Probability Lesson #2
Sample Space
• A sample space is the set of all possible outcomes of an
experiment.
• There are a variety of ways of representing or illustrating
sample spaces.
Listing Outcomes
• Example:
List the sample space of possible outcomes for:
(a.) tossing a coin
(b.) rolling a die
(a.) When a coin is tossed, there are two possible outcomes.
Therefore, sample space = {H, T}.
(b.) When a die is rolled, there are six possible outcomes.
Therefore, sample space = {1, 2, 3, 4, 5, 6}.
2-Dimensional Grids
• Example:
Illustrate the possible outcomes when 2 coins are tossed
using a 2-dimensional grid.
Each of the points on the grid represents one of the possible
outcomes: {HH, HT, TH, TT}.
Tree Diagrams
• The sample space in the previous example could also be
represented by a tree diagram. The advantage of tree
diagrams is that they can be used when more than two
operations are involved.
• Example 1:
Illustrate, using a tree diagram, the possible outcomes when:
(a.) tossing two coins
(b.) drawing two marbles from a bag containing a number of
red, green and yellow marbles.
(a.)
Each “branch” gives a different outcome and the sample
space can be seen to be {HH, HT, TH, TT}.
(b.)
• Example 2:
John plays Peter at tennis and the first to win two sets wins
the match. Illustrate the sample space using a tree diagram.
If J means “John wins the set” and P means “Peter wins the
set” then the tree diagram display is:
Note: The sample space is S = {JJ, JPJ, JPP, PJJ, PJP, PP}.
Theoretical Probability
• Consider the octagonal spinner below.
• Since the spinner is symmetrical, when it is spun, the arrowed
marker could finish with equal likelihood on each of the
sections marked 1 to 8.
• Therefore, we would say that the likelihood of obtaining a
particular number, for example, 4, would be:
1 chance in 8,
1
1
, 12 % , or 0.125
2
8
• This is a mathematical (or theoretical) probability and is
based on what we theoretically expect to occur.
• The theoretical probability of a particular event is a measure
of the chance of that event occurring in any trial of the
experiment.
• If we are interested in the event of getting a result of 6 or more
from one spin of the octagonal spinner, there are three
favourable results (6, 7 or 8) out of the eight possible results,
and each of these is equally likely to occur.
3
• So, the probability of a result of 6 or more is ,
8
3
P(6 or more) =
8
• In general, for an event E containing equally likely possible
results:
P(E) = the number of members of the event E
the total number of possible outcomes
• Example:
A ticket is randomly selected from a basket containing 3 green,
4 yellow and 5 blue tickets. Determine the probability of getting:
(a.) a green ticket
(b.) a green or yellow ticket
(c.) an orange ticket
(d.) a green, yellow or blue ticket
The sample space is: {G, G, G, Y, Y, Y, B, B, B, B, B} which
has 3 + 4 + 5 = 12 outcomes.
(a.) P(green) =
3
1
=
12
4
(b.) P(a green or yellow ticket) =
(c.) P(orange) =
0
=0
12
3+4
7
=
12
12
3+4+5
=1
12
• In this example, notice that in (c.), an orange result cannot
occur and the calculated probability is 0, which fits the fact that
it has no chance of occurring.
• Also notice in (d.), a green, yellow or blue result is certain to
occur. It is 100% likely which is perfectly described using a 1.
(d.) P(green, yellow or blue) =
• The two events of no chance of occurring with a probability of 0
and certain to occur with probability 1 are two extremes.
• Consequently, for any event E, 0 ≤ P(E) ≤ 1.
Complementary Events
• Example:
An ordinary 6-sided die is rolled once. Determine the chance
of:
(a.) getting a 6
(b.) not getting a 6
(c.) getting a 1 or 2
(d.) not getting a 1 or 2
The sample space of possible outcomes is {1, 2, 3, 4, 5, 6}.
(a.) P(6) =
1
6
(b.) P(not getting a 6) = P(1, 2, 3, 4, or 5) =
(c.) P(1 or 2) =
5
6
2
1
=
6
3
(d.) P(not getting a 1 or 2) = P(3, 4, 5, or 6) =
4
2
=
6
3
• In this example, we should notice that:
P(6) + P(not getting a 6) = 1 and
P(1 or 2) + P(not getting a 1 or 2) = 1.
• This should come as no surprise to us because getting a 6
and not getting a 6 are complementary events where one
of them must occur.
Notation
• If E is an event, then E’ is the complementary event of E.
• So, P(E) + P(E’) = 1.
• Another useful way to look at it is:
P(E not occurring) = 1 – P(E occurring)
Using Grids To Find Probabilities
• Two dimensional grids give us excellent visual displays of
sample spaces. From these we can count favourable outcomes
and therefore calculate probabilities.
• Example:
Use a two-dimensional grid to illustrate the sample space for
tossing a coin and rolling a die simultaneously. From this grid,
determine the probability of:
(a.) tossing a head
(b.) getting a tail and a 5
(c.) getting a tail or a 5
There are 12 members in the sample space.
(a.) P(head) =
6
1
=
12
2
1
12
7
(c.) P(tail or a ‘5’) =
{the enclosed points}
12
(b.) P(tail and a ‘5’) =