S3.amazonaws.com



Module 6 – Sample Spaces and Events

Definition: a random experiment is an experiment that produces an outcome that cannot be predicted in advance

Note: The word “experiment” here just means that “something is being done, and we don’t know in advance exactly what will happen” – it does NOT mean the kind of statistical study called an experiment (where there is random sample and the researcher applies a treatment to some of the subjects).

Definition: a sample space is the list of all possible simple outcomes of a random experiment.

Sample spaces are called S and are listed in set braces { … }

It is always important to consider in listing a sample space: Does order matter?

There is no rule for this. You must understand the situation and see if the order of the various events does matter or not for the particular experiment.

Examples: Answers on next page

List the sample space for these random experiments:

Ex. 1: Toss a die and see what lands on the top side (a die is a six-sided cube with dots on the faces – the plural is dice)

Ex. 2: Toss a coin three times

Ex. 3: In the family of Mom, Dad, Susie, Tran, and Grandma, randomly select two people to do the dishes today.

Ex. 4: The online question can be answered until you get it right, but with a maximum of 5 attempts.

Ex. 5: In the family of Mom, Dad, Susie, Tran, and Grandma, randomly select one person to wash the floor and one person to vacuum.

An event is something that could happen when the random experiment is done. An event might be a “simple outcome” or it might be more complicated (it might be a collection of some of the simple outcomes in the sample space).

A capital letter is used to denote an event, but not the letter S.

The following examples refer to the random experiments described above.

Ex. 1: Toss a fair die and see what lands on the top side

Consider these events, and list the simple outcomes of the sample space that give the event.

Event E = “get an even number” =

Event B = “get the number 2” =

Event H = “get a number greater than 4”

When the simple outcomes listed in a sample space are equally likely, then the probability of an event can be found using the Theoretical (Classical) approach. For an event A:

[pic]

For our last example, find the probability of each event.

Ex. 2: Toss a coin three times.

To find these probabilities, find the simple outcomes in the sample space that are in the event.

P(at least one Tail was tossed) =

P(at least one Tail and at least one Head are tossed) =

Ex. 3: In the family of Mom, Dad, Susie, Tran, and Grandma randomly select two people to do the dishes today.

Find the probability of these events.

Event A = both people selected are adults

Event J = both people chosen are adults or both are children

Ex. 4: The online question can be answered until you get it right, but with a maximum of 5 attempts.

Event T = it takes three attempts to get it right.

Can we find the probability of T?

First consider: are the simple outcomes in the sample space equally likely?

Ex. 5: In the family of Mom, Dad, Susie, Tran, and Grandma, randomly select one person to wash the floor and one person to vacuum.

Find the probability of these events.

Event U = both people selected are adults

Event V = both people chosen are adults or both are children

ANSWERS

List the sample space for these random experiments:

Ex. 1: Toss a die and see what lands on the top side (a die is a six-sided cube with dots on the faces – the plural is dice)

{1, 2, 3, 4, 5, 6} ( this is a complete list of the six ways the die might land

Ex. 2: Toss a coin three times

Let H be “the coin lands Heads on this toss” and T be “the coin lands Tails on this toss”.

The coin is definitely tossed three times, and the first time it lands one way (H or T), the second toss lands either H or T, and the third toss lands H or T.

Sample space = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Note 1: this is a case where the order of the H or T in an outcome makes a difference. For example: If you toss the coin 3 times and the first time it lands Heads then the next two tosses land Tails, that is NOT the same as tossing the coin 3 times and the first two times it lands Tails and the last time it lands Heads. Outcome HTT is not the same thing as TTH.

Note 2: the 8 items in the sample space could have been listed in a different order. But it is really good to be able to list them in an organized way so you don’t miss any.

Ex. 3: In the family of Mom, Dad, Susie, Tran, and Grandma, randomly select two people to do the dishes today.

Use M, D, S, T, and G to abbreviate the people’s names.

Note: In this case the order of two people selected makes no difference. If M is chosen first then D, then M and D do the dishes. That is the same result as when D is chosen first and M second – it is still the case that M and D do the dishes! So in the sample space, MD is listed once showing that those two people do the dishes. We don’t list DM separately since that would be the same result (the same two people doing the dishes).

Sample Space = { MD, MS, MT, MG, DS, DT, DG, ST, SG, TG}

Note: the 10 items in this list could have been written in a different order – but it is good to write them in an organized way.

Ex. 4: The online question can be answered until you get it right, but with a maximum of 5 attempts.

Let C = question answered correctly; W = question answered incorrectly.

Sample Space = { C, W C, W W C, W W W C, W W W W C, W W W W W }

Ex. 5: In the family of Mom, Dad, Susie, Tran, and Grandma, randomly select one person to wash the floor and one person to vacuum.

In this sample space, the first person listed will wash the floor and the second person listed will vacuum. So note that order matters. MD means Mom washes floor and Dad vacuums. That is a different thing than DM when Dad washes floor and Mom vacuums.

Sample Space = {MD, MS, MT, MG, DM, DS, DT, DG, SM, SD, ST, SG,

TM, TD, TS, TG, GM, GD, GS, GT }

There are 20 items in the sample space. Be organized about writing them.

ANSWERS continued

Ex. 1: Toss a fair die and see what lands on the top side

Consider these events, and list the simple outcomes of the sample space that give the event.

Event E = “get an even number” = { 2, 4, 6}

Event B = “get the number 2” = { 2 }

Event H = “get a number greater than 4” = {5, 6}

The events in the sample space are equally likely when the die is fair, so we can find the probabilities:

P(E) = 3/6, or 50% or .5 or ½

P(B) = 1/6

P(H) = 2/6, or 1/3 or 33.3%

Ex. 2: Toss a coin three times.

To find these probabilities, find the simple outcomes in the sample space that are in the event.

P(at least one Tail was tossed) = 7/8

Note: at least one Tail means 1 or 2 or 3 Tails.

Outcomes in this event are: HHT, HTH, HTT, THH, THT, TTH, TTT

Note: It might be easier to think of what is NOT in this event: HHH. So that means the other 7 items in the sample space must be in the event.

P(at least one Tail and at least one Head are tossed) =6/8

Outcomes in the event are: HHT, HTH, HTT, THH, THT, TTH

Ex. 3: In the family of Mom, Dad, Susie, Tran, and Grandma randomly select two people to do the dishes today.

Event A = both people selected are adults

Outcomes in this event are: MD MG DG, so P(A) = 3/10

Event J = both people chosen are adults or both are children

Outcomes in this event are: MD MG DG ST, so P(J) = 4/10

Ex. 4: The online question can be answered until you get it right, but with a maximum of 5 attempts.

Event T = it takes three attempts to get it right.

Can we find the probability of T? No

First consider: are the simple outcomes in the sample space equally likely? No, the simple outcomes are not equally likely. Suppose somebody was simply guessing – then it is not at all likely they’d get the question correct the first time. It is more likely that after trying a couple wrong answers they catch on and then get it right.

Ex. 5: In the family of Mom, Dad, Susie, Tran, and Grandma, randomly select one person to wash the floor and one person to vacuum.

Find the probability of these events.

Event U = both people selected are adults

Outcomes in this event are: MD MG DM DG GM GD.

So P(U) = 6/20, or .3 or 30%

Event V = both people chosen are adults or both are children

Outcomes in this event are: MD MG DM DG GM GD ST TS

So P(V) = 8/20 = .4 or 40%

Module 6 (Addition Rule)

1. Example – using the Theoretical (aka Classical) approach “aka” = “also known as”

|Example |

|Procedure is: a hat has10 pieces of paper in it, |

|each with one of these letters on it: |

|m, n, o, p, q, r, s, t, u, v |

|and the experiment (or procedure) is to randomly select one of the papers. |

What is the Sample Space?

Event A = get a vowel =

P(A) =

Event B = get one of letters in the word “stop” =

P(B) =

ANSWERS S ={m, n, o p, q, r, s, t, u, v} P(A) = P(get or u) = 2/10 = .2 or 20% P(B) = P( get s, t, o, or p) = 4/10 = .4 = 40%

2. Now consider the “compound event” of A or B

[Note: a compound event is a combination of some other events]

P(A or B) means P(event A occurs or event B occurs or they both occur)

Using the basic concepts of probability:

P ( A or B) = [pic]

• In the example above:

A or B occurs if get: s, t, o, p or u ( in 5 ways

Sample space has a total of 10 simple events

So P (A or B) = Answer: 5/10

• How does P(A or B) relate to the separate P(A) and P(B) ?

There is one item in the overlap of events A and B

(namely the letter “o”)

So P(A and B) = 1/10

Notice: P(A) + P(B) - P(A and B) = 2/10 + 4/10 - 1/10

= 5/10 = P(A or B)

| |

|“General Addition Rule” |

|P( A or B) = P(A) + P(B) - P(A and B) |

| |

|“Intuitive Addition Rule” |

|To find P(A or B), |

|find the sum of the number of ways event A can occur and the number of ways event B can occur, adding in such a way that every outcome |

|is counted only once. |

|P(A or B) is equal to that sum, divided by the total number of outcomes in the sample space. |

Extra Example: Toss penny, nickel, dime. List sample space to get this answer.

Event C= penny lands H.

Event D = get 2 Heads. This means exactly 2.

P(C or D)=

Answer to Extra Example: Let outcomes be listed first for the penny, then nickel, then dime.

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

P(C) = 4/8 P(D) = 3/8 P(C or D) = P(C) + P(D) – P(C and D) = 4/8 + 3/8 – 2/8 = 5/8

Note: P(C and D) = P(penny is H and there are 2 H) = P(HHT, HTH) = 2/8

3. Considering P(A and B)

P(A and B) = P(event A occurs and event B occurs)

A and B is the “overlap” of events A and B, or their “intersection”.

In the General Addition Rule, we need to subtract the probability of the “overlap” of A and B only if there is an overlap.

If A and B do NOT overlap,

they are called mutually exclusive or disjoint

Def: Events A and B are called disjoint or mutually exclusive if they cannot occur at the same time (that is, they do not overlap).

If events A and B are disjoint, then P(A and B) = 0.

Example: of disjoint events for the Procedure of randomly selecting one letter from m, n, o, p, q, r, s, t, u, v

C = get a letter in the word “run” P(C) = 3/10

B = get a letter in the word “stop” P(B) = 4/10

You can see that events C and B do not overlap.

C and B are disjoint (they are mutually exclusive).

So P(C and B) = 0

So P (C or B) = P(C) + P(B) – P(C and B)

= 3/10 + 4/10 - 0 = 7/10

4. When events A and B are disjoint (which means P(A and B) = 0),

Then the addition rule becomes: P(A or B) = P(A) + P(B)

This form of the addition rule, for disjoint events, can be extended to more than two events.

| |

|If events A, B, and C are disjoint: P(A or B or C) = P(A) + P(B) + P(C) |

| |

|NOTE: OLI textbook p. 104 is very important. |

|Do ALL of the questions and activities on this page. |

|The topic is distinguishing between disjoint events and independent events. |

5. Definition: When events A and B are independent means

the fact that one event has occurred does not affect the probability that the other event will occur.

If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent, or not independent.

Example 1: Toss a penny and a nickel.

Event A = the penny lands heads.

Event B = the nickel lands heads.

Are A and B independent events? _______

Yes, since the penny landing Heads does not affect how the nickel lands.

Example 2: My daughter put 2 bad batteries in the bag with the 4 good batteries. I reach in and grab one battery. Then I replace it in the bag (put it back in the bag), mix them up, and grab a battery.

Event A = the first battery I got was good.

Event B = the second battery I got was good.

Are A and B independent? ______

Yes. Event A happening or not doesn’t make any difference for the second battery’s probabilities since the first battery is replaced back in the bag. So the second battery is always chosen from the among the 2 bad and 4 good batteries, no matter whether the first had been good or bad.

Example 3: My daughter put 2 bad batteries in the bag with the 4 good batteries. I reach in and grab one battery and then reach in and grab a second battery. I do this without replacing the first battery in the bag – since I want to use both batteries.

Event A = the first battery I got was good.

Event B = the second battery I got was good.

Are A and B independent? _______

No. If Event A occurs then the first battery was good, and that battery is NOT put back in the bag. So the second battery is chosen from the remaining 3 good and 2 bad – so P(B) = 3/5. However, if Event A does NOT occur then the first battery was bad, and that battery is not in the bag when the second is chosen. So the second battery is chosen from the remaining 4 good and 1 bad – so P(B) = 4/5. Notice: the P(B) is different depending on whether or not A occurred.

Example 4: Toss a penny and a nickel.

Event A = the penny lands heads.

Event B = the penny lands tails.

Are A and B independent events? _______

No. If Event A occurs, then the penny landed heads. Then the P(B) = P(the penny lands tails) = 0 since the penny cannot land tails since it landed heads. However, if Event A does not occur then the penny did not land heads, so it must have landed tails, so P(B) = 1. The P(B) is different depending on whether or not A occurred. [Note: the penny and nickel are tossed only once in this “experiment”, so there is only one way that the penny lands.]

• For each example above, are the events disjoint?

(same asking: are the events mutually exclusive?)

Ex 1: Not disjoint (they can occur at the same time: penny lands heads and nickel lands heads too)

Ex 2: Not disjoint (both events can occur – I can get both good batteries)

Ex 3: Not disjoint (both events can occur – I can get both good batteries)

Ex 4: Disjoint (the penny cannot land heads and also land tails)

Note: instead of saying “disjoint” we could have said “mutually exclusive” – they mean the same thing.

Summarize results: Ex. 1: Independent; Not Disjoint

Ex. 2: Independent; Not Disjoint

Ex. 2: Not Independent; Not Disjoint

Ex. 2: Not Independent; Disjoint

Summary of our results above re Disjoint and Independent:

| |A, B Independent |A, B Not Independent |

|A, B Disjoint |DOES NOT EXIST |Example 4 above |

|A, B Not Disjoint |Examples 1 & 2 above |Example 3 above |

Why is “Disjoint and Independent” impossible?

If A and B are disjoint, it means that they do not overlap, meaning they cannot both occur at the same time. In that case, if one of them occurs, then the other one can NOT occur. So, if we know event A has occurred, that does affect the probability that event B occurs (since it makes P(B) = 0 ). And thus A and B are Not independent in this case when they are disjoint.

Summary:

• If A and B are Disjoint, then they are Not Independent.

• If A and B are Independent, then they are Not Disjoint.

• A and B might be neither Disjoint not Independent (i.e., A and B might be Not Disjoint and Not Independent).

• IA and B can never be both Independent and Disjoint.

6.

| |

|The Multiplication Rule for Independent Events: |

|If A and B are two independent events, then P(A and B) = P(A) • P(B). |

Quiz Example: Consider a Quiz with two questions. The first is a true-false question. The second question is multiple choice with 5 choices.

If a person randomly guesses the answers, find P(both answers are correct).

This is the same as saying to find

P(first answer is correct and second answer is correct).

Think of: event A = first answer is correct

event B = second answer is correct

To find the probability, first consider: are these events A and B independent? ______

Yes they are independent. One of the events occurring (one answer is guessed correctly) does not affect the probability that the other answer is guessed correctly, since both are simply random guesses that don’t affect each other.

P(A and B) = ?

P(A and B) = P(A)•P(B) since they are independent = ½ • (1/5) = 1/10

College Major Example: Suppose a college has these proportions of students in various types of majors:

|Type of Major |Science |Arts-Humanities |Health Services |

|Proportion of students |0.35 |0.20 |0.45 |

Two students are chosen simultaneously and at random. What is the probability that both have the same type of major? Note: we must assume this college has a large number of students, so that we can treat the selecting of two students as independent events. (i.e., selecting one student without replacement can be treated as selecting with replacement since there are so very many students.) See page below re “Determining if two events are independent…”.

Answer is at end of these notes – try it on your own first (it’s complex, but not tricky).

Module 6

7. Complements

The Complement of event A consists of all the outcomes that are NOT in event A.

Event A and the Complement of event A, put together, comprise all possible outcomes. Therefore:

P(A) + P(not A) = 1. This leads to:

The Rules for Complements:

P(A) = 1 – P(not A)

P(not A) = 1 – P(A)

• Sometimes it is easier to find the probability of a complement than the original event – not always but sometimes. In those cases, we can find the easy probability of the complement, and use that to figure out the probability of the event.

- Often the complement’s probability is easier to get if the original event was about “at least one” of something. Or if the original event was about “at most” some number of something.

Probability of “at least one”

“at least one” means 1 or 2 or 3 or … to the highest # possible

If an event is “get at least one of some characteristic Q”,

Its complement is “get none of that same characteristic Q”

Poodle Example

Suppose that in one (strange) town, 20% of the dogs are poodles. If 6 dogs are randomly selected from the town, what is the probability that at least one is a poodle?

Facts given:

P(when you select one dog, it is a poodle) = 20% = .2

P(when you select one dog, it is not a poodle) = _____ Answer: .8

When 6 dogs are randomly selected, we want to find

P(at least one dog selected is a poodle).

Let event A = at least one dog selected is a poodle.

So A = 1 or 2 or 3 or 4 or 5 or 6 dogs selected are poodles.

The complement of A = no dogs selected are poodles

So event “not A” = the six dogs selected are NOT poodles.

P(not A) = P(all six dogs are not poodles)

= P(1st is not poodle AND 2nd is not poodle AND

3rd is not poodle AND 4th is not poodle AND 5th is not poodle

AND 6th is not poodle)

= P(1st not poodle)•P(2nd not poodle)•P(3rd not poodle)•

P(4th not poodle)•P(5th not poodle) •P(6th not poodle)

= ( )• ( )• ( )• ( )• ( )• ( ) = ________

P(A) = 1 - P(not A) = ________

In this example we used the Multiplication Rule for Independent Events.

WERE these events independent? Yes, since we assume the town has a lot of poodles and so selecting one poodle doesn’t change the probability of selecting another poodle.

ANSWER for this problem: P(not A) = (.8)(.8)(.8)(.8)(.8)(.8) = (.8)^6 = .262144

P(A) = 1 - .262144 = .737856 = .7379 rounded to 4 places

Determining if two events are independent or dependent

Sometimes you need to determine whether two events are independent or not. Often you need to just figure this out from context. Here are some ideas that might help.

1. If items are selected “with replacement” then the events are independent - since the group of items to select from for the second item is the same no matter what was chosen first since that first item was put back into the group before the second item was selected.

2. If items are selected “without replacement” then the events are dependent – since the original group now has the first item missing from it, so the group of items left to select the second item from depends on what was selected first that is now missing from the group.

3. If “two different” items are to be randomly selected, that means the selection is without replacement. Why? Because to select “different” items, the first one cannot be put back in (if it were put back in, then it might be chosen again – but the problem said to randomly select two different items). HOWEVER, if the group that the two items are selected from is very large, then … read the next item…

4. If a group of items is very large, then it doesn’t matter if the selection is made with or without replacement – we can always treat it as if it is “with replacement” (that is, independent events).

Why? Because the answers come out almost exactly the same – close enough for us.

Example: If there are 100,000 people in a city and 40,000 of them have green eyes, and we randomly select two different people (so this is without replacement), find the probability that both have green eyes. The most exact answer, using “without replacement” is: (40,000/100,000)∙(39,999/99,999) = 0.159976, which rounds to 0.1600

-However if we treat this “as if with replacement” the answer is (40,000/100,000)∙(40,000/100,000) = 0.1600

-So the two answers come out the same for at least four decimal places – and that is enough decimal places for our purposes.

5. When the given information simply states the percentage of a population that has a certain characteristic, but does not say how many are in the population … then we must assume the population is large and so the selection is “as if with replacement”. Why? If a population is not large (e.g. if it is only 100 or fewer) then we cannot do the problem unless we know how large it is. Since we are not told how large the population is, we assume it is large so that we can answer the question.

ANSWER to

College Major Example: Suppose a college has these proportions of students in various types of majors:

|Type of Major |Science |Arts-Humanities |Health Services |

|Proportion of students |0.35 |0.20 |0.45 |

Two students are chosen simultaneously and at random. What is the probability that both have the same type of major?

Note 1: we must assume this college has a large number of students, so that we can treat the selecting of two students as independent events. (i.e., selecting one student without replacement can be treated as selecting with replacement since there are so very many students.) See page re “Determining if two events are independent…”.

Note 2: You wouldn’t need to write all these steps, but you must think all of them and write some:

P(both have same type major)

There are three ways to have the same type of major

= P(both are Science or both are Art-Hum or both are Health Services)

Use the addition rule for P(A or B) and note that these events are disjoint (they have no overlap since one can’t have 2 people of one major and also 2 people of a different major)

= P(both are Science) + P(both are Art-Hum) + P(both are Health Services)

P(both are Science) = P(1st is Science and 2nd is Science)

Notice that the two events(of 1st is Science and of 2nd is Science) are independent ( as noted above – since there are many students in a college). So we can use the multiplication rule for independent events. …

= P(1st is Science) • P(2nd is Science)

= (0.35) • (0.35)

Similar reasoning applies to the other types of majors, so

P(both are Science) + P(both are Art-Hum) + P(both are Health Services)

= (0.35) • (0.35) + (0.20) • (0.20) + (0.45) • (0.45)

= .001225 + ,04 + .2025

= .243725

P(both have the same type of major) = .2437 rounded to four decimal places.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download