Problem solving and investigationsmathsbooks.net/Fitzpatrick Data and Reasoning/7 Problem Solving a… · 208 PROBLEM SOLVING AND INVESTIGATIONS 3 Arrival times 2:00 ----- Two friends

CHAPTER.

7

Problem solving and investigations

a Keno draw.

b Tattslotto draw.

206 PROBLEM SOLVING AND INVESTIGATIONS

This chapter consists of a number of problems and investigations relating to probability theory as well as permutations and combinati.ons. They should be attempted after the study of Chapters 2 to 5. They are intended for group discussion.

7 .1 Problems

1 The birthday problem Assume that there are n people in a room. What is the probability that at least two people in that group have the same birthday, i.e. the same day and month of the year, but not necessarily the same year? Would you be surprised that there is about a 50-50 chance when n = 23? Enquire among the members of your maths class or among friends at a party. Find out the birthday of the players in your favourite football team. How large would the group have to be to be almost certain that at least two had the same birthday? Let us first find the probability that, in a group of n people, no two have the same birthday. Since there are 365 days in the year (ignoring a leap year) there are 365n possibilities for the birthdays of the n people, each of which is equally likely. If the n people are to have different birthdays, then the first person can be born on any one of the 365 days, the second person on any one of the remaining 364 days, the third person on any one of the remaining 363 days and so on. So, using the multiplication principle, there are 365 x 364 x 363 . .. (365 - n + 1) ways then people can have different birthdays. The probability that no two people have the same birthday is:

p (X = 0) = 365 X 364 X 363 . .. X (365 - n + 1)r

365n

The probability that at least two people have the same birthday is:

Pr(X � 2) = l _ Pr(X = 0) = l _ 365 X 364 X 363365

.n X (365 - n + 1)

If n = 10 Pr(X :::e: 2) = l _ 365 X 364 x 363 ... x 356' r

3�

= 0.1169 (using a calculator) Copy and complete the following table:

n 5 15 20 23 30

Pr(X ;l? 2)

40 50 70 90

PROBLEM SOLVING AND INVESTIGATIONS 207

The program below calculates the probabilities for groups of various size, N.

1 0 REM BIRTHDAY PROBLEM

20 CLS

30 PRINT "PROGRAM TO FIND PROBABILITY THAT AT LEAST TWO MEMBERS OF

A GROUP HAVE THE SAME BIRTHDAY."

40 PRINT "Enter number of people": INPUT N

50 PR = 1

60 FOR I = 1 TO N

70 D=365

80 Q = 365 -I + 1

90 PR=PR*(Q/D): PROB= 1 -PR

1 00 IF I = N THEN 1 20

110 NEXT I

120 PRINT "NUMBER OF PEOPLE= "N, "PROB OF SHARED BIRTHDAY= "PROB

130 END

If N = 366, we are certain that at least two people will have the same birthday (excluding leap years.)

Estimate the number of people necessary for the probability of shared birthdays to be 10% certain, 20% certain ... 90% certain and 99% certain.

Check your estimates by using the program.

2 Bertrand's problem

A

Figure 7-1

If a chord of a circle is chosen at random, what is the probability that its length is greater than the length of the side of an equilateral triangle inscribed in the circle? Present arguments to show that the required probability may be:

a ½, by considering arcs of the circle

b ½, by considering the distance of chords from the centre

c ¼, by considering midpoints of chords.

Discuss 'equally likely events' in these cases.


3 Arrival times 2:00 �-------------�

Two friends AH and J edda plan to meet at a particular place, some time between 1 p.m. and 2 p.rn. They decide that, upon

arrival, each will wait for 10 minutes only and then leave if the other has not turned up. Draw a grid, as shown on the right,

and shade a region corresponding to Ali and Jedda meeting. Show that the

probability they will meet is ;!-

1 :50

i 1:40 "O Ql

� 1:30

·E 1 :20 <(

1:10

1 :00 1:10 1 :20 1 :30 1 :40 1 :50 2:00

Figure 7-2 Arrival time (Ali)

4 System entries for Tattslotto An entrant for Tattslotto is required to mark any six numbers out of 45 on a frame, for a cost of 25 cents. However, an entrant may mark 7, 8, 9 ... 20 numbers. This is called a 'system entry'. For a system 7, the cost is $1.75 (plus commission). Why? When you have

answered this, calculate the cost for each of the systems, 8 to 20 (ignore the commission).

Figure 7-3

5 Bufton needle problem

11111111 A.AA.A.A.A.A.A.

1 2 3 4 5 6 7 8 vvvvvvvv A A A. A A. A A A 9 10 11 12 13· 14 15 16

vvvvvvvv A. A. A A A A A A. 17 18 19 20 21 22 23 24 vvvvvvvv A. A A A A A A A25 26 27 28 29 30 31 32 vvvvvvvv A A A A A A A A 33 34 35 36 37 38 39 40 vvvvvvvv

414243«.is T vvvvv L...li..J

This problem involves the probability that a needle, shorter than the width of a plank, will fall across the crack between two adjacent planks.

Figure 7-4


Consider a needle of a certain length, e.g. 2 units, and planks wider than the length of the needle, e.g. 4 units. What is the probability that the needle will fall across a crack, if it is dropped at random on the plank? We can define the position of the needle at each throw by noting the position of the midpoint of the needle and the angle between the needle and the crack. The previous diagram (Figure 7-4) shows three possible positions of th� needle. We can show the various possible positions of the needle by means of a rectangl!:! of width, 4 units, as before. The length represents the angle in terms of 1r.

4

3

1o

0

Figure 7-5

30

on T

n

This rectangle, whose area is 41r, represents all possible positions in which the needle can fall. Tpe shaded area of this rectangle corresponds to those positions of the centre of the needle in which the needle crosses a crack. This can be calculated using trigonometry and the area of the shaded region can be calculated by calculus. However, as a project, you should do this calculation graphically to scale. The probability of the needle falling on a crack is the

ratio of the shaded regions to the area of the rectangle, i.e. ;� , where I is the length of the

needle and dis the width of the plank. So if/ = 2 and d = 4, the probability is l. 7!'

In 1901, Lazzarini performed the experiment 3408 times and came up with 1r = 3.1415929 (a very good result!). It is probable that he stopped the experiment when he had a good value for 1r.

6 Keno

Figure 7-6

MARK SPOT NUMBER

� m lfil lfil 0 lfil @J [Q]

IT] ID] � @.!] � � � !Bl

@]B]��

��§ m��§

[fil��

[filB] ��I@]��

[zJ II] � @1] � � � �

@JI!§][@]��[@]��

@JB]��

l!Q]�@Q]HQ]�[@��

Keno is a game of chance iri which you invest $1 and then select at least three and no more than IO n.umbers between 1 and 80 inclusive, Each selection is called a 'Spot'. So, if you select five numbers you are playing a 'Spot 5' game. Twenty (20) numbers are drawn at random from the 80 and, to win, your numbers must be among the 20 drawn. So if all five


numbers in your Spot 5 entry are among the 20 drawn, you have a 'Match 5', and win $800 as shown in the table below.

Tp.e prizes and the chances of winning vary according to which Spots you play and how many numbers you match. The 'Spot 3' prize is only $35, but your chances of winning are much better, and the cost is the same whichever Spots you play. The following table lists all the Spots and the prizes for matching the drawn numbers in the Keno games conducted by Tattersall's in Victoria.

SPOT MATCH

How many How many of your numbers chosen numbers are you choose. among the 20 drawn.

3 3 4 4 5 5 6 6

7 7

8 7

8 8 9 7

9 8 9 9

10 0 10 8 10 9 10 10-Jackpot

WIN

The value of your prize.

$35 $170 $800

$4000 $20 000

$2 000 $35 000

$250 $8 000

$100 000 $5

$1 000 $10 000

Jackpot announced daily -minimum$250 000

Of interest to us is the probability associated with each of the Spots and the expected gain ( or loss) in each case as a result of investing $1 to play.

Assuming that each of the 80 numbers is equally likely to be drawn, the number of ways

of selecting 20 numbers is (�g). So, for each Spot:

Pr =

Number of favourable selections(�g)

For Spot 3, for example, consider the 80 numbers to be made up of two kinds of numbers, namely the three you have drawn and 77 other numbers from which 17 have been drawn.

G) GDPr(winning Spot 3) = (�g)

= 0.013875

This means that a player expects to win $35 with a probability of 0.013875, and to lose the $1 invested with a probability of 1 - 0.013875.

Expected gain = $35 x 0.013875 - $1 x 0.986125 = - $0.5005 (a loss of about 50 cents)


Show that the probability associated with each of the other Spots and the expected loss in each case are as shown in the table below. (It is advisable to calculate the probabilities to about six decimal places or more. Why?)

Spot Match Probability Expected loss ($)

3 3 0.013875 0.5005

4 4 0.003063 0.4762

5 5 0.000645 0.4834

6 6 0.000129 0.4839

7 7 0.0000244 0.5119

8 7 0.0001605 -

)-- 0.5268

8 8 0.0000435

9 7 0.0005917

1-9 8 0.0000326 0.5185

9 9 0.000000724

10 0 0.045791

-

10 8 0.0001354

- 0.5010

10 9 0.00000612

-

10 10 0.00000011

7 Craps Craps is a popular game at casinos. A player throws two dice and, if the score (sum of the two numbers uppermost) is 7 or 11, the player wins. If a 2, 3 or 12 is scored, the player loses. If any of the other combinations is scored, i.e. 4, 5, 6, 8, 9 or 10, the player throws the dice again and continues throwing until a 7 is scored (in which case he/she loses), or until he/ she repeats the score received on the first throw (in which case he/she wins). Calculate the probability that: a the player wins on the first throw, (�)

b the player loses on the first throw, (i) c the player scores 4 on the first throw and wins, (}6)

d the player scores 8 on the first throw and wins, (3�6)

e the player wins, (;!1) . Represent this situation on a probability tree diagram with two sets of branches. Show that, if a player were to outlay $100 in playing craps, he/she would expect to lose about $1.41.


8 The Australian Football League final series After the AFL home-and-away games are completed, the final series consists of matches between the top five teams as illustrated below.

4.5

2.3

Figure 7-7

L = Loss W=Win B =Bye

a If all teams are equally likely to win each of the matches they play, and the results of the six finals matches are determined independently, calculate the probability that: (i) Team 4 or Team 5 wins the grand final.

(ii) Team 2 or Team 3 wins the grand final.(iii) Team 1 wins the grand final.

b Repeat Question Sa for the case where the likelihood of winning is proportional to the number of points won in the season, as shown in the following table.

Team Points

1 72

2 64

3 62

4 56

5 56

So, if Team 2 were to play Team 4, the probability that Team 2 will win is:

64 64 S 64 + 56 = 120 = 15·

c Repeat Question Sb for the situation where, each time a team wins, it has four points added to its total before the new probabilities are computed. A team with a bye has two points added to its total. (This model allows for the advantage of a week's rest for the team with a bye.)

d The model for predicting probabilities used in Questions Sb and Sc does not seem to produce sufficiently diverse probabilities - surely Team 1 has a higher probability than 0.56 of beating Team 5. Calculate the mean number of points for the five teams in Question Sb and the standard deviation from the mean of each team's points. Multiply this deviation by five and add the result to the number of points for each team.


(i) Use these new points to compute the probabilities of each team winning the grandfinal.

(ii) Is this a better model for predicting the result? How could you find out?e (i) Devise a diagram similar to Figure 7-7 for a final series involving:

• 6 teams • 7 teams.

(ii) How many matches are played in these cases?(iii) How many matches are played if n teams are in the final series? (Give formulae for

the cases where n is even and n is odd.)(iv) How many weeks are needed for a final series with n teams?

9 A medical problem X and Yare two fatal diseases which give rise to identical symptoms. Of people with these

symptoms, 850Jo have disease X and 150Jo have disease Y. The one test available to

distinguish between these two diseases is only 800Jo reliable. The treatment for disease X will cure a person suffering from disease X but shorten that person's life if he or she has disease Y. Similarly, a person suffering from disease X, given the treatment for disease Y, will die

sooner, but a person with disease Y, given the correct treatment, will be cured.

a Find the probability that a person with the symptoms has disease X and is tested anddiagnosed to have: (i) disease X.

(ii) disease Y.

b Repeat Question 9a for a person with disease Y.

c If the test tells you that you have disease Y, which treatment if any would you request?

10 A business person's problem Each trading day, a retailer purchases perishable articles for $3 each and sells them to

customers at $5 each. Each article sold entails an overhead expense to the retailer of 50 cents. Articles unsold at the end of the day are returned to the supplier who refunds $1 per article to the retailer. The daily demand, X, varies randomly from day to day, ranging from 7 to 10 as shown in the following table:

Demand (X) 7 8 9 10

Proportion of days 0.10 0.40 0.40 0.10

In order to estimate the standing daily order that should be placed, the retailer calculates

the conditional daily profit, P, which varies according to the number of articles, n, ordered and the demand, X.

Insert the elements of the following 4 x 4 matrix of conditional daily profits.

Conditional daily profit P

7 8 9 10

j[ j Calculate the expected daily profit over a period of time for each value of n and so determine which value of n would yield the greatest profit.


11 A game of chance A person pays two dollars to play a game of chance in which he or she is issued with one of three vouchers A, B or C. The person then draws one ball at random from an urn which contains 6 red, 3 white and 1 blue ball. The matrix below shows the value in dollars of the prize the player receives if he or she is the holder of a certain voucher and draws a particular coloured ball from the urn.

A

Red [

2 White 5 Blue 0

B C

nJ Calculate the player's expected gain (or loss) per game for holding vouchers, A, Band Cif the game is played many times.

D 7 .2 Investigations1 Doran and Marcia play a game with the following rules:

(i) A stake is decided and each player puts this on the table.(ii) A die is rolled. If the numbers 1, ,2 or 3 come up, Doran wins the moneyon the table. If a 5 or a 6 comes up, Marcia wins. If a 4 comes up, it is a drawand the die is rolled again for the same stake.Marcia knows the probabilities in this game, so, at her suggestion, rules (iii) and(iv) are added:(iii) Marcia can choose the stake for each game.(iv) Marcia can decide, and announce in advance, how many games will beplayed.Marcia has exactly $20, and cannot borrow any more money, so play will stop if she runs out of money, even if the decided number of games is not played. Before you read further, can you see a strategy by which Marcia may win? Ifso, investigate it carefully, calculating the expected gain for Marcia. Under what circumstances will she lose? Marcia works to the following system: She says that 200 games will be played. She bets 10 cents on the first game. If she wins, she bets 10 cents on the next game. If she loses, she bets 20 cents on the next game. If that loses, she bets 40 cents on the next game, continuing to double until she wins. Each time she wins, she bets 10 cents on the next game. Notice that this means that each time Marcia wins, even after a string of losses, she adds 10 cents to her-winnings. In 200 games, she can expect to win 80 of them and, therefore, win $8. However, she will lose if she finishes on a run of losses. How many games can Marcia lose in a row and still have money with which to play? Does this number of games depend on how many games have been played? Calculate, using a tree diagram, Marcia's winnings after four games and, therefore, her expected gain after four games? What is her probability of winning something after four games? The following computer program enables you to simulate this game and to keep track of Marcia's progress. It prints out the total Marcia wins and the maximum stake she must put up and, finally, the average winnings for the run of trials.


10 INPUT "NUMBER OF GAMES"; N

11 INPUT "NUMBER OF TRIALS"; K

1 2 FOR J = 1 TO K

14 REM: S = STAKE IN CENTS, SM = MAXIMUM STAKE 15 S = 1 0:SM = 0

16 W = 0: REM :W = WINNINGS

20 FOR I = 1 TO N

30 X = RND (1) * 6: REM : X IS A RANDOM NUMBER BETWEEN 0 AND 6

40 IF INT (X) < 3 GOTO 100

50 IF INT (X) = 3 GOTO 30

60 IF INT (X) > 3 GOTO 200

1 00 W = W - S:S = 2 * S

120 IF S> 2000 + W THEN PRINT"*";: GOTO 400

130 REM : MARCIA HAS ONLY $20

150 IF S>SM THEN SM = S

170 GOTO 300

200 W = W + S:S = 10: GOTO 250

250 GOTO 300

300 NEXTI

400 PRINT W;" "; SM

450 T = T + W

500 NEXT J

600 PRINT T / K

For small numbers of games, Marcia often loses and her average winnings will

be negative. For a larger number of games, e.g. 200, she appears to win more

often, with a positive average winnings. Try running the program a number of

times to check this out. Can you explain it? Make an estimate of the minimum number of games Marcia should elect to play so that her probability of winning

is 0.5. Investigate this with the computer program.

Modify the program to enable you to keep a count of how often Marcia cannot complete the elected number of games before she goes broke. How does this

number vary with the number of games played?

2 A competition is held in which a list of 30 cricketers is provided and competitors are required to pick a team of eleven (in batting order), 12th man, captain and

vice-captain, to match a team selected by a panel of experts.

a How many possible different entries are there?*

b What assumptions are made if it is stated 'The probability of one entry being

correct is 1 divided by N' where N is the answer to Question 2a?

c Since all possible teams are not equally likely to be correct (who would

realistically pick 11 bowlers?), what is a more realistic figure for the

probability discussed in Question 2b. What assumptions are made now?

3 'The Poisson distribution may be used to approximate the binomial distribution

when n is large, pis small and np is moderate, where n is the sample size and pis the probability of success'.

This rule is frequently used but is rather vague. What does 'np is moderate' really mean?

*(Answer: ��; x 11 x 10 = 4.557 x 10 18)

'J


Suppose we have a random sample of 1000 people from a population in which 0.1 % have a particular disease. Then: the probability that there are x people in the sample with the disease is

given by (1�0) (0.00l)x (0.999) 1000 - x using a binomial distribution.

Using a Poisson distribution, u = 1, the expected number of people with the disease and the probability that x people in the sample of 1000 have the disease is:

e- 1 .1 x

1 . . . --, - = -, , which is easier to calculate. x. ex.

Investigate the quality of the approximation for different values of n and p, over a range of events (or values of x). Tabulate these and rephrase the above rule more precisely.

4 In many card games, such as Bridge, Solo or 500, all the cards are dealt among four players and it becomes important to a player's success that he/ she knows the distribution of the cards of a particular suit (club, hearts, diamonds or spades). For example, a player may know where eight of the 13 diamonds are (perhaps holding eight in his/ her hand, or having seen some played in the course of the deal) and needs to know which players hold the other five - if another player holds at least four of them, the first player will lose. What is the probability of this? We can compute this exactly using counting techniques from Chapter 3. Call the five diamonds D1, D2, D3, D4, D5 and the three other players A, Band C. Each

card has a probability of½ of being in each hand, so there are 243 ( = 3 5)

ways of distributing the five cards among the three players. One of these ways has all five cards with player A, one has all five cards with player B and one has all five cards with player C. So there are three ways out of the 243 where all five cards lie with one player. Player A could hold exactly four diamonds in five ways (any one of D1, D2, D3, D4 or D5 could be missing). The fifth card could be with B or C - two possibilities. So, of the 243 ways, 10 of them give player Afour diamonds, and a total of 30 of them give any one player four diamonds (10 ways for B to have four diamonds, another 10 for C to have four diamonds). The total number of ways that a player can have 4 or more diamonds is:

3 + 30 = 33

and the probability of this is ]}3 = ��. This problem is relatively simple, compared to other information which may be useful to a good player. (The problem is somewhat harder, even if three or more diamonds are needed in one hand, or if a particular card, for example a Queen, is needed with the four or five diamonds.) But we can find good approximations to the probabilities in this kind of problem, using a computer. Remember that probability is the long-term ratio of an event happening. We can simulate the event using random numbers on a computer and repeat the experiment any number of times.


The following program simulates the above problem.

10 K = 0 20 FOR I = 1 TO 1 000 30 A= O:B = O:C = 0 40 FOR J = 1 TO 5 1 50 X = RND( 1 ) 160 IF 3*X � 1 GOTO 300 170 IF 3*X � 2 GOTO 200 1 80 C = C + 1 : GOTO 400 200 B = B + 1 : GOTO 400 300 A= A+ 1 400 NEXT J 500 IF A � 4 THEN K = K + 1 510 IF B � 4 THEN K = K + 1 520 IF C � 4 THEN K = K + 1 600 NEXT! 700 PRINT "PROBABILITY THAT AT LEAST 4 CARDS ARE IN ONE HAND

= "; K/1000

Note:

(i) Line 50 assigns X a random number between 0 and 1.(ii) Lines 160-300 deal a single card into one of three hands and count how many

cards are in the hands.(iii) Lines 500-520 record the result of a single 'deal' counting the number of

'deals' where at least four cards end up in one hand.

a Run the program several times. Decide whether 1000 'deals' is enough, too many or just right. 1 Compare the computer's answer with �1.

b Modify the program at lines 20 and 700 to allow 100, 500, 2000 and 10 000 'deals'. Compare the results of several runs.

c Modify the program at lines 500, 510 and 520 so that it counts the number of hands where there are least three cards in a hand. Try to follow an argument such as the one above to find the mathematical probability and compare it with the value given by the computer.

d Playing Solo, you hold eight diamonds (including the Ace, King, Queen and 10) and another Ace. You need to hold nine cards in order to win 'ForeignAbundance'. This will happen provided the Jack is not held by an opponentwho has three or four other diamonds in his/her hand. Before the player bids'Foreign Abundance', he/she would like to know the probability of success.Modify the previous program: Replace line 40 with:

40 X = RND(1) 50 IF 3*Y� 1 GOTO 90 60 IF 3*Y� 2 GOTO 80 70 C = 1 0 : GOTO 1 00 80 B = 1 0 : GOTO 1 00 90 A= 10

1 00 FOR J = 1 TO 4


This will identify the hand which has the Jack. Replace lines 500-700 with:

500 IF A ;;;i: 13 THEN K = K + 1

510 IF B ;;;i: 13 THEN K = K + 1

520 IF C ;;;i: 13 THEN K = K + 1

600 NEXTI

700 PRINT 'PROBABILITY THAT JACK IS HELD WITH AT LEAST 3 OTHERS

= -";K/1000

Follow through the program and see that, if the Jack has been put in a hand with three or four diamonds then one of A, B or C will be 13 or 14 and K (the counter) is increased by one.

Type in and run the program.

Compute the probability using counting techniques and compare the two probabilities.

5 Replace the Queen in the hand in the previous question with the Jack. Now you

will succeed unless the Queen is held with at least two other diamonds in the one opposition hand.

a Modify the program to compute the probability of this.

b Calculate the probability using counting techniques.

6 Replace the 'outside' Ace in the hand in the previous question with a low

diamond -you now hold Ace, King, Jack, ten, and five more diamonds. You succeed if three of the remaining four diamonds, including the Queen, are not

in one hand. What is the probability of this?

7 In Solo, to succeed at 'misere', you must win no round. It is important to hold the lowest cards in each suit; you will lose if one of your opponents holds the

lowest card in a suit and more cards in that suit than your other opponents. So,

if you have six diamonds, but not the two, you will almost certainly lose if the distribution between your three opponents of the other seven diamonds is 4-2-1,

4-3-0, 5-2-0, 5-1-1, 6-1-0 or 7-0-0, and the two is in the longest hand. Modify

the above program to find the probability of this. What distributions lose for youif you hold seven diamonds without the two? What is the probability of losingthen?

Documents

Problem solving and investigationsmathsbooks.net/Fitzpatrick Data and Reasoning/7 Problem Solving a… · 208 PROBLEM SOLVING AND INVESTIGATIONS 3 Arrival times 2:00 ----- Two friends