Chapter 8: Further Topics in Algebra

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Chapter 8 Further Topics in Algebra

• Section 8.5 Probability

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Introduction to Probability Theory

• Probability Theory is the branch of mathematics
  that deals with the likelihood (probability) that
  events otherwise left to chance actually will or
  will not occur. Probability theory has
  far-reaching consequences in daily living, and
  its applications are practically innumerable.
• In order to study probability from a mathematical
  perspective and with mathematical precision, we
  need to define the concept of probability in a
  precise manner. To do so, we borrow from
  combinatorics the concept of an event, and
  introduce a new concept, the sample space.
• As in combinatorics, an event is simply an
  occurrence, and the term event can be used
  loosely to suit our purposes.
• A sample space is the set of all possible
  outcomes of an event. Often, we call the sample
  space S for mnemonic purposes. An example of a
  sample space is the set S which indicates all
  possible outcomes of a triple coin toss, S
  HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

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Introduction to Probability Theory (contd)

• Notice that the sample space S above has eight
  elements, and we say that the size (or
  cardinality) of S is 8, denoted n(S) 8. Notice
  also that while S was simple to enumerate (list
  out all the elements) in its totality, the
  Fundamental Counting Principle predicts a size of
  8 for S. (How?)Often, for more complicated
  sample spaces, we must use combinatorial methods
  to determine the size of the space.
• Example 1 Consider again the case of a
  5-question, multiple-choice quiz where each
  question has four possible answers A, B, C, or D.
  Let S be the sample space for this quiz,
  consisting of all possible outcomes of this
  quiz. (For example, one element of S might be
  ABCDB). Calculate n(S), the size of this sample
  space.
• Solution Since a student can answer any
  question in one of five ways (i.e., he/she can
  also choose not to answer certain questions),
  there are (5)(5)(5)(5)(5) 3125 possible
  outcomes for this quiz. That is, n(S) 3125.
• With these concepts under our belt, we can
  proceed to precisely define what we mean by the
  probability of an event.

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Definition of Probability of an Event

• Probability of an Event
• If the outcomes of a finite sample space S are
  equally likely, and if E is an event in S, then
  the probability of E is given by P(E)
  n(E)/n(S), where n(E) and n(S) represent the
  number of outcomes in E and S, respectively.
• An easier way to look at the definition of the
  probability of an event is that this probability
  P(E) of an event E actually occurring is equal to
  the number of ways in which E does occur out of
  the total number of outcomes possible.
• Example 2 A fair coin is tossed three times.
  What is the probability of obtaining at least one
  head?
• Solution Since the sample space is of size 8,
  we know that n(S) 8. We must determine n(E),
  which is done by counting the number of trials
  that include one, two, or three heads. There
  are 7 such trials, so n(E) 7. Thus, P(E)
  n(E)/n(S) 7/8. Our odds are 7 out of 8 of
  obtaining at least one head.
• Notice that we often express probabilities in
  decimal or percentage form. For example 2 above,
  we would have said that the probability of at
  least one head is 0.875, or that there is an
  87.5 chance of obtaining at least one head.
• It is important to note that, because of the way
  probability is defined, P(E) for any event E will
  always be between 0 and 1. If P(E) 0, then E is
  impossible. If P(E) 1, then E will definitely
  occur.

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Computing Probabilities

• Example 3 Suppose, as before, that a class
  consists of 10 students, each of which is
  expected to give a brief oral presentation on
  some topic in mathematics. Jean-Pierre, a student
  in the class, is experiencing considerable
  anxiety about the possibility of his being the
  first to have to present. In order to alleviate
  his stress, he attempts to ascertain the
  probability that he will have to go first. What
  is this probability?
• Solution Following the definition of the
  probability of an event, we must determine n(E),
  the number of orders which place Jean-Pierre
  first, as well as n(S), the total number of
  orders in which the presentations can be
  delivered. We previously determined that n(S)
  3,628,800. To determine n(E), we place
  Jean-Pierre first and count the remaining orders.
  There are 9! 362,880 orders which place J.-P.
  first thus, n(E) 362,880. Therefore P(E)
  n(E)/n(S) 362,880/3,628,800 0.1. There is a
  10 chance that J.-P. will have to go first.

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Computing Probabilities (contd)

• Example 4 (text p. 637, Example 1). One card is
  drawn from a standard deck of 52 cards. Find the
  probability that the card is an ace.
• Solution Since there are C(52, 1) 52 ways to
  choose one card out of 52, there are 52 possible
  outcomes. That is, n(S) 52. To compute n(E),
  we note that there are four aces in the deck,
  and thus 4 ways in which we may choose an ace.
  (We may choose an ace of hearts, clubs, diamonds,
  or spades). Therefore n(E) 4. Hence, the
  probability of drawing an ace is P(E)
  n(E)/n(S) 4/52 1/13 0.077 (approximately).
  We have a 7.7 chance of drawing an ace.
• Sometimes we wish to compute the probability that
  an event E will not occurthat is, that its
  complementary event E will occur. E is called
  the complement of E, and we compute P(E) in the
  following manner
• Probability of a Complement
• Let E be an event and E be its complement. If
  the probability of E is P(E), then P(E) 1-P(E).

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Probability of a Complement

• The definition given above makes sense in the
  real world as well. Given any event E, either E
  will occur or it will not occur. (Principle of
  Excluded Middle). If E does not occur, then E
  occurs (since E is the event signaling that E
  doesnt occur). That is, the probability that
  either E or E will occur is 1 (i.e., 100).
  Thus the probability that E will occur must
  simply be the leftover probability after weve
  considered the likelihood of E.
• Notice that the probability that E and E will
  occur is necessarily 0.
• Example 5 Since Jean-Pierre has discovered that
  there is a 10 chance that he will have to go
  first, he easily determines that there is a 90
  chance (1 1/10 9/10 0.9 90) that he will
  not have to go first. Jean-Pierre finds this
  very reassuring.
• Example 6 A card is drawn from a standard deck
  of 52 cards. Find the probability that the card
  is not an ace.
• Solution Since the probability that the card
  is an ace is 1/13, the probability that the card
  is not an ace is given by 1 P(E) 1 1/13
  12/13 0.923 (approximately). There is a 92.3
  chance that the card drawn will not be an ace.

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Probability of a Complement (contd)

• Example 7 Suppose, yet again, that a
  5-question, multiple choice quiz is given, and
  that for each question there are four
  possibilities of answer. Suppose further that
  Marie-Hélène went out drinking last night, didnt
  study, has a hangover, and simply guesses at each
  question on the quiz. What is the probability
  that Marie-Hélène will make something less than
  100 on the quiz?
• Solution In order to make something less than
  100, M.-H. must make anything but 100. Let E be
  the event that signals M.-H. making 100 on the
  quiz. Then we are looking for E (not E ), and
  P(E ) will be given by P(E) 1 P(E). So we
  must first determine P(E). Since there are
  (5)(5)(5)(5)(5) 3125 possible quiz outcomes,
  n(S) 3125 and since only one such quiz will
  signal a score of 100, n(E) 1. Thus P(E)
  1/3125. Then P(E) 1 1/3125 0.99968. There
  is a 99.968 chance that M.-H. will not score
  100 on the quiz.

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Unions and Intersections

• Two important set-theoretic concepts,
  particularly useful in probability studies, are
  the concepts of union and intersection.
• The union of two sets is the set containing all
  members of either set. That is, an element a will
  be in the union of sets A and B if a is in A or
  in B. (It is often useful to think of a union of
  sets as an or condition, and to read the union
  symbol as or).
• Example 8 Let A -3, 0, 1, 2, B 1, 5, 9.
  Then the union of A and B is given by
• The intersection of two sets is the set
  containing only those members which are in both
  sets. That is, an element a will be in the
  intersection of sets A and B if a is in A and a
  is in B. (It is often useful to think of an
  intersection of sets as an and condition, and
  to read the intersection symbol as and).
• Example 9 Let A and B be defined as above. Then
  the intersection of A and B is given by

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Independent Events and Compound Probabilities

• When two events do not influence each other in
  any way, they are called independent events. An
  example of independent events would be the triple
  coin toss. No single toss is affected in any way
  by the toss that precedes or follows it.
• When finding the probability that both of two
  independent events will occur, we compute the
  probability of their intersection as follows
• Probability of Independent Events
• If E1 and E2 are independent events, then the
  probability that E1 and E2 will both occur is
  given by
• An easier way of thinking about this compound
  probability is that the probability of both of
  two independent events occurring is simply the
  product of the probabilities of the individual
  events.

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Compound Probabilities (contd)

• Example 10 Reconsider our triple coin toss.
  What is the probability that the first and third
  tosses will be heads?
• Solution Since no toss will be affected by
  other tosses, we may consider each toss as an
  independent event. Let E1 denote the first toss
  and let E2 denote the third toss. (Note that,
  since the tosses are independent of each other,
  we may completely ignore the second toss in this
  problem). Now, for E1 there are only two
  possible outcomes (i.e., heads or tails), so
  that if S1 is the sample space for E1, n(S1)
  2. Clearly, n(E1) 1 (there is only way to toss
  a head), so that P(E1) ½ 0.5. Similarly,
  P(E2) 0.5, and so
• Notice that we can easily confirm this
  particular probability by inspecting the sample
  space S for the entire triple toss. As noted
  above, S HHH, HHT, HTH, HTT, THH, THT, TTH,
  TTT, and the number of triple tosses which
  signal a first and third toss of H is 2/8 ¼
  0.25. There is a 25 probability that the first
  and third tosses will be heads.

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Compound Probabilities (contd)

• Sometimes, we wish to compute the probability
  that one or another event will occur (or both).
  In order to do so, we compute the probability of
  the union of the events in question. (Remember
  that a union can be viewed as an or condition).
• Probability of the Union of Two Events
• For any two events E1 and E2,
• Note that if E1 and E2 are mutually exclusive
  events, then
• by necessity (it is impossible for both events to
  occur), and so

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Compound Probabilities (contd)

• Example 11 (text p. 641, Example 6). A single
  card is drawn from a standard deck of 52 cards.
  Find the probability that it is either an ace or
  a king.
• Solution Since only one card is being drawn,
  and that card cannot be both an ace and a king,
  we know immediately that . We
  must compute P(E1) (the event that signals
  drawing an ace) and P(E2) (the event that
  signals drawing a king). We have already
  determined that P(E1) 1/13 (since there are
  4/52 ways to draw an ace). Similarly, since
  there are four kings in the deck, P(E2) 1/13.
  Then
• That is, there is about a 15.4 chance of
  drawing either an ace or a king.

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Compound Probabilities (contd)

• Example 12 What is the probability of drawing a
  flush on the first deal of a 5-card hand out of a
  standard deck of 52 playing cards?
• Solution The is a compound probability problem.
  A flush consists of 5 cards all of the same suit,
  and there are 4 suits in a standard deck. Thus we
  may look at the problem as determining the
  probability of drawing 5 hearts or 5 diamonds or
  5 clubs or 5 spades. That is, we seek the
  probability of a union of events E1, E2, E3, E4,
  where E1 signals drawing five hearts, E2 five
  diamonds, E3 five clubs, and E4 five spades. To
  determine the number of ways we can draw five
  hearts, consider that there are 13 cards in the
  suit of hearts. Thus there are C(13, 5) 1287
  ways to draw five hearts. Thus n(E1) 1287.
  Similarly, there are 1287 ways to draw five
  diamonds, 1287 ways to draw five clubs, and 1287
  ways to draw five spades. Therefore n(Ek) 1287
  for k 1, 2, 3, 4. Since there are, as
  previously noted, C(52, 5) 2,598,960 ways to
  draw any five cards out of 52, we have n(S)
  2,598,960. Hence, P(E1) n(E1)/n(S)
  1287/2598960 0.000495 (approximately).
  Similarly, P(Ek) 0.000495 for k 1, 2, 3, 4.
  Clearly, the events Ek are mutually exclusive, so
  that we neednt concern ourselves with
  intersections. Then

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Dependent Events and Compound Probabilities

• When an event is influenced in some way by a
  preceding event, we say that the events in
  question are dependent events. Computing the
  probability of dependent events is similar to
  computing the probability of independent events,
  except that we must know or determine the
  conditional probability of the second event,
  given that the first event occurred. That is
• Probability of Dependent Events
• If E1 and E2 are dependent events, then
• The expression is read the
  probability of E2 given that E1 occurred, and is
  called the conditional probability of E2 given
  E1.

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Dependent Events and Compound Probabilities
(contd)

• Example 13 (text p. 644, Example 10). Find the
  probability of drawing two hearts from a standard
  deck of 52 cards, when the first card is not
  replaced.
• Solution Clearly, all possibilities for the
  first card are equally likely. There are C(52,
  1) 52 ways to draw a single card from a deck of
  52 cards, so n(S) 52. Since we want to know
  the probability that the card is a heart, we
  note that there are 13 hearts in the deck. That
  is, n(E1) 13, and so P(E1) n(E1)/n(S)
  13/52. Since we do not replace the first card,
  however, the second card must be drawn from a
  different deck this one consisting only of 51
  cards and containing only 12 hearts. Thus P(E2)
  12/51, given E1. Then
• Thus, there is about a 5.9 chance of drawing
  two hearts when the first card is not replaced.
• Homework for this section