PROBABILITY

Document Sample
PROBABILITY Powered By Docstoc
					                                                                                              1
An Introduction to Probability


If you go to a supermarket and select 5 lbs of apples at $0.79 per lb, you can easily predict
the amount you will be charged (5 ∙ 0.79 = $3.95) at the checkout counter. The amount
charged for such purchases is a Deterministic Phenomenon. It can be predicted exactly on
the basis of the information given.

On the other hand, consider the problem faced by the produce manager of the supermarket,
who must order enough apples to have on hand each day without knowing exactly how
many pounds customer will buy during the day. The customer‟s demand is an example of
Random Phenomenon. The study of probability is concerned with such random
phenomenon. Even though we can‟t be certain, whether or not a given result will occur, we
can often obtain a good measure of its likelihood or probability.

The theory of mathematical probability originated from game of chance.


Experiment:        An experiment is the process by which an observation is obtained.
                   (note - the observation need not be numerical). Each repetition of an
                   experiment is called Trial.

Examples: 1.       Recording the test grades.
          2.       Tossing a coin and observing the side that appears.
          3.       Drawing a card from a standard deck.
          4.       Inspecting a toy to determine whether it is safe or not.

A random experiment has the following properties.

1.    It is performed according to well defined set of rules.

2.    It is repetitive in nature, theoretically infinite time.

3.    The result of each trial depend “on chance” and can therefore not be uniquely
      predicted.



Sample Space:      The Sample Space is the set of all the possibilities that can occur when
                   an experiment is performed. The Sample set is denoted by S and the
                   element of S is called Sample Point.
                                                                                                  2


Outcome:               Each individual occurrence of an experiment.


Event (E):             An event is a specific collection of sample points denoted by E. Also, E
                       is a subset of S.


Note: The Sample Space could be finite or infinite. For example, in tossing a pair of coins,
      there are 4 possible outcomes { HH, HT, TH, TT}, representing a finite sample
      space. Where as, the sample space of the experiment of measuring the strength of
      cotton is infinite because outcome may be any positive number within a certain
      range. Same holds in measuring the lifetime of a light bulb, or the height and weight
      of a person.


Some other examples of the sample space are:

   1.         Rolling a pair of dice (one red and one green). Total 36 possibilities in the
              sample space.

                              Green
                   1         2        3        4      5        6
          1      (1,1)     (1,2)    (1,3)   (1,4)   (1,5)   (1,6)
          2      (2,1)     (2,2)    (2,3)   (2,4)   (2,5)   (2,6)
        R 3      (3,1)     (3,2)    (3,3)   (3,4)   (3,5)   (3,6)
        e 4      (4,1)     (4,2)    (4,3)   (4,4)   (4,5)   (4,6)
        d 5      (5,1)     (5,2)    (5,3)   (5,4)   (5,5)   (5,6)
          6      (6,1)     (6,2)    (6,3)   (6,4)   (6,5)   (6,6)

        The above table is called Product Table.


   2.         A swimming team consists of four members: {A, B, C, D}. Choose two team
              members at random: {AB, AC, AD, BC, BD, CD}, six possibilities in the
              sample space.


In most practical problems, we are not so much interested in the individual outcomes, but
whether an outcome belong (or does not belong) to a certain set A (event) of the outcomes.
                                                                                            3
Notation for Probability: P denotes a probability; A, B, etc denotes specific events; and
P(A) denotes the probability of event A occurring.


Three different ways to define probability:

Rule 1: Classical definition of Mathematical probability

      Assume that a given experiment has n different simple events, each of which has an
      equal chance of occurring (called equally likely outcomes), Then probability P(A)
      of an event A is:

                  number of ways A can occur        s
      P(A) =                                      
               total number of simple events in S   n



      Another way: The outcomes that makeup the event A is called favorable outcomes
      and the remaining outcomes are called unfavorable outcomes to the event.
      If all outcomes are equally likely,

               number of favorable outcomes to A
      P(A) =
                total number of simple events in S



Example:    Roll a single Die. There are 6 equally likely outcomes in the sample space
            S = {1, 2, 3, 4, 5, 6}.

      a.    Probability of rolling a “2” P(roll a 2) = 1 / 6

      b.    Probability of rolling an odd number, P(A) = 3/6 = 1/2 = 0.5
            Where favorable outcomes to event A = {1, 3, 5}

      c.    Probability of rolling a number greater than 6, P(number > 6) = 0/6 = 0
            Such events are called impossible events.

      d.    P(whole number between 1 and 6 inclusive) = 6/6 = 1
            Such events are called certain events.
                                                                                                   4


In most cases this classical definition of probability is entirely appropriate still it has some
limitations.

        1.      Outcomes in the sample space have to be equally likely
        2.      The sample space has to be finite.

In these cases, Relative Frequency approach of probability is used.

Rule 2:         Empirical Probability (Relative Frequency Approximation)
Conduct (or observe) an experiment a large number of times, and count the number of
times event A actually occurs. Then P(A) is estimated (or approximated) as

                        number of times A occurred
        P(A) =
                  number of times exp eriment was repeated




Rule 3:         Subjective Probability

P(A), the probability of an event, is found by simply guessing or estimating its value based
on the information given. Weather forecasting is a good example of subjective probability.


Note: It is very important to note that the classical definition requires equally likely
     outcomes. If the outcomes are not equally likely, use the Empirical rule to
     estimate the probability.


Law of Large Numbers: As the N, the number of times experiment is repeated, is
increased; the Empirical probability (approximation) tends to get closer to the actual
probability.


Basic Properties of Probability

   1.        The probability of an impossible event is 0.
   2.        The probability of a certain event is 1.
   3.        For any event 0 ≤ P(A) ≤ 1
                                                                                               5
Complementary Event

The complement of an event A, denoted by A or A' , consists of all outcomes in which
event A does not occur.

      Since        P(A) + P( A ) = 1
                   P( A ) = P(not A) = 1 – P(A)



Example 1: Roll a pair of dice. Total 36 equally likely possibilities.

      P(rolling a double) = 6/36 = 1/6
      P(not rolling a double) = 1 – P(rolling a double) = 1 – 1/6 = 5/6


Example 2: Flip three different coins or flip one coin three times.

      Sample Space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
      Eight equally likely possibilities.

      P(all heads) = p( HHH ) = 1/8
      P(not getting all heads) = 1 – 1/8 = 7/8



Probability of “at least one”
Example 1: A coin is tossed five times; find the probability of getting at least one “Head”.

           Since each coin turns up H or T, there are 25 = 32 equally likely outcomes.
           If event A: at least one “Head”,
           then complement of A (not A): no “Head”  all “Tails”
           Since only one outcome will have all Tails; P(all Tails) = 1 /32
           Using complementary rule;
                   P(at least one “Head”) = 1 – P(all Tails)
                                          = 1 – 1/32 = 31/32


Note: In general, the complementary rule is useful when the event of interest can be
     expressed as “at least one”.
                                                                                             6
Example 2: If a family has six children, find the probability that at least one boy in the
     family?

       There are 26 = 64 equally likely outcomes.
       Since the complement of “at least one boy” is “all girls”
       P(at least one boy) = 1 – P(all girls)
                          = 1 – 1/26 = 1 – 1/64 = 63/64



ODDS

If an event is twice as likely to occur than not to occur, we say that the odds are 2 to 1.
In betting, the word “Odds” is also used to denote the ratio of the wager of one party to that
of another. If a gambler says that he/she will give 3 top 1 odds on the occurrence of an
event, the gambler means that he/she is willing to bet $3 against $1 (or perhaps $30 against
$10) that the event will occur. That is, if you win, you make $3 for each $1 bet.

If E is an event, then
                                number of outcomes in E         n( E )     P( E )
      Odds in favor of E =                                            
                              number of outcomes not in E       n( E )     P( E )


      It is represented as n(E): n( E )

                          number of outcomes not in E       n( E )     P( E )
      Odds against E =                                            
                            number of outcomes in E         n( E )     P( E )


      Note: Only whole numbers are used in Odds. Odds represent a ratio of two
      whole numbers.


The Payoff Odds against event E represent the ratio of net profit (if you win) to the
amount bet. Payoff Odds against event E = (net profit) : (amount bet)


Example: Let event E is rolling a pair of dice and getting a double.

                                    n( E )   6 1
      Odds in favor of a double =          =   or 1: 5
                                    n( E ) 30 5

                                    30 5
      Odds against of a double =       or 5 : 1
                                    6 1
                                                                                               7
                  6 1
      P(double) =     0.16
                  36 6

Note: Probability and odds are two different quantities. Odds represent a ratio of two
numbers not a fraction. Where as, probability is a fraction and could be also represented as
a decimal number.

Review Examples for Probability and Odds

Example 1: A card is drawn at random from a deck of 52 cards. Find the following
           probabilities and odds of the events:

      a.    P(red card) =

            where as Odds in favor of red card =

      b.    P(a face card) =
            P(not a face card) = 1 – P(a face card) =

            where as Odds against of a face card =

      c.    P(a face card and a club) =
                  ↑
              K, Q, J of club

      d.    P(a face card or a club) =
                   ↑
              12 face cards + 10 other clubs

      Note in each case 0 ≤ P(A) ≤ 1 is true.


Example 2: Roll a pair of dice and let Sum represent the sum of two faces. The sample
           space can be represented as:

          Sum: 2           3       4      5    6   7      8     9      10    11     12
Number of ways: 1          2       3      4    5   6      5     4      3     2      1

      a. P(rolling a sum 9) =

      b. P(rolling a sum 7 or 11) =

      c. P(rolling a sum 7 and 11) =
                                                                                            8
Sec 3.3: Addition Rule


P(A or B) = P(A) + P(B) – P(A and B)
    ↑                          ↑
 either A or B or both A and B   A and B both occur at the same time

If event A and B are disjoint or mutually exclusive (can not both occur at the same time)
then P(A and B) = 0, and P(A or B) = P(A) + P(B)


Example 1: Draw one card at random from a deck of 52 cards. Find the following
           probabilities:

      a.       Probability of drawing a Club or a Heart

               Using Classical definition: P(Club or Heart) = 26/52 = ½

               Using Addition rule:
                 P(Club or Heart) = P(club) + P(Heart) – P(Club and Heart)
                                   = 13 / 52 + 13 / 52 – 0 / 52
                                   = 26 /52 = ½

      Note: Here, event drawing a Club and drawing a Heart are mutually exclusive
      events.


      b.       Probability of drawing a Club or a Jack

               Using Classical definition: P(Club or Jack) = 16/52

               Using Addition rule:
                 P(Club or Jack) = P(club) + P(Jack) – P(Club and Jack)
                                 = 13 / 52 + 4 / 52 – 1 / 52
                                 = 16 / 52


Example 2: A company has 66 employees in the following departments:
           Maintenance      14          Office Staff      8
           Management       7           Production        37

      If one employee is chosen at random,
      a. What is the probability that employee works in Maintenance or Production?
                                                                                            9
      The events are mutually exclusive because an employee cannot be in maintenance
      and in production at the same time.
      Using Addition Rule: P(Maintenance or Production) =

      b. What is the probability that employee works in Maintenance and Production?
          P(Maintenance and Production) =


Example 3: Out of 36 people applying for the job, 20 are men and 16 are women. Eight of
           the men and 12 of the women have Ph.D.‟s. If one person is selected at random
           for the first interview, find the probability that

             First organize the information:              M            W      Total
                                              Ph.D.       8            12     20
                                          Not Ph.D.       12           4      16
                                              Total       20           16     36
             a.    the one chosen has a Ph.D.
                         P(PhD) =

             b.    the one chosen is a woman and has a Ph.D.
                         P(W and PhD) =

             c.    the one chosen is a woman or has a Ph.D.
                   P(W or PhD) = P(W) + P(PhD) – P(W and PhD)
                                =

Example 4: If 47% of the people read a newspaper (N) every day, 62% of the people watch
           the TV news, and 25% do both. Find the probability of selecting a person at
           random and getting someone who reads the newspaper or watches the news on
           TV.
           Using Addition Rule:     P(N or TV) = P(N) + P(TV) – P(N and TV)


Example 5: Of the 20 television programs to be aired this evening, Marc plans to watch
            one, which he will pick at random by throwing a dart at TV schedule. If 8 of
            the programs are educational, 9 are interesting, and 5 are both educational and
            interesting, find the probability that the show he watches will have at least one
            of these attributes.
            If E represent “educational” and I represent “interesting”, then
            P(E) = 8/20, P(I) = 9/20, and P(E and I) = 5/20
            P(E or I) =
Multiplication Rule
                                                                                              10


For Multiplication rule the key word is AND. P(A and B) represent the probability that
both event A and B occur.

Independent Events: Two events A and B are independent if knowing that one occurs
does not change the probability that the other occurs.
If event A and B are independent then, P(A and B) = P(A) P(B)

1.     A coin is flipped and a die is rolled, find the probability of getting a Head on the
       coin and a 4 on the die.

      Sample Space = {(H, 1), (H, 2),….,(H, 6), (T, 1), (T, 2),….., (T, 6)}
      Twelve equally likely outcomes.

      Using Classical Definition       P(H, 4) = 1/12

                                                                   1 1 1
      Using Multiplication Rule        P(H and 4) = P(H) P(4) =      
                                                                   2 6 12



2.    A box contains 3 red balls, 2 blue balls, and 5 white balls. A ball is selected at
      random, its color noted, and then replaced back in the box. A second ball is selected
      and its color is noted (called with replacement case which implies independent
      events because there is no effect of the first selection on the second selection). Find
      the probability of each of the following events.

a.    Selecting 2 blue balls.

      P(select a blue and then, another blue) = P(B) P(B) =


b.    Selecting a blue ball and then a white ball.

      P(select a blue and then, a white) = P(B) P(W) =
                                                                                          11


3.   The probability that a person selected at random did not graduate from high school is
     0.25. If three people are selected at random, find the probability that

a.   all three do not have a high school diploma.

     Since each person is independent ,
     P(all three do not have a high school diploma) =


b.   all three have a high school diploma.

     P(all three have a high school diploma) =
Note: Events in Part a and b above are not complement of each other.

c.   at least one has high school diploma.

     P(at least one has HS diploma) = 1 – P(none have HS diploma)
                                   =


4.   The probability that a medical test will show positive is 0.12. If four people are
     tested at random, find the probability that

a.   all four will show positive.

     P(all positive) =
b.   at least one show negative.

     P(at least one show negative) = 1 – P(all positive) =
c.   at least one show positive.

     P(at least one show positive) = 1 – P(all negative) =
                                                                                                    12
Many times events are not independent; the occurrence of the first event affects or changes
the probability of the occurrence of the second event. Such events are called dependent
events or not independent events.
Now, consider the example 2 above with some changes;

          A box contains 3 red balls, 2 blue balls, and 5 white balls. A ball is selected at
          random, its color noted, and then without replacing it back in the box; a second ball
          is selected and its color is noted (called without replacement case which implies
          dependent events because the first selection will effect the probability of the second
          selection). Find the probability of each of the following events.

a.        Selecting 2 blue balls.

          P(select a blue and then, another blue) = P(B) P(B / B at the first )
                                                 =


          P(B / B at the first ) is called conditional probability of selecting a blue ball given
          that a blue ball was selected at the first draw.
          Knowing the fact that a blue ball was selected at the first draw, probability of
          selecting another blue ball at the second draw is changed to 1/9.

b.        Selecting a blue ball and then a white ball.

          P(select a blue and then, a white) = P(B) P(W / B) =


          Knowing the fact that a blue ball was selected at the first draw, probability of
          selecting a white ball at the second draw is 5/9.


Some other examples of dependent events.

     1.      Being a lifeguard and getting a suntan.

     2.      Having high grades and getting a scholarship.

     3.      Parking in a no-parking zone and getting a parking ticket.
                                                                                             13
General Multiplication Rule:

When two events are dependent            P(A and B) = P(A) P(B/A)
                           Or            P(B and A) = P(B) P(A/B)

Where P(B/A) represent the conditional probability of event B given that event A has
already occurred.
Given that event A has occurred, the relevant sample space is no longer S but consists of
outcomes in A only.

1.     In a shipment of 25 microwave ovens, 2 are defective. If two ovens are randomly
       selected and tested without replacing the first one back, find the probability that

a.    both are defective

      P(both defective) = P(1st defective) P(2nd defective / 1st defective)
                        =



b.    first one is defective but second is not.

      P(first defective but second not) = P(1st def.) P(2nd non def. / 1st def.)
                                        =



c.    one is defective and one is non defective

      Since the order is not given, consider all possible cases.

     P(1st def) P(2nd non def / 1st def) or P(1st non def) P(2nd def / 1st non def)
                                                                                             14
2.   The World Wide Insurance Company found that 53% of the residents of a city had
     homeowner‟s insurance with the company. Of these clients, 27% also had car
     insurance with the company. If a resident is selected at random, find the probability
     that the resident has both homeowners and car insurance with the World Wide
     Insurance Company.

     Given: P(homeowner‟s insurance) =
            P(car insurance / homeowner‟s insurance) =
            P(homeowners and car insurance) =



3.   Box one contains 2 red balls and 1 blue ball. Box two contains 3 blue balls and 1 red
     ball. A coin is flipped. If it falls heads up, box one is selected and a ball is drawn. If
     it falls tails up, box two is selected and a ball is drawn. Find the probability of
     selecting a red ball.

     Use a tree diagram to organize the given information

                          2/3     R


      1/2     B1
                                  B
                            1/3

                   1/ 4           R
      1/2     B2

                      3/ 4
                                  B




     P(R) =
                                                                                        15
From General Multiplication Rule P(A and B) = P(B) P(A/B)
                                                             P( A and B)
The Conditional probability of A given B is P(A/B) =
                                                                 P( B)
                       number of outcomes in A and B
      Also P(A/B) =
                          number of outcomes in B


NOTE: If P(A/B) = P(A)  event A and B are independent.

4.    A recent survey asked 100 people if they thought women in the armed forces should
      be permitted to participate in combat. The results of the survey are shown in the
      table.

      Gender                yes         No             Total
      Male                  12          38              50
      Female                42          8               50

Find the probability that

a.    the respondent answered “Yes”, given that the respondent was a female

Using the above formula           P(yes / F) = 42 / 50
                                                         P(Y and F ) 42 / 100 42
      Using the probability form        P(yes / F) =                         
                                                            P( F )    50 / 100 50



b.    the respondent was a female, given that the respondent answered “Yes”

      P(F / Yes) =

c.    the respondent was a male, given that the respondent answered “No”
      P(M / NO) =



Note: 5% rule for independence

If a sample size is no more than 5% of the size of the population, treat the selection as
being independent (even if the selections are made without replacement).
                                                                                              16
Sec 3.7      Counting Methods


1.    By Listing: Simply by identifying all the different items in a list and counting
                  them.

Example 1: Flipping three coins: S = {HHH, HHT, ………., TTT}, 8 possibilities.

Example 2: Rolling a pair of dice, there are 36 possibilities.

Example 3: Consider a team of 4 people: T = {A, B, C, D} all equally likely

      a.     How many ways they can select a Captain for the team?

             Since all team members are equally likely, there are 4 ways.

      b.     Number of ways they can select a Captain and a Leader (no one can hold both
             positions)

             Use a product table:                           Leader
                                       A             B            C             D

                                 A     –             AB           AC            AD
                   Captain       B     BA            –            BC            BD
                                 C     CA            CB           –             CD
                                 D     DA            DB           DC            –

             Total 12 possibilities.
             Here AB represent that A is a captain and B is a leader, where as
                  BA represent that B is a captain and A is a leader

                    AB  BA         order is important (Permutation)

             Can also use Slot filling method to find out total possibilities in this case.

             4 ways to select a Captain and then only 3 ways to select the Leader
                  4 · 3 = 12 ways      (and translate to multiplication)


      c.     Select a sub team of 2 members for a game (order is not important ie. AB =
             BA) can be done in 6 ways: AB, AC, AD, BC, BD, and CD.
                                                                                           17
2.    FUNDAMENTAL COUNTING PRINCIPLES


If event M can occur in m ways, after that, event N can occur in n ways, then event M
followed by event N can occur in m.n ways.
                                             OR

If first choice can be made in m ways and a second choice in n ways then the two choices
can be made together in m.n ways.

Note: This can be extended to any number of options.

EXAMPLES
1.    Local telephone number consists of 3 digits followed by 4 digits. How many
      numbers can be assigned before the telephone company runs out of numbers?

a.    When repetition of digits are allowed.

            10 · 10 · …… · 10 = 107 ways

b.    When repetition of digits are not allowed.

            10 · 9 · …… · 4

2.    In Kutztown, all telephone numbers start with 683 or 684. How many numbers can
      be assigned before the telephone company needs a new exchange.

            1 · 1 · 1 · 10 · 10 · 10 · 10 OR 1 · 1 · 1 · 10 · 10 · 10 · 10 = 20,000
           „6‟ „8‟ „3‟                      „6‟ „8‟ „4‟



3.    If a test consists of 12 true - false questions, in how many different ways can a
      student mark the test paper with one answer to each question?



4.    In how many ways a teacher could randomly assign the grades A, B, C, D, and F to
      40 students.
                                                                                    18
5.    Given the set of digits {0, 1, 3, 4, 5}, note set containing digit 0.

a.    How many three-digit numbers can be formed?

      Using slot filling:


b.    how many three-digit numbers can be formed
i.    if number must be even?




ii.   if number must be even and no repetition of digits is allowed?

            even numbers ending with „0‟ OR even numbers ending with „4‟



6.    Given the set of digits {1, 3, 4, 5, 6}, note set does not contain digit 0.

a.    How many three-digit numbers can be formed?

      Using slot filling:


b.    how many three-digit numbers can be formed
i.    if number must be even?




ii.   if number must be even and no repetition of digits is allowed?

            even numbers ending with „4‟ OR even numbers ending with „6‟
                                                                                            19


7.    Assuming that in a mid size car, the driver and one passenger sit in the front seat and
      three passengers sit in the back seat.

a.    In how many ways can you and your four friends seat themselves in this car?



b.    Suppose that the owner of the car insists on driving. How many arrangements are
      possible now?



c.    Suppose the owner drives and Mary Smith, having short legs, is forced to sit in the
      middle of the back seat, How many arrangements are possible under these
      conditions?




Note: In example 7, there was only one set of items (5 people). We arranged all of them
      and the order is important. This represents another counting method called
      Permutation.


2.    PERMUTATION :              Ordered arrangements of elements

The number of permutations (or arrangements) of r items selected from n available items
is:   n Pr or P(n, r) ; where r  n


Note: Calculate n Pr directly from the calculator or use slot filling method.
                                                                   n!     n!
When r = n (arranging all the items)                n Pn   =               n!
                                                               ( n  n) ! 0 !

1.     How many four letter words can be formed, using each of the letters {m, a, t, h}
       exactly once per word.

             4 · 3 · 2 · 1 = 4! = 24 or             4 P4   = 4! = 24 ways

2.    How many different ways can seven students be seated in seven seats?
                                                                                                   20
                                           n!
         When r  n              n Pr =
                                        (n  r ) !

3.        In how many ways can three different office positions be filled, if there are 7
          applicants who are qualified for all three positions? No one may hold more than one
          position.
                      7 P3 = 210 ways    or 7 · 6 · 5 = 210 ways

4.       If seven people board an airplane and there are 9 aisle seats, in how many ways can
         they be seated if they all choose aisle seats?

5.       A corporation will select two sites from 10 available sites under consideration for
         building two plants. If one plant will produce flashbulbs and the other camera, in
         how many ways can the selection be made?



In many situations, ordering is not important (team problem part c), and we are interested
only in the number of possible combinations. This results in a third counting method.

                                                            n!
3.       Combination:            n Cr   or C(n, r) =                      rn
                                                       (n  r ) !  r !
     A selection of r objects from a set of n distinct objects without regard to their ordering.

1.       A scientist must select three rats from 10 available rats to be used as a control group.
         In how many ways can the control group be selected?
                             10!
                10 C3   =            = 120 ways
                            7 !  3!


2.       A sociologist is interested in drawing a random sample of 6 individuals from a group
         consists of 12 males and 10 females.
a.       How many different samples of size 6 are possible?
                              22 !
                22 C6   =              = 74,613 different samples
                            16 !  6 !

b.       How many samples would consists of all males?

c.       How many samples would consists of all females?

d.       How many samples would consists of 3 males and 3 females?
                                                                                              21
            WORK SHEET ( PERMUTATION AND COMBINATION )

1.    In how many ways can the committee select two of the five candidates and award

a.    one $800 and one $400 scholarship?

b.    two equal $600 scholarships?


2.    A club has a eight-member executive board. In how many ways can the executive
      board choose
a.    a president, a vice president, and a secretary of the club (no one person can hold
      more than one seat)?

b.    a three member committee that will serve on the club council?


3.   If 10 women compete in a marathon, in how many ways can the first three places be
     taken?



WORK SHEET :             COUNTING WITH PROBABILITY

A swim coach must choose 4 swimmers from a roster of 9 ( 5 girls and 4 boys ) to
participate in today‟s race. Assume that the members are equally well qualified.

1.    How many different relay teams are possible?

2.    How many such team consist of girls only?

3.    What is the probability that a relay team , chosen at random, consists of girls only?

4.    How many team will consist of boys only? What is the probability ?

5.    What is the probability that a relay team consist of 2 boys and 2 girls?
            Selecting 2 boys and 2 girls =

            P(2 boys and 2 girls) =

6.    What is the probability that all the team members are of the same sex?
            P(same sex team) = P(all girls team) or P(all boys team)
7.    What is the probability that at least one boy is chosen for the team?
                                                                                                 22
             P(at least one boy) =


8.    What is the probability that at most one girl is chosen for the team?
            P(at most one girl) = P(all boys) or P(1 girl and 3 boys)




Arrangements with Duplicate Items

All the work so far dealt with arrangement of different items. But some times, duplicate
items are involved, for example, arranging the letters of the word zoo.

       Order of letters        Letters
        1, 2, 3                  zoo
        1, 3, 2                  zoo
        2, 1, 3                  ozo
        2, 3, 1                  ooz
        3, 1, 2                  ozo
        3, 2, 1                  ooz

Although there are P(3, 3) = 3! = 6 arrangements, some are indistinguishable from others
(because of the duplicate letter o). For any arrangement that is picked, onother one just like
it can be written by reversing the two o‟s. Since two o‟s can always be arranged in 2!
Different ways, the number of truly different (or distinguishable) arrangements is
3! / 2! = 3 (zoo, ozo, ooz)

Definition: If there are n items with n1 of them alike, n2 of them alike,….., nk of them
                                                                                n!
alike, the number of distinguishable arrangements of n items is
                                                                        n1!  n 2 !    nk !

Example:     How many ways we can arrange the letters of the word

      a.     MISSISSIPPI
             since there are 11 letters but there are 4 I‟s, 4 S‟s, and 2 P‟s, total arrangements
                         11!
             are                 = 34,650 ways
                    4!  4!  2!

      b.     STATISTICS

REVIEW - PROBABILITY WORK SHEET
                                                                                            23
The probability that Harvard will win the Ivy league title in basketball is 0.4 and the
probability that Kentucky will win the South-eastern title is 0.8 . If these events are
independent, find the probability that

1.    Harvard does not win its league title.

2.    Kentucky does not win its title.

3.    Both Harvard and Kentucky win their respective titles.

4.    Harvard wins its title but Kentucky does not.

5.    Kentucky wins its title but Harvard does not.

6.    Neither Harvard nor Kentucky wins their respective titles.

7.    Exactly one of the two teams wins its title.

8.    At least one of the two teams wins its title.

9.    Odds in favor of both team winning their respective titles.



Work Sheet – Counting & Probability

Suppose a club consists of 10 members (6 males and 4 females).

1.    Let us suppose that club members arrive one by one at a meeting

      a.     In how many different orders can they arrive?

      b.     In how many such orderings will John arrives first, Sue arrives third, and Amy
             arrives last?

      c.     Compute the probability that the event described in part (b) will occur.

      d.     Compute the probability that the event described in part (b) will not occur.

      e.   What is the probability that the members arrive in alphabetical order?
2.    Now consider a committee of 4 members chosen at random.
                                                                                            24
      a.     How many different committees are possible?

      b.     How many such committees will contain Amy?

      c.     What is the probability that Amy will be on the committee?

      d.     How many committees will contain 2 males and 2 females?

      e.     What is the probability of the event described in part (d)?


      f.     How many such committees will contain all females, and what is the
             probability of such committee?


      g.     What is the probability that at least one male is chosen?


      h.     What is the probability that all the members are of the same sex?




Duplicate Items & Probability

In a state lottery, four digits are drawn at random one at a time with replacement from 0 to 9
(repetition of digits allowed). Suppose that you win if any order of your selected digits is
drawn.

      a.      How many different lottery numbers are possible?


      b.      What is the probability of winning if you select 2, 5, 8, 9?


      c.      What is the probability of winning if you select 2, 2, 8, 9?



      d.      What is the probability of winning if you select 2, 2, 8, 8?

Answer Sheet
                                      25


Probability Work Sheet

   a.      0.6
   b.      0.2
   c.      0.32
   d.      0.08
   e.      0.48
   f.      0.12
   g.      0.56
   h.      0.88
   i.      32 : 68



Work Sheet – Counting & Probability

   1a.      10!
    a.      7!
    b.      7! / 10! = 1/720
    c.      719/720
    d.      1 / 10!


   2a.      210
    b.      84
    c.      84 / 210 = 2/5
    d.      90
    e.      90/210 = 3/7
    f.      1, 1/210
    g.      209/210
    h.      16/210


Duplicate Items & Probability

    a.      104 = 10,000
    b.      4!/10000 = 24/10000
    c.      12/10000
    d.      6/10000

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:21
posted:10/28/2011
language:English
pages:25