October 29, 2008
October 28, 2008
Statistics Notes — Probability Examples using the Addition Rule and Conditional Probability
Addition Rule:
If events A and B are mutually exclusive (disjoint), then P(A or B) = P(A) + P(B). Otherwise, P(A or B) = P(A) + P(B) – P(A and B)
Example 1: mutually exclusive
In a group of 101 students 30 are freshmen and 41 are sophomores. Find the probability that a student picked from this group at random is either a freshman or sophomore.
Note that P(freshman) = 30/101 and P(sophomore) = 41/101. Thus P(freshman or sophomore) = 30/101 + 41/101 = 71/101
Example 2: not mutually exclusive
In a group of 101 students 40 are juniors, 50 are female, and 22 are female juniors. Find the probability that a student picked from this group at random is either a junior or female.
Note that P(junior) = 40/101 and P(female) = 50/101, and P(junior and female) = 22/101. Thus P(junior or female) = 40/101 + 50/101 – 22/101 = 68/101
Not sure why? When we add 40 juniors to 50 females and get a total of 90, we have overcounted. The 22 female juniors were counted twice; 90 minus 22 makes 68 students who are juniors or female.
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Conditional Probability
Recall that the probability of an event occurring given that another event has already occurred is called a conditional probability.
In a card game, suppose a player needs to draw two cards of the same suit in order to win. Of the 52 cards, there are 13 cards in each suit. Suppose first the player draws a heart. Now the player wishes to draw a second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a deck of 51 cards. So the conditional probability P(Draw second heart|First card a heart) = 12/51.
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Suppose an individual applying to a college determines that he has an 80% chance of being accepted, and he knows that dormitory housing will only be provided for 60% of all of the accepted students. The chance of the student being accepted and receiving dormitory housing is defined by
P(Accepted and Dormitory Housing) = P(Accepted) × P(Dormitory Housing|Accepted) = (0.80) × (0.60) = 0.48
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To calculate the probability of the intersection of more than two events, the conditional probabilities of all of the preceding events must be considered. In the case of three events, A, B, and C, the probability of the intersection P(A and B and C) = P(A) × P(B|A) × P(C|A and B).
Consider the college applicant who has determined that he has 0.80 probability of acceptance and that only 60% of the accepted students will receive dormitory housing. Of the accepted students who receive dormitory housing, 80% will have at least one room mate. The probability of being accepted and receiving dormitory housing and having no room mates is calculated by:
P(Accepted and Dormitory Housing and No Roommates) = P(Accepted) × P(Dormitory Housing|Accepted) × P(No Room mates|Dormitory Housing and Accepted) = (0.80) × (0.60) × (0.20) = 0.096. The student has about a 10% chance of receiving a single room at the college. Not happening.
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The probability that event B occurs, given that event A has already occurred is
P(B|A) = P(A and B) / P(A)
This formula comes from the general multiplication principle and a little bit of algebra. I showed you on an earlier post.
Since we are given that event A has occurred, we have a reduced sample space. Instead of the entire sample space S, we now have a sample space of A since we know A has occurred. So the old rule about being the number in the event divided by the number in the sample space still applies. It is the number in A and B (must be in A since A has occurred) divided by the number in A. If you then divided numerator and denominator of the right hand side by the number in the sample space S, then you have the probability of A and B divided by the probability of A.
The question, “Do you smoke?” was asked of 100 people. Results are shown in the table.
………….Yes No Total
Male ..19 41 ..60
Female 12 28 ..40
Total ..31 69 100
What is the probability of a randomly selected individual being a male who smokes?
This is just a joint probability. The number of “Male and Smoke” divided by the total = 19/100 = 0.19
What is the probability of a randomly selected individual being a male?
This is the total for male divided by the total = 60/100 = 0.60. Since no mention is made of smoking or not smoking, it includes all the cases.
What is the probability of a randomly selected individual smoking?
Again, since no mention is made of gender, this is a marginal probability, the total who smoke divided by the total = 31/100 = 0.31.
What is the probability of a randomly selected male smoking?
This time, you’re told that you have a male – think of stratified sampling. What is the probability that the male smokes? Well, 19 males smoke out of 60 males, so 19/60 = 0.3167
What is the probability that a randomly selected smoker is male?
This time, you’re told that you have a smoker and asked to find the probability that the smoker is also male. There are 19 male smokers out of 31 total smokers, so 19/31 = 0.6129
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Statistics Notes — Probability Examples using the Addition and Multiplication Rules
Try the following examples. If you hold the left click of your mouse down and slide the mouse under the choices, you will see the answers.
1. According to National Center for Health Statistics, between 1988 and 1991, 33.3% of U.S. adults 20 years of age and over were overweight. If two adults 20 years of age and over in the U.S. are selected at random, what is the probability that both are overweight?
a. P(both overweight) = 11.1%
b. P(both overweight) = 88.9%
c. P(both overweight) = 0.0%
d. P(both overweight) = 100.0%
P(2 overweight) = P(overweight) × P(overweight) = 0.333 × 0.333 = 0.110889 = 11.0889% (a)
2. A bag of M & M’s in my office had the following color break down, 6 red, 4 blue, 4 green, 7 yellow, 1 orange, and 24 brown. If an M & M is randomly selected, what is the probability that the M & M is either red or blue?
a. P(red or blue) = 0.130
b. P(red or blue) = 0.087
c. P(red or blue) = 0.152
d. P(red or blue) = 0.217
P(red or blue) = P(red) + P(blue) = 6/46 + 4/46 = 10/46 = 0.2173913043 (d)
3. Given a well shuffled deck of standard playing cards, what is the probability of randomly selecting either a red card or a queen?
a. P(red card or queen) = 0.577
b. P(red card or queen) = 0.038
c. P(red card or queen) = 0.462
d. P(red card or queen) = 0.538
P(red or queen) = P(red) + P(queen) – P(red queen) = 26/52 + 4/52 – 2/52 = 28/52 = 0.5384615385 (d)
4. Given a well shuffled deck of standard playing cards, what is the probability of randomly selecting an even numbered card?
a. P(even numbered card) = 0.385
b. P(even numbered card) = 0.423
c. P(even numbered card) = 0.0770
d. P(even numbered card) = 0.462
P(even numbered card) = P(2) + P(4) + P(6) + P(8) + P(10) = 4/52 + 4/52 + 4/52 + 4/52 + 4/52 = 20/52 = 0.3846153846 (a)
5. According to The Digest of Education Statistics 1996, 78.0% of U.S. 13-year-olds were able to perform numerical operations and beginning problem solving. If five 13- year-olds are randomly selected, what is the probability that none of them can perform numerical operations and beginning problem solving?
a. P(none can perform beginning problem solving) = 0.0005.
b. P(none can perform beginning problem solving) = 0.9995.
c. P(none can perform beginning problem solving) = 1.0000.
d. P(none can perform beginning problem solving) = 0.7800.
P(none of 5) = 0.22 × 0.22 × 0.22 × 0.22 × 0.22 = 0.0005153632 (a)
6. According to the Labor Force Statistics from the Current Population Survey, in 1996, 52.8% of families have both the husband and wife employed. If 3 famililes are randomly selected, what is the probability that they all have both spouses employed?
a. P(both employed) = 47.2%.
b. P(both employed) = 14.7%.
c. P(both employed) = 85.3%.
d. P(both employed) = 52.7%.
P(3) = 0.528 × 0.528 × 0.528 = 0.147197952 (b)
7. According to National Center for Health Statistics, 39% of the births occurring in the last 5 years to women who were betweeen 15 – 44 years of age in 1998, were unintended. (SOURCE: “Unintended Pregnancy and Childbearing” by Linda J. Piccinino, Lynn Wilcox, and James Marks, edits, From Data to Action.) If 6 new mothers are randomly selected, what is the probability that at least one of their births was unintended?
a. P(at least one is unintended) = 94.8%.
b. P(at least one is unintended) = 99.6%
c. P(at least one is unintended) = 0.35%.
d. P(at least one is unintended) = 5.15%
P(at least one) = P(1) + P(2) + P(3) + P(4) + P(5) + P(6). This is not as easy as it seems. It is not at all like the card problem earlier. The best way and the only way, right now, is to use the idea of complements.
P(at least one) is the complement to P(none). P(at least one) = 1 – P(none) = 1 – (0.61 × 0.61 × 0.61 × 0.61 × 0.61 × 0.61) = 0.9484796256 (a)
8. When answering questions involving the multiplication rule of probability, what is the “key” question that you should ask yourself?
a. Are the two events mutually exclusive?
b. Are the two events independent?
c. Do I care about these two events?
d. None of the above.
Multiplication Rule — the concern is dependent and independent events.
Addition Rule — the concern is mutually or non-mutually exclusive events.
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Statistics Notes — Summary of basic probability rules for events A and B
A statistical experiment or statistical observation is any random activity that results in a recordable outcome. The sample space is the set of all possible distinct outcomes. Events are subsets of the sample space.
1. P(entire sample space) = 1
2. For any event A: 0 ≤ P(A) ≤ 1
3. Complement of A: P(not A) = 1 – P(A)
4. Events A and B are independent events if P(A) = P(A, given B)
5. Multiplication Rules
If A and B are independent events: P(A and B) = P(A) × P(B)
If A and B are dependent events: P(A and B) = P(A) × P(B | A)
Equivalently; P(B | A) = P(A and B)/P(A)
6. Events A and B are mutually exclusive if P(A and B) = 0
7. Addition Rules
If A and B are mutually exclusive events: P(A or B) = P(A) + P(B)
If A and B are not mutually exclusive: P(A or B) = P(A) + P(B) – P(A and B)
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Statistics Notes — Common probability examples using the Addition Rule 2
An event is a set of outcomes. It is a subset of the sample space for an activity or experiment.
Event: Drawing a black card from a deck of standard cards.
Probability of this event = 26/52 = 1/2
When an event corresponds to a single outcome of the activity, it is often called a simple event.
Simple Event: Drawing the queen of spades from a deck of standard cards.
Probability of this event = 1/52
Two events that have NO outcomes in common are called mutually exclusive. These are events that cannot occur at the same time.
Example: A pair of dice is rolled. The events of rolling a 6 and of rolling a double have the outcome (3,3) in common. These two events are NOT mutually exclusive.
A pair of dice is rolled. The events of rolling a 9 and of rolling a double have NO outcomes in common. These two events ARE mutually exclusive.
For any two mutually exclusive events, the probability that an outcome will be in one event or the other event is the sum of their individual probabilities.
If A and B are mutually exclusive events, P(A or B) = P(A) + P(B)
For any two events which are not mutually exclusive, the probability that an outcome will be in one event or the other event is the sum of their individual probabilities minus the probability of the outcome being in both events.
If events A and B are NOT mutually exclusive, P(A or B) = P(A) + P(B) – P(A and B)
Example 1: A pair of dice is rolled. What is the probability that the sum of the numbers rolled is either 7 or 11?
Six outcomes have a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
P(7) = 6/36
Two outcomes have a sum of 11: (5,6), (6,5)
P(11) = 2/36
The sum of the numbers cannot be 7 and 11 at the same time, so these events are mutually exclusive.
P(7 or 11) = P(7) + P(11) = 6/36 + 2/36 = 8/36 = 2/9 = 0.2222
Example 2: A pair of dice is rolled. What is the probability that the sum of the numbers rolled is either an even number or a multiple of 3?
Of the 36 possible outcomes, 18 are even sums.
P(even) = 18/36 = 1/2
Sums of 3, 6, 9, and 12 are multiples of 3. There are 12 sums that are multiples of 3.
P(multiple of 3)= 12/36 = 1/3
However, some of these outcomes appear in both events. The sums that are even and a multiple of 3 are 6 and 12. There are 6 ordered pairs with these sums.
P(even AND a multiple of 3) = 6/36 = 1/6
P(even OR a multiple of 3) = P(even) + P(multiple of 3) – P(even and a multiple of 3) = 18/36 + 12/36 – 6/36 = 24/36 = 2/3 = 0.6667
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Georges-Louis Leclerc, Comte de Buffon — Buffon’s Needle
Buffon’s Needle
Buffon’s Needle is a problem first posed by the French naturalist and mathematician, the Comte de Buffon.
Georges-Louis Leclerc, Comte de Buffon (September 7, 1707 – April 16, 1788) was a French naturalist, mathematician, biologist, cosmologist and author. Buffon’s views influenced the next two generations of naturalists, including Jean-Baptiste Lamarck and Charles Darwin. Darwin himself, in his foreword to the 6th edition of the Origin of Species, credited Aristotle with foreshadowing the concept of natural selection but also stated that “the first author who in modern times has treated it in a scientific spirit was Buffon.”
Georges-Louis Leclerc or Compte de Buffon was born to an aristocratic family in Montbard, France on September 7, 1707. A man of many interests and talents, Buffon first studied law and then medicine, botany, and mathematics.
Buffon’s greatest scientific work was Histoire Naturelle, (Natural History), published in 36 volumes between 1749 and 1789. This work is one the first scientific treatments of natural history, giving analyses of geology, zoology, and botany without reference to the Bible.
Buffon died on April 16, 1788 in Paris.
In mathematics, Buffon’s famous coin problem and needle problem are considered to be among the first problems in geometric probability, and greatly stimulated the development of the subject.
Buffon’s needle experiment is a very old and famous random experiment. Consider a plane, ruled with equidistant parallel lines, where the distance between the lines is D. A needle of length L is tossed onto the plane. The experiment consists of dropping a needle on a hardwood floor. The main event of interest is that the needle crosses a crack between floorboards. Strangely enough, the probability of this event leads to a statistical estimate of the number pi.
The Experiment: We assume that the floorboards are uniform and that each has width of 1. We will also assume that the needle has length L less than 1 so that the needle cannot cross more than one crack. Finally, we assume that the cracks between the floorboards and the needle are line segments.
Use the following applet with the default settings.
http://www.math.uah.edu/stat/applets/BuffonNeedleExperiment.xhtml
Notice the value given for pi in the far right window. It may not very close. Change the Stop 10 to Stop 1000 and run the experiment again. Notice the value for pi change. It may get close, and then get farther away, but the final value is closer to pi than the first value you saw. If you ran this experiment longer, you would see the value, which is the results of a probability experiment, get closer to pi. Outstanding.
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October 27, 2008
Statistics Notes — Common probability examples using the Addition Rule
Common probability examples using the Addition Rule
1. You are going to pull one card out of a deck. Find P(Ace or King).
The addition rule says we need to find P(Ace) + P(King) – P(both).
P(Ace) = 4/52
P(King) = 4/52
P(both at the same time) = 0
P(Ace or King) = 4/52 + 4/52 – 0 = 8/52 = 2/13
2. You are going to roll two dice. Find P(sum that is even or sum that is a multiple of 3).
The addition rule says we need to find P(even) + P(multiple of 3) – P(both).
P(even) means how many ways to roll 2, 4, 6, 8, 10, or 12.
P(even) = 18/36
P(multiple of 3) means how many ways to roll 3, 6, 9 or 12.
P(multiple of 3) = 12/36
P(both) means what is the overlap. Notice that 6 and 12 occur in both places and have been counted twice. We need to subtract those out. This subtraction eliminates the overlap.
P(both) = 6/36
So P(sum that is even or a multiple of 3) = 18/36 + 12/36 – 6/36 = 24/36 = 2/3.
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Try the following examples. If you hold the left click of your mouse down and slide the mouse beside the problems, you will see the answers.
Example: The first three problems involve drawing a card from an ordinary deck of cards.
1a. P(two or red) = P(two) + P(red) – P(two that is red) = 4/52 + 26/52 – 2/52 = 28/52 = 0.5385
1b. P(three or jack) = P(three) + P(jack) – P(three that is a jack) = 4/52 + 4/52 – 0 = 8/52 = 0.1538
1c. P(club or four) = P(club) + P(four) – P(club that is a four) = 13/52 + 4/52 – 1/52 = 16/52 = 0.3077
Example: The last two problems involve rolling two dice.
2a. P(sum that is odd or sum that is even) = P(sum that is odd) + P(sum that is even) – P(sum that is odd and even) = 18/36 + 18/36 – 0 = 36/36 = 1
2b. P(sum of 4 or sum that is a multiple of 3) = P(sum of 4) + P(sum that is a multiple of 3) – P(sum of 4 that is also a multiple of 3) = 3/36 + 12/36 – 0 = 15/36 = 0.4167
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Lesson Plans — Week 11 — October 27 – 31, 2008
Lesson Plans — Week 11 — October 27 – 31, 2008
Pre-Calculus
Standards: Models for Real World Phenomena; Algebraic Functions; Trigonometric Functions; Sequences and Series
Monday, October 20: 3.7 Use the Remainder and Factor Theorems; Use the Rational Zeros Theorem to list the potential rational zeros of a polynomial function; Find the real zeros of a polynomial function; Solve polynomial equations
Tuesday, October 21: 3.8 Utilize the conjugate pairs theorem to find the complex zeros of a polynomial; Find a polynomial function with specified zeros; Find the complex zeros of a polynomial function
Wednesday, October 22: 3.8 Utilize the conjugate pairs theorem to find the complex zeros of a polynomial; Find a polynomial function with specified zeros; Find the complex zeros of a polynomial function
Thursday, October 23: Review
Friday, October 24: TEST
Geometry
Standards: Number and Operations; Algebra; Geometry; Measurement; Data Analysis and Probability
Monday, October 20: 4.1 Recognize congruent figures and their corresponding parts
Tuesday, October 21: 4.2 and 4.3 Prove two triangles congruent using the SSS and SAS Postulates; Prove two triangles congruent using the ASA Postulate and the AAS Theorem
Wednesday, October 22: 4.1, 4.2, and 4.3 Recognize congruent figures and their corresponding parts; Prove two triangles congruent using the SSS and SAS Postulates; Prove two triangles congruent using the ASA Postulate and the AAS Theorem
Thursday, October 23: 4.4 Use triangle congruence and CPCTC to prove that parts of two triangles are congruent
Friday, October 24: 4.4 Use triangle congruence and CPCTC to prove that parts of two triangles are congruent
Statistics
Standards: Experimental Design; Data Analysis
Monday, October 20: 3.2 Find the probability of an event given that another event has occurred; Distinguish between independent and dependent events; Use the Multiplication Rule to find the probability of two events occurring in sequence; Use the Multiplication Rule to find conditional probabilities
Tuesday, October 21: 3.2 Find the probability of an event given that another event has occurred; Distinguish between independent and dependent events; Use the Multiplication Rule to find the probability of two events occurring in sequence; Use the Multiplication Rule to find conditional probabilities
Wednesday, October 22: 3.3 Determine if two events are mutually exclusive; Use the Addition Rule to find the probability of two events
Thursday, October 23: 3.3 Determine if two events are mutually exclusive; Use the Addition Rule to find the probability of two events
Friday, October 24: 3.3 Determine if two events are mutually exclusive; Use the Addition Rule to find the probability of two events
For all classes.
Activities: Lecture, Board work, Classroom participation
Material: Book problems and Teacher made handouts
Assessment: Daily assignments graded for accuracy. Unit test planned.
In order to address the Physical Activity requirement… My classes and I will enjoy a 15 minute walk once a week.
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