Chapter 6: Section 6-3 Sample Spaces - Creighton University · Chapter 6: Section 6-3 Sample Spaces...

Chapter 6: Section 6-3Sample Spaces

D. S. MalikCreighton University, Omaha, NE

D. S. Malik Creighton University, Omaha, NE ()Chapter 6: Section 6-3 Sample Spaces 1 / 22

Experiment

Experiment: An activity or occurrence with an observable result. Thefollowing 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 4green marbles and recording the color.

3 Making a phone call and observing whether the call was answered bya person.

4 A family has three children. Record the number of boys and girls.

Definition(i) A repetition of an experiment is called a trial.(ii) The results of a trial are called outcomes.


Sample Space

DefinitionThe set of all possible outcomes of an experiment is called a samplespace.

Example(i) A coin is tossed and the top face is observed. Let H stand for the headand T stand for the tail. Then the sample space, S , is the set:S = {H,T}.(ii) A six-faced die is rolled and the number rolled is recorded. The samplespace is: S = {1, 2, 3, 4, 5, 6}.(iii) A coin is tossed three times. An experiment consists of recording thenumber of heads and tails. The sample space is:

S = {3H and 0T , 2H and 1T , 1H and 2T , 0H and 3T}.


ExampleA coin is tossed three times. An experiment consists of recording the orderin which heads and tails occurred. Suppose that HHH means that all threetosses are heads, HTH means that the first toss is a head, the second tossis a tail, and the third toss is a head, and so on. The sample space is:

S = {HHH, HHT , HTH, HTT , THH, THT , TTH, TTT}.

RemarkFrom the preceding examples, it is clear that the sample space of anexperiment depends on the types of outcomes that we want to observe.For example, suppose that the experiment is tossing a coin three times. Ifwe are interested in counting the number of heads and tails, then eachelement of the sample space consists of number of heads and tails. On theother hand, if we are interested in recording the order in which the headsand tails occur, then the sample space consists of elements in the previousexample.


Event

DefinitionLet S be a sample space of an experiment. An event of the experiment orthe sample space is a subset of the sample space S .

ExampleA coin is tossed twice. Then the sample space is:

S = {HH,HT ,TH,TT}.

Let E be the event that the second toss is a tail. Then

E = {HT ,TT}.

Let F be the event that each outcome come consists of at least one head.Then

F = {HH,HT ,TH}.


ExampleA six-faced die is rolled. Then the sample space is:

S = {1, 2, 3, 4, 5, 6}.

Let E be the event that the number rolled is 2. Then

E = {2}.

Note that n(E ) = 1.Let F be the event that the number rolled is a prime number. Then

F = {2, 3, 5}.

Note that n(F ) = 3.


ExampleA card is drawn from a well-shuffl ed deck of 52 cards. The sample space isthe set of 52 cards. Let E be the event that the drawn card is an ace or ared 10. Let H stand for a heart, D stand for a diamond, S stand for aspade, and C stand for a club. Also 1C means the ace of clubs, and so on.Then

E = {1H, 1D, 1S , 1C , 10H, 10D}.Note that n(E ) = 6.Let F be the event that the drawn card is a club or a queen. Then

F = {1C , 2C , 3C , 4C , 5C , 6C , 7C , 8C , 9C ,10C , JC , QC , KC , QH, QD, QS}.



ExampleTwo six-faced dice are rolled and the numbers rolled are recorded. Thenthe sample space is:

S =

(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),

.

Let E be the event that the first die shows a 2 and the second die showsan odd number. Then E = {(2, 1), (2, 3), (2, 5)}. Note that n(E ) = 3.Let F be the event that the sum of the numbers rolled is 8. Then

F = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}.



ExampleA bag contains 3 red and 4 green marbles. Suppose that the red marblesare 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 eventthat the drawn marble is red. Then

E = {R1,R2,R3}.


Sample Spaces and Tree Diagrams

Suppose that a coin is flipped twice. The outcomes of this experimentcan be determined using a tree.

When the coin is flipped the first time, it has two outcomes —a heador a tail, which are represented by two branches as shown in thefollowing figure.

At the end of the branch we write the outcome.

Suppose that the first flip resulted in a head. So we are at the end ofthe top branch. The coin is flipped the second time and the outcomesare —a head or a tail. So we again draw two branches from thisoutcome and label the end of these branches with H and T . Now ifthe first outcome was a tail, then for the second flip, we draw twobranches out of the bottom branch and label the ends with H and Trespectively. Next, in the tree, we traverse each path from left toright and write the outcome resulting from that path, see thefollowing figure.


HH

(a)(b)

H

T

(c)

First FlipHT

Second Flip Outcome

HT

HT TT

TH

H

T

First FlipHT

Second Flip

HT

H

T

First Flip


ExampleA 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 thebox, and then selecting a marble from the bag. The tree diagram of thisexperiment is:

aRa

b

Selecta Bag R

GB

Select aMarble Outcome

RB

aGaB

bRbB

YW

bYbW

The sample space is: S = {aR, aG , aB, bR, bB, bY , bW }.D. S. Malik Creighton University, Omaha, NE ()Chapter 6: Section 6-3 Sample Spaces 12 / 22

ExampleA bag contains red, green, and blue marbles. An experiment consists ofrolling 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 beconsidered two-step experiment. In the first step, a die is rolled and in thesecond step a marble is drawn.When a die is rolled, the number rolled has 6 possibilities. So the first stepcan be completed in 6 ways. Next, the bag contains three types ofmarbles—red, green, and blue. So when a marble is drawn, there are threepossibilities for the color. Thus, the second step can be completed in 3ways. Now for each choice of the first step, there are three choices for thesecond step. Hence, the entire experiment can be completed in

6 · 3 = 18 ways.

Hence, n(S) = 18.


Theorem(Multiplication Principle) Suppose that a multistep experiment consistsof k (independent) steps. Suppose that the first step can be completed inn1 ways, the second step can be completed in n2 ways regardless of theresults of step 1, the third step can be completed in n3 ways regardless ofthe 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 first i − 1steps. Then the entire experiment can be completed in

n1 · n2 · · · · · nk

ways.


Example(i) A bakery offers 3 types of cookies and 4 types of cakes. A special dealconsists of one type of cookie and one type of cake. An experimentconsists of creating a special deal.This is a two-step experiment. In the first step, we select a cookie, and inthe second step, we select a cake. There are 3 choices for the first stepand 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 adie. Let S be the sample space of this experiment.This is a four-step experiment. In the first 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 ofways this experiment can be done is:

2 · 2 · 2 · 6 = 48.

Hence, n(S) = 48.D. S. Malik Creighton University, Omaha, NE ()Chapter 6: Section 6-3 Sample Spaces 15 / 22

Probability Theory: Some Basic Concepts

Throughout we assume that the sample space is finite.

Suppose that a die is rolled. Then the sample space is:

S = {1, 2, 3, 4, 5, 6}.

Suppose that you are asked the following question: when a die isrolled, what are the chances (or likelihood) that the number rolled isa 2.

Let E be the event that the number rolled is 2. Then E = {2}.We can now reformulate our question as follows: What are thechances that the event E will occur?

In mathematical terms, this question is posed as follows: What is theprobability that the number rolled is a 2 or what is the probabilitythat the even E will occur?


Probability Theory: Some Basic Concepts

DefinitionLet S be a (finite) sample space of an experiment. Suppose that alloutcomes of the experiment are equally likely. Let E be an event of theexperiment. Then the probability that the event E will occur, writtenPr[E ], is

Pr[E ] =n(E )n(S)

.

ExampleSuppose that a fair die is rolled. Let E be the event that the rollednumber is odd. Then E = {1, 3, 5}, so n(E ) = 3. If S is the sample spaceof this experiment, then n(S) = 6. Hence,

Pr[E ] =n(E )n(S)

=36=12.


Remark(Drawn at random) While discussing probabilities, very often we usephrases such as “a card is drawn at random”or “a marble is drawn atrandom.”When we use words “drawn at random,” then it should beunderstood that all outcomes of the sample space are equally-likely. So alloutcomes have the same probability.

ExampleSuppose that a card is drawn at random from a deck of 52 cards. Let S bethe sample space. Then n(S) = 52.(i) Let E be the event that the drawn card is an ace. Then n(E ) = 4.Hence, Pr[E ] = n(E )

n(S ) =452 =

113 .

(ii) Let F be the event that the drawn card is a heart or an ace. There are13 hearts and 4 aces. However, one of the aces is also a heart. Thus,n(F ) = 13+ 4− 1 = 16. Hence,

Pr[F ] =n(F )n(S)

=1652=413.


ExampleA bag contains 4 red, 6 blue, and 7 green marbles. A marble is drawn atrandom. 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 redmarbles and if the drawn marble is red, then it can be any one of these redmarbles. Thus, n(E ) = 4. Hence,

Pr[E ] =n(E )n(S)

=417.

(ii) Let F be the event that the drawn marble is red or blue. There are 4red marbles and 6 blue marbles. Thus, n(F ) = 10. Hence,

Pr[F ] =n(F )n(S)

=1017.


TheoremLet S be a (finite) 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 ′] = 1− Pr[E ].

Notation: Let S be a sample space and E be an event in S . Supposethat E = {a}, i.e., the number of elements in E is 1. Thensometimes we write Pr[a] for Pr[E ].

Therefore, if a die is rolled, then Pr[1] means the probabilitythat the number rolled is 1. Similarly, if a card is drawn froma deck of 52 cards, then Pr[K ] means the probability thatthe drawn card is a king. When a coin is tossed, then Pr[H ]means the probability that the top face is a head.


Exercise: Suppose that two dice are rolled. Find the following events.(a) The sum of the numbers rolled is 7; (b) The product ofthe numbers rolled is 6; (c) The sum of the numbers rolled is13.

Solution:

a. Let E be the event that the sum of the numbers rolled is 7. Then

E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}.Here the ordered pair (a, b) means that the first die rolls an a and thesecond die rolls a b. For example, (2, 5) means that the first die rollsa 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 = {(1, 6), (2, 3), (3, 2), (6, 1}.c. Let G be the event that the sum of the numbers rolled is 13. Thelargest number on each die is a 6. So the largest sum of the numbersrolled is 12. Hence,

G = ∅.D. S. Malik Creighton University, Omaha, NE ()Chapter 6: Section 6-3 Sample Spaces 21 / 22

Exercise: An urn contains 30 marbles including 6 red and 7 blue. Amarble is drawn at random. Find the following probabilities.

(a) The drawn marble is blue; (b) The drawn marble is not red; (c) Thedrawn 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.Thus, Pr[E1] = 7

30 .

b. Let E2 be the event that the drawn marble is not red. Thenn(E2) = 30− 6 = 24. Thus, Pr[E2] = 24

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,

Pr[E3] =1730.

d. Let E4 be the event that the drawn marble is red or blue. Thenn(E4) = 6+ 7. Thus,

Pr[E4] =1330.


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