outcomes $=4+13-1=16$=4+13−1=16

$P\left(A\cup B\right)$P(A∪B)	$=$=	$\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$number of favourable outcomestotal possible outcomes​
	$=$=	$\frac{16}{52}$1652​
	$=$=	$\frac{4}{13}$413​

Alternative, calculation:

$P\left(A\cup B\right)$P(A∪B)	$=$=	$P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$P(A)+P(B)−P(A∩B)
	$=$=	$\frac{1}{4}+\frac{1}{13}-\frac{1}{52}$14​+113​−152​
	$=$=	$\frac{4}{13}$413​

Practice questions

QUESTION 1

QUESTION 2

Question 3

Question 4