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Chapter 6: Section 6-3
Sample Spaces
D. S. Malik
Creighton University, Omaha, NE
D. S. Malik Creighton University, Omaha, NE ()Chapter 6: Section 6-3 Sample Spaces
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Experiment
Experiment: An activity or occurrence with an observable result. The
following are some examples of experiments.
1
Tossing a coin and observing the top face.
2
Selecting a marble from a box that contains 5 red, 3 blue, and 4
green marbles and recording the color.
3
Making a phone call and observing whether the call was answered by
a person.
4
A family has three children. Record the number of boys and girls.
De…nition
(i) A repetition of an experiment is called a trial.
(ii) The results of a trial are called outcomes.
D. S. Malik Creighton University, Omaha, NE ()Chapter 6: Section 6-3 Sample Spaces
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Sample Space
De…nition
The set of all possible outcomes of an experiment is called a sample
space.
Example
(i) A coin is tossed and the top face is observed. Let H stand for the head
and T stand for the tail. Then the sample space, S, is the set:
S = fH, T g.
(ii) A six-faced die is rolled and the number rolled is recorded. The sample
space is: S = f1, 2, 3, 4, 5, 6g.
(iii) A coin is tossed three times. An experiment consists of recording the
number of heads and tails. The sample space is:
S = f3H and 0T , 2H and 1T , 1H and 2T , 0H and 3T g.
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Example
A coin is tossed three times. An experiment consists of recording the order
in which heads and tails occurred. Suppose that HHH means that all three
tosses are heads, HTH means that the …rst toss is a head, the second toss
is a tail, and the third toss is a head, and so on. The sample space is:
S = fHHH, HHT , HTH, HTT , THH, THT , TTH, TTT g.
Remark
From the preceding examples, it is clear that the sample space of an
experiment depends on the types of outcomes that we want to observe.
For example, suppose that the experiment is tossing a coin three times. If
we are interested in counting the number of heads and tails, then each
element of the sample space consists of number of heads and tails. On the
other hand, if we are interested in recording the order in which the heads
and tails occur, then the sample space consists of elements in the previous
example.
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Event
De…nition
Let S be a sample space of an experiment. An event of the experiment or
the sample space is a subset of the sample space S.
Example
A coin is tossed twice. Then the sample space is:
S = fHH, HT , TH, TT g.
Let E be the event that the second toss is a tail. Then
E = fHT , TT g.
Let F be the event that each outcome come consists of at least one head.
Then
F = fHH, HT , TH g.
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Example
A six-faced die is rolled. Then the sample space is:
S = f1, 2, 3, 4, 5, 6g.
Let E be the event that the number rolled is 2. Then
E = f2g.
Note that n (E ) = 1.
Let F be the event that the number rolled is a prime number. Then
F = f2, 3, 5g.
Note that n (F ) = 3.
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Example
A card is drawn from a well-shu- ed deck of 52 cards. The sample space is
the set of 52 cards. Let E be the event that the drawn card is an ace or a
red 10. Let H stand for a heart, D stand for a diamond, S stand for a
spade, and C stand for a club. Also 1C means the ace of clubs, and so on.
Then
E = f1H, 1D, 1S, 1C , 10H, 10D g.
Note that n (E ) = 6.
Let F be the event that the drawn card is a club or a queen. Then
F
= f1C , 2C , 3C , 4C , 5C , 6C , 7C , 8C , 9C ,
10C , JC , QC , KC , QH, QD, QS g.
Note that n (F ) = 16.
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Example
Two six-faced dice are
the sample space is:
8
(1, 1),
>
>
> (2, 1),
>
>
>
<
(3, 1),
S=
(4, 1),
>
>
>
>
>
(5, 1),
>
:
(6, 1),
rolled and the numbers rolled are recorded. Then
(1, 2),
(2, 2),
(3, 2),
(4, 2),
(5, 2),
(6, 2),
(1, 3),
(2, 3),
(3, 3),
(4, 3),
(5, 3),
(6, 3),
(1, 4),
(2, 4),
(3, 4),
(4, 4),
(5, 4),
(6, 4),
(1, 5),
(2, 5),
(3, 5),
(4, 5),
(5, 5),
(6, 5),
(1, 6),
(2, 6),
(3, 6),
(4, 6),
(5, 6),
(6, 6),
9
>
>
>
>
>
>
=
>
>
>
>
>
>
;
.
Let E be the event that the …rst die shows a 2 and the second die shows
an odd number. Then E = f(2, 1), (2, 3), (2, 5)g. Note that n (E ) = 3.
Let F be the event that the sum of the numbers rolled is 8. Then
F = f(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)g.
Note that n (F ) = 4.
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Example
A bag contains 3 red and 4 green marbles. Suppose that the red marbles
are labeled as R1 , R2 , and R3 ; and the green marbles are labelled as G1 ,
G2 , G3 , and G4 . One marble is drawn from the bag. Let E be the event
that the drawn marble is red. Then
E = fR1 , R2 , R3 g.
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Sample Spaces and Tree Diagrams
Suppose that a coin is ‡ipped twice. The outcomes of this experiment
can be determined using a tree.
When the coin is ‡ipped the …rst time, it has two outcomes – a head
or a tail, which are represented by two branches as shown in the
following …gure.
At the end of the branch we write the outcome.
Suppose that the …rst ‡ip resulted in a head. So we are at the end of
the top branch. The coin is ‡ipped the second time and the outcomes
are – a head or a tail. So we again draw two branches from this
outcome and label the end of these branches with H and T . Now if
the …rst outcome was a tail, then for the second ‡ip, we draw two
branches out of the bottom branch and label the ends with H and T
respectively. Next, in the tree, we traverse each path from left to
right and write the outcome resulting from that path, see the
following …gure.
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First Flip
First Flip
H
H
T
T
(a)
Second Flip
H
T
H
T
(b)
Second Flip
H
First Flip
H
T
Outcome
HH
T
HT
H
TH
T
TT
(c)
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Example
A box contains two bags–a and b. Bag a contains three marbles–red,
green, and blue. Bag b contains four marbles–red (R ), blue (B ), yellow
(Y ), and white (W ). An experiment consists of selecting a bag from the
box, and then selecting a marble from the bag. The tree diagram of this
experiment is:
Select
a Bag
a
b
Select a
Marble
R
G
Outcome
aR
aG
B
aB
R
bR
B
Y
W
bB
bY
bW
The sample space is: S = faR, aG , aB, bR, bB, bY , bW g.
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Example
A bag contains red, green, and blue marbles. An experiment consists of
rolling a die followed by drawing a marble and noting its color.
Let S be the sample space. We want to determine n (S ). This can be
considered two-step experiment. In the …rst step, a die is rolled and in the
second step a marble is drawn.
When a die is rolled, the number rolled has 6 possibilities. So the …rst step
can be completed in 6 ways. Next, the bag contains three types of
marbles–red, green, and blue. So when a marble is drawn, there are three
possibilities for the color. Thus, the second step can be completed in 3
ways. Now for each choice of the …rst step, there are three choices for the
second step. Hence, the entire experiment can be completed in
6 3 = 18 ways.
Hence, n (S ) = 18.
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Theorem
( Multiplication Principle) Suppose that a multistep experiment consists
of k (independent) steps. Suppose that the …rst step can be completed in
n1 ways, the second step can be completed in n2 ways regardless of the
results of step 1, the third step can be completed in n3 ways regardless of
the results of steps 1 and 2, and so on. In general, suppose that step i,
can be completed in ni ways, regardless of the results of the …rst i 1
steps. Then the entire experiment can be completed in
n1 n2
nk
ways.
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Example
(i) A bakery o¤ers 3 types of cookies and 4 types of cakes. A special deal
consists of one type of cookie and one type of cake. An experiment
consists of creating a special deal.
This is a two-step experiment. In the …rst step, we select a cookie, and in
the second step, we select a cake. There are 3 choices for the …rst step
and 4 choices for the second step. Thus, by the multiplication principle,
the number of ways this experiment can be done is:
3 4 = 12
(ii) An experiment consists of tossing a coin 3 times followed by rolling a
die. Let S be the sample space of this experiment.
This is a four-step experiment. In the …rst three steps, a coin is tossed.
Each of these steps have 2 choices. In the fourth step, a die is rolled,
which has 6 choices. Thus, by the multiplication principle, the number of
ways this experiment can be done is:
2 2 2 6 = 48.
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Probability Theory: Some Basic Concepts
Throughout we assume that the sample space is …nite.
Suppose that a die is rolled. Then the sample space is:
S = f1, 2, 3, 4, 5, 6g.
Suppose that you are asked the following question: when a die is
rolled, what are the chances (or likelihood) that the number rolled is
a 2.
Let E be the event that the number rolled is 2. Then E = f2g.
We can now reformulate our question as follows: What are the
chances that the event E will occur?
In mathematical terms, this question is posed as follows: What is the
probability that the number rolled is a 2 or what is the probability
that the even E will occur?
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Probability Theory: Some Basic Concepts
De…nition
Let S be a (…nite) sample space of an experiment. Suppose that all
outcomes of the experiment are equally likely. Let E be an event of the
experiment. Then the probability that the event E will occur, written
Pr[E ], is
n (E )
.
Pr[E ] =
n (S )
Example
Suppose that a fair die is rolled. Let E be the event that the rolled
number is odd. Then E = f1, 3, 5g, so n (E ) = 3. If S is the sample space
of this experiment, then n (S ) = 6. Hence,
Pr[E ] =
n (E )
3
1
= = .
n (S )
6
2
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Remark
( Drawn at random) While discussing probabilities, very often we use
phrases such as “a card is drawn at random” or “a marble is drawn at
random.” When we use words “drawn at random,” then it should be
understood that all outcomes of the sample space are equally-likely. So all
outcomes have the same probability.
Example
Suppose that a card is drawn at random from a deck of 52 cards. Let S be
the sample space. Then n (S ) = 52.
(i) Let E be the event that the drawn card is an ace. Then n (E ) = 4.
n (E )
4
1
Hence, Pr[E ] = n (S ) = 52
= 13
.
(ii) Let F be the event that the drawn card is a heart or an ace. There are
13 hearts and 4 aces. However, one of the aces is also a heart. Thus,
n (F ) = 13 + 4 1 = 16. Hence,
Pr[F ] =
16
4
n (F )
=
= .
n (S )
52
13
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Example
A bag contains 4 red, 6 blue, and 7 green marbles. A marble is drawn at
random. Let S be the sample space. Then n (S ) = 17.
(i) Let E be the event that the drawn marble is red. There are 4 red
marbles and if the drawn marble is red, then it can be any one of these red
marbles. Thus, n (E ) = 4. Hence,
Pr[E ] =
4
n (E )
= .
n (S )
17
(ii) Let F be the event that the drawn marble is red or blue. There are 4
red marbles and 6 blue marbles. Thus, n (F ) = 10. Hence,
Pr[F ] =
n (F )
10
= .
n (S )
17
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Theorem
Let S be a (…nite) sample space such that all outcomes are equally likely,
and let E be an event in S.
(i) Pr[∅] = 0.
(ii) Pr[S ] = 1.
(iii) 0 Pr[E ] 1.
(iv) Pr[E 0 ] = 1 Pr[E ].
Notation: Let S be a sample space and E be an event in S. Suppose
that E = fag, i.e., the number of elements in E is 1. Then
sometimes we write Pr[a] for Pr[E ].
Therefore, if a die is rolled, then Pr[1] means the probability
that the number rolled is 1. Similarly, if a card is drawn from
a deck of 52 cards, then Pr[K ] means the probability that
the drawn card is a king. When a coin is tossed, then Pr[H ]
means the probability that the top face is a head.
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Exercise: Suppose that two dice are rolled. Find the following events.
(a) The sum of the numbers rolled is 7; (b) The product of
the numbers rolled is 6; (c) The sum of the numbers rolled is
13.
Solution:
a. Let E be the event that the sum of the numbers rolled is 7. Then
E = f(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)g.
Here the ordered pair (a, b ) means that the …rst die rolls an a and the
second die rolls a b. For example, (2, 5) means that the …rst die rolls
a 2 and the second die rolls a 5.
b. Let F be the event that the product of the numbers rolled is 6. Then
F = f(1, 6), (2, 3), (3, 2), (6, 1g.
c. Let G be the event that the sum of the numbers rolled is 13. The
largest number on each die is a 6. So the largest sum of the numbers
rolled is 12. Hence,
G = ∅.
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Exercise: An urn contains 30 marbles including 6 red and 7 blue. A
marble is drawn at random. Find the following probabilities.
(a) The drawn marble is blue; (b) The drawn marble is not red; (c) The
drawn marble is neither red nor blue; (d) The drawn marble is red or blue.
Solution: Let S be the sample space. Then n (S ) = 30.
a. Let E1 be the event that the drawn marble is blue. Then n (E1 ) = 7.
7
.
Thus, Pr[E1 ] = 30
b. Let E2 be the event that the drawn marble is not red. Then
24
n (E2 ) = 30 6 = 24. Thus, Pr[E2 ] = 30
= 45 .
c. Let E3 be the event that the drawn marble is neither red nor blue.
Then n (E3 ) = 30 6 7 = 17. Thus,
17
.
30
d. Let E4 be the event that the drawn marble is red or blue. Then
n (E4 ) = 6 + 7. Thus,
13
.
Pr[E4 ] =
30
Pr[E3 ] =
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