Contents
9.1 Combining probabilities
9.2 Discrete probability distributions 9.3 Expected values
9.4 Binomial probability distribution Chapter review
9
Data models
—Part A
Syllabus subject matter
Syllabus guide chapter 9
Introduction to models for data
■ Basic probability rules of complements and unions
■ Odds as an application of probabilities
■ Probability distributions and expected values for a discrete variable
■ Uniform discrete distribution; random numbers
■ Identification of binomial situations, binomial expected values and binomial probabilities using tables or calculators
■ Discrete data that could be uniform or binomial; comparison of relative frequencies with
9.1
Combining probabilities
You have already studied some probability this year in Chapter 6. This elective expands the notions of probability to cover more situations. Many complex situations can be broken down to combinations of simpler cases. You have already seen how to do this with tree diagrams and grids. The multiplication principle arises naturally with tree diagrams, but applies to other situations where the probability of two events is combined.
Multiplication principle
The probability of two events occurring together is the product of the probabilities of the individual events. This is written as
P(A ∩ B) = P(A and B) = P(A) × P(B) In mathematics, the sign ∩ is read as ‘and’, meaning both A and B.
!
What is the probability of getting 2 sixes when two normal dice are thrown?
Solution
Write symbols for the events. Let A = ‘The first die shows a six’
Let B = ‘The second die shows a six’
Write the probabilities. P(A)=
P(B)=
Use the multiplication principle. P(A and B)= P(A) × P(B)
= × =
Write the answer. The probability of getting 2 sixes is ≈ 2.8%. 1
6
---1 6
---1 6
--- 1
6
--- 1
36
---1 36
---Example
1
In some cases, the probabilities of the events affect each other, so the calculation of the individual probabilities is a little more complicated.
We may be interested in finding the combined probability of either of two events. A diagram of the sample space is quite useful for these cases.
Diagrams like the one in Example 3 are called Venn diagrams and are useful for showing combined probabilities.
A bag contains 2 red and 4 green marbles. Two marbles are taken out without replacement. What is the probability that the first is red and the second is green?
Solution
Write symbols for the events. Let R = ‘The first marble is red’
Let G = ‘The second marble is green’
Write the probabilities. P(R)= =
There are now 4 green and 1 red marbles left
in the bag. P(G)=
Use the multiplication principle. P(R and G)= P(R) × P(G)
= ×
=
Write the answer. The probability of getting a red then a
green marble is ≈ 26.7%.
2 6
--- 1
3
---4 5
---1 3
--- 4
5
---4 15
---4 15
---Example
2
A bag contains 3 large purple marbles, 2 small purple marbles, 4 large orange marbles and 3 small orange marbles. One marble is taken at random. What is the probability that it is a small marble or an orange marble?
Solution
Show the marbles in a diagram.
It is easier to draw a rectangle than to draw the bag around the marbles.
Put a labelled circle around the small marbles and another around the orange marbles.
Use the diagram to find the probability. P(small or orange)=
=
Write the answer. The probability that the marble is small
or orange is = 75%.
Orange marbles Small
marbles
9 12
---3 4
---3 4
From Example 3, and from looking at the Venn diagram, we can formulate a rule for the probabilities of combined events.
In Example 3, P(small or orange)= P(small) + P(orange) − P(small and orange) = + −
=
= 75%
Venn diagrams
For a probability situation, a Venn diagram shows the whole sample space as an enclosing rectangle. An event within the sample space is shown by a circle. Overlapping circles show where different events share common elements of the sample space.
The combined event ‘A and B’, written as A ∩ B, is shown by shading the overlapping part of the A and B circles. It is also called the intersection of A and B. Both A and B occur.
The combined event ‘A or B’, written as A ∪ B, is shown by shading all of the A and B circles. It is also called the union of A and B. At least one of A and B occurs.
The complement of the event A, written as A′or , is shown by shading all the part outside the A circle. The complement is the same as not A. An event A and its complement A′ make up the sample space.
P(A) + P(A′) = 1
A and B are called mutually exclusive when P(A and B) = 0, so they have no overlap and no sample points in common.
A
A
B A B A B
A ∩ B A ∪ B A′=notA= A and B
mutually exclusive A
A B
A′
!
Addition rule
For events A and B,
P(A or B)= P(A) + P(B) − P(A and B) or P(A ∪ B)= P(A) + P(B) − P(A ∩ B) The probability of A or B is the probability of A plus the probability of B minus the probability of A and B.
A ∩ B
A B
!
5 12 --- 7
12 --- 3
12
---9 12
---A card is drawn from a normal pack of playing cards. What is the probability that it is a spade or a court card (jack, queen, king)?
Solution
A normal deck has 52 cards separated into 4 different suits (♠, ♣, ♦, ♥). The 13 cards in each suit are 2, 3, 4, …, 10, jack, queen, king, ace.
Venn diagrams are useful for visualising and simplifying sample spaces. Write the number of spades. n(spade)= 13
Write the number of court cards (J-Q-K). n(court)= 12
Write the number of spade court cards. n(spade and court)= 3
Calculate the probability of each event. P(spade)=
P(court)=
P(spade and court)=
Use the addition rule. P(spade or court)= P(S) + P(C) − P(S and C)
Substitute values. = + −
Simplify.
=
=
Write the answer. The probability of a spade or court card
is ≈ 0.423.
13 52
---12 52
---3 52
---13 52
--- 12
52
--- 3
52
---22 52
---11 26
---11 26
---Of 30 students in an English class, 18 take Chemistry, 16 study Physics and 20 do Biology; 6 students take all three sciences; 11 do Physics and Chemistry, 10 do Physics and Biology, and 2 do only Chemistry. What is the probability that a student selected at random:
a does Chemistry and Biology? b does none of these sciences?
Solution
By drawing the intersections on a diagram as shown, we can work out the required numbers. Draw a Venn diagram with three intersecting circles.
Label the circles C, P and B.
Put the 6 students who do all three in the overlap of the three circles.
Since 11 do Physics and Chemistry, put 5 in the other part of the C–P intersection.
Since 10 do Physics and Biology, put 4 in the other part of the B–P intersection.
Put 2 in the ‘Chemistry only’ part of C.
16 do Physics and 5 + 6 + 4 = 15 are accounted for, so there must be 1 doing only Physics.
18 do Chemistry and 2 + 5 + 6 = 13 have been put in place, so there must be 5 more in the other part of the C–B intersection.
20 do Biology and 5 + 6 + 4 = 15 are in the
intersections, so there must be 5 doing only Biology. There are 30 students, and 28 are included in the circles, so there must be 2 outside the circles
C = 18
B = 20 P = 16
2 5 6
4
C
B P
2 5
6 4 5
2
1
5
When people gamble, they use probability. However, the odds in gambling are stated a little differently from ordinary probabilities. Odds show the ratio of amounts to be won by each side in a wager. The winner gets their own money back, together with the money bet by the other person. It is usual for the money bet by each side to be held by an independent
stakeholder who would then give all the money to the winner. Odds may be stated in reverse
as odds on. Bookmakers most commonly offer bets at odds.
Fair odds reflect probabilities. If you gamble at fair odds, you would expect to win and
lose about the same amount in the long run. Of course, bookmakers and casinos do not set fair odds because they need to make money to cover expenses and make a profit. You can work out fair odds for simple situations using the probabilities of winning and losing. The
complement is used for this purpose. The odds will be the reverse of the ratio of wins:losses,
so that the amount won compensates for the number of losses.
a Write the rule for P(C and B). P(C and B)=
5 + 6 = 11 students do Chemistry and
Biology, shown in the overlap of the circles. = Evaluate and round off. ≈ 0.367
b Write the rule for P(no science). P(no science)=
2 students take none of these sciences. =
Evaluate and round off. ≈ 0.067
n C and B( )
n sample space( )
---11 30
---n ---no scie---nce( )
n sample space( )
---2 30
---The probability of getting at least 3 heads from 4 tosses of a coin is 0.3125. What are the fair odds for betting on at least 3 heads from 4 tosses of a coin?
Solution
Write the probability of winning. P(win)= 0.3125
Write the probability of losing. P(loss)= 1 − P(win)
= 0.6875
Write the ratio of wins : losses. Wins : losses= 0.3125 : 0.6875
Simplify. = 3125 : 6875
= 5 :11
Reverse to find the fair odds. The fair odds are 11: 5.
Example
6
A bookmaker offers odds that are not fair. The bookmaker is free to change the odds offered as he or she accepts bets. In fact, bookmakers set much lower odds on the favourite in a race than on other horses. This isn’t because they think the favourite will necessarily win, but because more people are betting on that horse. A bookmaker tries to set the odds so that no matter which horse wins, the bookmaker will still make money. Work in groups of three or four to simulate bookies’ problems.
In your group, run a 6-horse race, with runners Sheelberite, Gophabroke, Farstanag,
Ubuedee, Dooatun and Levdbeind. To start with, set odds of 4 :1 for every horse. Then
whichever horse wins, you should make money. Draw up a table with these headings:
Now throw a die 15 times and note the numbers. For each 2, add $100 to the amount bet on horse 2. For each 3, add $100 to the amount bet on horse 3, and so on. Work out the total bet and the payout for each horse (don’t forget the bet). For any horse with a payout that exceeds the amount that has been bet, reduce the odds to make the new odds. For any horse with a very low payout figure, increase the odds.
Draw up another table with the new odds and throw 15 times again. For the new payout figure, you will need to add the payout to the old payout figure because the old bets have already been accepted. When you have done this a few times, discuss the process as a class group. Find out what ‘laying off’ means.
Horse Odds Amount bet Payout New odds
1 Sheelberite 4 :1
2 Gophabroke 4 :1
3 Farstanag 4 :1
Exercise 9.1
Combining probabilities
1 A bag contains 4 yellow and 5 blue marbles. Two marbles are chosen at random. Find
the probability that:
a both are yellow b the first is yellow and the second is blue c the first is blue and the second is yellow d they are different colours.
2 Three cards are dealt from a well-shuffled pack of cards. Find the probability that they
are all kings.
3 A card is drawn from a deck of 52 normal playing cards. Find the probability that it is:
a a queen b a red card
c a red queen d a queen or a red card
e not a queen f not a heart
g neither a five nor a spade.
4 Two normal (6-sided) dice are rolled and the total is calculated. Find the probability that: a the total is greater than 8 b the total is not greater than 8
c the total is even and less than 7 d the total is even or less than 7
e the total is 6 or 8 f the total is neither even nor less than 7.
Modelling and problem solving
5 In Brisbane, the probability of an afternoon storm on any summer’s day is about 0.07.
What is the probability in summer that:
a there will not be an afternoon storm on a certain day? b there will be no afternoon storms for a week (7 days)? c there will be an afternoon storm at least once in a week?
Additional exercise
6 A good pool player pots the ball she has chosen about 75% of the time. What is the
probability that she has a run of 10 goes (that is, pots the chosen ball 9 times in a row)?
7 The following data relates to the causes of business bankruptcy in Australia in the 2000/01
financial year.
A business is chosen at random from those that went bankrupt in the 2000/01 financial year. What is the probability that the bankruptcy was:
a due to lack of capital or lack of business ability? b not due to economic and seasonal conditions?
8 A caterer prepares two cheesecakes—one chocolate and one passionfruit. Each cake is cut
into 6 slices. Each slice of cake is placed onto a coloured plate—4 of each colour red, green and blue. If a guest randomly selects a plate, what is the probability that:
a it is red?
b it has a piece of chocolate cheesecake on it?
c it is a red plate holding a piece of chocolate cheesecake?
d it is blue or it holds a piece of passionfruit cheesecake ? e it is not blue and it holds a piece of passionfruit cheesecake?
9 In a class of 28 students, 12 wore no ear-rings or bracelets, 5 wore both ear-rings and
bracelets, and 12 wore bracelets. Draw a diagram and find the probability of choosing a student wearing:
a ear-rings b only ear-rings c only bracelets.
Cause % of bankruptcies
Lack of capital 9.9
Lack of business ability 19.5
Economic and seasonal conditions 33.2
Bad debts, high interest and excessive drawings 10.4
Personal and other reasons 27.0
9.2
Discrete probability distributions
Suppose we toss a fair coin 4 times and count the number of successive tails. The possible outcomes for this experiment and their probabilities are as follows.
Event tails in a rowNumber of Probability
HHHH 0 = 0.0625
HHHT, HHTH, HTHH, THHH, THHT, THTH, HTHT 1 = 0.4375
HHTT, HTTH, TTHH, THTT, TTHT 2 = 0.3125
HTTT, TTTH 3 = 0.125
TTTT 4 = 0.0625
10 A group of runners were comparing footwear: 12 wore white joggers and the others
different-coloured or multicoloured joggers; 15 wore Adidas and the others wore different brands; 4 wore white joggers that were not Adidas; and 14 did not wear either Adidas or white joggers. Draw a diagram and use it to find:
a the total number of runners
b the probability of choosing an Adidas wearer at random c the probability of choosing a runner with white joggers
d the probability of choosing a runner without Adidas
e the probability of choosing a runner with non-white joggers.
11 The probability of getting 2 heads from 6 tosses of a fair coin is 0.234 375. What are the
fair odds for betting on 2 heads from 6 tosses?
12 An American roulette wheel has 38 numbers (1–36 and two uncoloured 0s). Casinos offer
odds of 35: 1 to bet on a number and even odds for black or red. Half the numbers from 1 to 36 are red and half are black. What are the fair odds for:
a betting on a number? b betting on red?
13 A poker machine has 3 wheels. Each is marked with an ace, king, queen and lemon. To
win, you must get 3 aces, 3 kings or 3 queens.
a What is the probability of 3 aces? b What is the probability of 3 kings? c What is the probability of 3 queens? d What is the probability of a win? e What are the fair odds?
f If it costs 50 cents to play, what should the average payout be?
14 A dice game is played by throwing two dice. Players may bet on any total except 7 against
the thrower. What are the fair odds for each total, apart from 7?
15 A coin-throwing game at a fair has circles drawn on a table. To play, you throw a dollar
coin onto the table from behind a rope about 3 m away. You win if your dollar lands inside one of the coloured circles. There are 625 circles altogether; 25 are green and 5 are red. Assuming that landing in a coloured circle is at random:
a what are the fair odds for landing in each colour? b what would a fair payout be for each colour?
1 16
---7 16
---5 16
---1 8
---The number of successive tails is a discrete variable because it can have only specific values. Remember that a continuous variable can have any value within a range.
In Chapter 6, you constructed probability histograms for a number of different cases, including ones where the widths of the columns were different. In these cases, the heights of columns were adjusted so that the total area of the histogram was 100% (or 1 for decimals) because the total of all the probabilities must be 1. The properties are summarised below.
Probability functions
A discrete random variable is a discrete variable with a numerical value determined by the outcome of a random experiment. It is denoted by a capital letter. Specific values of the variable are denoted by a lower case letter.
A discrete probability distribution lists all the probabilities of a discrete random variable.
A probability distribution is also called a probability function and is written as
P(X = x) or p(x) for the variable X.
!
Discrete probability distribution properties
• The values of P(X = x) are between 0 and 1. • The total of all the values of P(X = x) is 1 (= 100%).
• The area of the histogram of any probability distribution must equal 1. The area of the histogram over an interval is the probability that the variable occurs in that interval.
!
Show the probability distribution of T, the number of successive tails from 4 tosses of a coin, as a histogram. Comment on the total of all values of P(T = t).
Solution
List the probabilities in a table.
Draw the histogram.
It is convenient to use the Σ
notation for the total.
ΣP(T = t)= p(0) + p(1) + p(2) + p(3) + p(4)
= 0.0625 + 0.4375 + 0.3125 + 0.125 + 0.0625
= 1
Make a comment. The total of all the probabilities is 1, so one of the possible values 0–4 is certain to occur.
t 0 1 2 3 4
p(t) 0.0625 0.4375 0.3125 0.125 0.0625
Successive tails from 4 tosses
0 1 2 3 4 T
0.3 0.5
0.4
0.2
0.1
0 p(t)
The probability distribution for throwing a fair die shown in Example 9 is an example of a very important type of probability distribution.
A particular function may be stated as
p(x) = x = 1, 2, 3, 4, 5
a Construct a table of values of the function.
b Could the function be a probability distribution function?
Solution
a Calculate the values.
b State obvious conditions. Every value is between 0 and 1.
Calculate total. ΣP(X = x)= + + + + = 1
State conclusion. All conditions are satisfied, so p(x) could be a probability distribution function. x
15
---x 1 2 3 4 5
p(x) 1 ≈ 0.067 ≈ 0.133 = 0.2 ≈ 0.267 ≈ 0.333 15
--- 2 15
--- 3 15
--- 4 15
--- 5 15
---1 15
--- 2
15
--- 3
15
--- 4
15
--- 5
15
---Example
8
Draw a histogram of the probability distribution for the random number D obtained by throwing a normal die and comment on your result.
Solution
Write the distribution.
Draw the histogram.
Comment on the shape. The histogram is the same height for all values of D, because the probability is the same for all values of
D. The probability function is a constant, p(d) =
d 1 2 3 4 5 6
p(d) 1
6 --- 1
6 --- 1
6 --- 1
6 --- 1
6 --- 1
6
---Outcomes of throwing a die
1 2 3 4 5 D
0.2
0.1
0 p(d)
6
1 6
---1 6
---.
Example
9
Uniform probability distribution
A uniform probability distribution has a constant value, so all the probabilities are the same. The histogram of a uniform probability distribution is rectangular.
A family has 4 children. Assuming that boys and girls are equally likely to be born:
a draw a histogram of the probability distribution of G, the number of girls in the family b find the probability of having from 1 to 3 girls.
Solution
a Using a tree diagram, we can obtain the
distribution. List the probabilities in a table.
Draw the histogram.
b Find the area for 1 to 3 girls. P(1 g 3)= area
= 1 × 0.25 + 1 × 0.375 + 1 × 0.25
= 0.875
Write the answer. The probability of having from 1 to 3 girls is 0.875.
g p(g)
0 = 0.0625
1 = 0.25
2 = 0.375
3 = 0.25
4 = 0.0625 1
16
---1 4
---3 8
---1 4
---1 16
---G = GGGG B = GGGB G = GGBG B = GGBB G = GBGG B = GBGB G = GBBG B = GBBB G = BGGG B = BGGB G = BGBG B = BGBB G = BBGG B = BBGB G = BBBG B = BBBB G
G
G
B
G
B
G
B
G
B B
G
B B
Girls in a 4-child family
0 1 2 3 4 G
0.2
0.1
0 0.3 p(g) 0.4
Example
10
1 Classify each of the following variables as discrete or continuous.
a The number of people in a car b The distance a car travels
c The number of journeys by a car in a week d The time a car is out of the garage e The number of cylinders in a car f The shoe sizes of men in a bus
g The heights of women in a bus h The prices of tracksuits
Exercise 9.2
Discrete probability distributions
Additional exercise
2 Which of the following could represent probability distributions?
3 A normal pack of cards is shuffled and the top card is turned over. The card is then replaced
and the pack reshuffled. The number of clubs from three cards, C, is noted.
a Construct the probability distribution for C. b Draw a histogram of the probability distribution.
4 A pair of normal dice are rolled and the sum of the numbers on the upper faces, S, is noted. a Construct the probability distribution for S.
b Draw a graph of the distribution. c Is it uniform?
Modelling and problem solving
5 The numbers of tracks on some randomly selected CDs were as follows:
11 13 19 20 13 11 15 13 14 16 16 14 12 18 20 12 14 14 16 20 20 14 14 12 13
a Use a frequency table to construct the probability distribution function for the number
of tracks, T.
b What fraction of the CDs have 12 tracks? c What fraction have 15 or more tracks?
6 A pair of normal dice are rolled and the following are noted:
The difference, D, of the numbers on the upper faces The smaller, S, of the numbers on the upper faces
a Construct the probability functions for both D and S.
b Draw graphs of the distributions. Is either distribution uniform? c The graphs are from the same data. Comment on the differences.
7 A coin is tossed 4 times. Let X be the random variable representing the largest number
of successive heads that occur in 4 tosses.
a Construct the probability distribution for X. b Draw a graph of the distribution.
8 Give two examples of situations that produce uniform probability distributions.
a x 10 15 20 b y −1 0 1 c x 1 2 3
p(x) 0.4 0.3 0.3 p(y) 0.15 0.75 0.10 p(x) 0.7 −0.2 0.5
Lists of Gold Lotto numbers are published from time to time in the newspaper. They are also available at www.goldencasket.com/gold_lotto/results.asp in the Golden Casket website. Use published results or results from the Internet site for this investigation into the distribution of Gold Lotto numbers.
Use your list of numbers to determine:
1 whether there are more even than odd numbers (using the ratio) 2 whether there are more numbers from 10 to 20 than from 30 to 40
3 the nature of the distribution by drawing a graph of the frequencies of all the numbers.
Discuss your findings as a class group.
9.3
Expected values
In many cases we are interested in the average value obtained from a theoretical or experimental probability situation.
In Example 11, we can also use the probabilities to calculate the average number of red lights Susan would expect to get on her way to work. The probability of 0.25 for 5 lights means that Susan should get 5 red lights (on average) a quarter of the time. This gives 5 × 0.25 = 1.25 red lights. Adding the red lights for the other three-quarters of the time should give the overall average.
0×0.00+1×0.10+2×0.10+3×0.05+4×0.35+5×0.25+6×0.10+7×0.00+8×0.05 =4.1
So this does work.
We define this average as the expected value for probability distributions in general, not just for those arising from experimental probabilities.
Susan drives through 8 sets of uncoordinated traffic lights on her way to work. She obtains the following numbers of red lights on 20 successive days on her way to work:
4 5 4 5 5 1 3 4 5 8 2 4 2 6 4 4 1 5 4 6
a Construct the probability distribution for R, the number of red lights obtained. b Calculate the average number of red lights.
Solution
a Make a frequency table and calculate the
probabilities as relative frequencies.
b Calculate the average (mean). Total number= 4 + 5 + 4 + … + 6 = 82
Average number= 82 ÷ 20 = 4.1
Write the answer. There is an average of 4.1 red lights on each trip. Red lights, r Frequency P(R = r)
0 0 0.00
1 2 0.10
2 2 0.10
3 1 0.05
4 7 0.35
5 5 0.25
6 2 0.10
7 0 0.00
8 1 0.05
Totals 20 1.00
Example
11
Expected value
The expected value, µ or E(X), of the probability distribution of the random variable
X is the mean value of X. For a discrete probability distribution, it can be calculated
using the formula
µ= E(X) =Σx × p(x)
The expected value of a distribution can be used to work out returns from games of chance.
By changing the amounts paid for winning and losing, the person running a game can make sure they will win in the long run. The proportion of bets a casino chooses to win in the long run is called the house percentage. It is the expected value of the operator, and naturally is negative for the person playing against the house.
A dice game is played so that a pair of dice are rolled and the total of the upper faces is counted. If the total is 11 or 12, you win $10. If the total is 2, 3 or 4, you win $4. For any other total, you lose $2. Is the game fair?
Solution
First calculate probabilities using a grid.
P(11 or 12)= =
P(2, 3 or 4)= =
P(others)= =
Write out the probability distribution. Let X be the random variable for the amount won.
Calculate the expected value. E(X) = 10 × + 4 × +−2 ×
= 0
Write the answer. The game is fair because you win $0 (and lose $0) in the long run, so you ‘break even’.
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
3 36
--- 1
12
---6 36
--- 1
6
---27 36
--- 3
4
---x 10 4 −2
P(X = x) 12---1 1
6
--- 3
4
---1 12
--- 1
6
--- 3
4
---Example
12
A game consists of a wheel divided into twelve equal sectors numbered 1 through 12. A $2 bet may be placed on any number. If your number comes up, you get $20 back. That is, you win $18 and get your $2 back.
a What is the house percentage?
b How much should you win for the game to be fair?
A graphics calculator can be used to calculate expected values.
Solution
Write out the probability distribution. You win $18 or lose $2.
b This is most easily calculated by considering the expected values of the wins and losses. a Choose the random variable. Let X be the amount to be won.
Calculate the expected value. E(X) = 18 × +−2 × ≈−0.33
Write as an amount. You expect to lose 33c on each bet of $2.
Calculate as a percentage. House percentage= × 100% ≈ 16.7%
Write a variable for the amount won. Let win amount= $A
Write the expected value of losses. Expected losses= 2 × =
Write the expected value of wins. Expected wins= A ×
Write an equation for a fair game. Expected wins= expected losses
A × =
Divide to find A. Use your calculator. A= ÷ = 22
Write the answer. For the game to be fair, you should win $22, so you should get back $24.
x +18 −2
P(X = x) 12---1 11
12
---1 12
--- 11
12
---0.33 2.00
---11 12
--- 11
6
---1 12
---1 12
--- 11
6
---11 6
--- 1
12
---A company is considering an investment and believes that the return in the following year will be a random amount X having the following probability distribution, where x is shown as thousands of dollars. Use a graphics calculator to find the expected return.
Solution
For the Casio and Texas Instruments calculators, the variable is entered in List 1 and the probabilities in List 2 and 1-variable statistics are calculated.
Casio CFX-9850GB PLUS
Use the STAT menu from the main menu.
Clear any data from List 1 and List 2 if necessary using
( ) (DEL A) (YES).
Enter the x values in List 1 and the P(X = x) values in List 2.
Enter the CALC submenu and SET the 1Var XList to List 1 and the 1Var Freq to List 2 using
( ) (CALC) (SET).
x (return) 100 250 300 350 400
P(X = x) 0.15 0.35 0.25 0.15 0.10
F6 F4 F1
F6 F2 F6
Example
14
Press to return to the CALC submenu and
then press to perform the calculations.
Texas Instruments TI-83
Press to enter the data.
Clear any data from L2 and L2 by pressing
.
Enter the x values in L1 and the P(X = x) values in L2.
Perform the 1-Var calculations by pressing 1
2 .
All calculators
Sharp EL-9650
The Sharp instructions are given on the CD-ROM.
The value of shown in the statistics is 270.
The value of shown in the statistics is 270.
Write the answer. The expected return is $270 000.
EXE
F1
x
STAT ENTER
CLEAR ENTER
STAT ENTER 2nd
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2nd ENTER
x
Calculator instructions
1 The results of a test out of 10 are shown below:
5 8 9 1 2 4 5 5 4 3 6 7 2 3 5 5 6 8 1 10 3 7 7 5 5 8 3 5 6 8 10 2 6 Find the expected value of the variable T for the test results.
2 Calculate E(X) for the following probability distribution.
3 Calculate E(X) for the following probability distribution.
4 If X is a random variable representing the number when a fair cubic die numbered 1–6
is rolled, calculate E(X).
5 Two dice are thrown and the sum of the numbers on the dice is noted. Calculate the
expected value.
x −2 0 2 4
p(x) 0.25 0.125 0.5 0.125
x −10 −5 5 10
p(x) 0.15 0.25 0.3 0.3
Modelling and problem solving
6 A dice game is played according to the following rules: Two dice are thrown. If the total
is less than 6 or greater than 9, the player wins $10. Otherwise the player loses $5.
a How much can you expect to win each time you play the game? b Is the game fair?
c If the game is unfair, how could it be altered to make it fair?
7 A coin is weighted so that P(H) = and P(T) = The coin is tossed 3 times and the results are recorded. If X is a random variable representing the greatest number of successive heads that occur:
a construct the probability distribution for X b find the expected value of X.
8 A particular game that operates from
newsagencies involves scratching the special coating away from a panel to reveal 6 amounts of money. If 3 matching amounts are revealed, the player wins that amount. In a certain series of the game, 10 million tickets are prepared and prizes are allocated as follows:
1 @ $100 000 2 @ $25 000 25 @ $1000 25 000 @ $100 25 000 @ $10 1 200 000 @ $2
If it costs $1 to play the game, calculate:
a the probability of winning a prize b the expected payout on each ticket c the house percentage.
9 A local supermarket has 4 checkout lanes. The manager has determined that the number of
lanes in use at 9:00 am is a random variable having the following probability distribution.
Calculate the expected number of lanes in use at 9:00 am at the supermarket.
10 New cars that come off an assembly line are fitted with 5 new tyres (including the spare).
The probability distribution for the number of defective tyres on each car is given as:
a Find the expected number of defective tyres per car. b What is the most likely number of defective tyres per car?
11 In a particular business venture, a person can make a profit of $10 000 with a probability of
0.2, a profit of $5000 with a probability of 0.2 and a loss of $1000 with a probability of 0.6. If X is the profit or loss that will result for the person, calculate:
a the expected value of X b the most likely value of X.
x 0 1 2 3 4
p(x) 0.10 0.15 0.15 0.20 0.40
x 0 1 2 3 4 5
p(x) 0.95 0.03 0.015 0.003 0.0015 0.0005
3 4
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9.4
Binomial probability distribution
In many probability situations, there are only two possible outcomes, such as heads or tails, boy or girl, etc. In many other situations, we can classify the outcomes so that there are only two possibilities, such as drawing an ace or not drawing an ace from a pack of cards. In these cases we classify the two outcomes as success and failure. Trials of this type are called Bernoulli trials (after Jacob Bernoulli, 1654–1705) and the distribution of the number of successes is called the binomial distribution.
Although the probabilities for the binomial distribution can be calculated using tree diagrams, the distribution is so common that tables of values have been produced. A binomial probability table is included on pages 462–65 in Appendix 1.
12 Two teams, A and B, play in a tournament. The probability that A wins any game in the
tournament is The first team to win 2 games in a row or a total of 3 games wins the tournament. Find the expected number of games in the tournament.
13 Two dice are loaded to increase the probability of throwing a 6. The actual probabilities of
the results are shown below.
A dice game is played where the players move round the board using the total, T, shown on two dice.
a Work out the probability function and expected value if one crooked die is used with
a normal die.
b Repeat the process if two crooked dice are used.
Number 1 2 3 4 5 6
Probability 0.05 0.15 0.15 0.15 0.15 0.35 1
2 ---.
Binomial distribution
A Bernoulli trial has the following characteristics:
• There are only two possible outcomes, usually called success and failure. • The probability of success ( p) and the probability of failure (q) do not change. • p + q = 1
• The result of one trial has no effect on other trials.
A binomial probability distribution is the distribution of the number of successes (x) from a fixed number n of Bernoulli trials, where the order of the successes does not matter.
!
State whether each of the following situations gives rise to a binomial distribution.
a A coin is tossed until 12 heads are obtained.
b The numbers of fair-haired children in families of 3 children are counted.
When using the binomial probability table, make sure you write down what you mean by success and failure, probability of success, number of trials and number of successes.
When you are using the table for probabilities greater than 0.50, the table is read from the bottom and right side instead of the top and left.
Solution
a Write the characteristics. There are only two possible outcomes.
The probability of success (heads) is the same for every trial.
The coin must land heads or tails, so p + q = 1.
The result of one toss does not affect others. We are interested in the number of successes. There may be different numbers of trials. The order of successes (heads) does not matter.
Write the answer. They are Bernoulli trials, but it is not a binomial distribution because there is not a fixed number of trials.
b Write the characteristics. There are only two possible outcomes.
The probability of success (fair) is the same for every trial.
A child must be fair or not fair, so p + q = 1.
The colour of one child’s hair does not affect others. We are interested in the number of successes. There are 3 trials—a fixed number.
The order of successes (fair) does not matter.
Write the answer. They are Bernoulli trials and it is a binomial distribution.
Example
16
A bag of 20 marbles contains 7 green ones. A marble is taken out and replaced. Use the binomial probability table to find the probability that 3 green marbles are obtained from 10 such trials.
Solution
Find the value from the table.
Find p for ‘success = green marble’. p = = 0.35 Write n and x. n = 10, x = 3
p
x 0.01 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
n = 10 0 1 2 3 4 5 0.9044 0.0914 0.0042 0.0001 0.5987 0.3151 0.0746 0.0105 0.0010 0.0001 0.3487 0.3874 0.1937 0.0574 0.0112 0.0015 0.1969 0.3474 0.2759 0.1298 0.0401 0.0085 0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 10 9 8 7 6 5
Write the answer. The probability of 3 green marbles is 0.2522. 7
20
---You can use a graphics calculator to find binomial probabilities.
Example
17
Find the binomial probability of 8 successes from 14 trials, where the probability of success is 0.77.
Solution
Read the table from the bottom and right, using p = 0.75, the closest value to 0.77. Write the values of p, n and x. p = 0.77, n = 14, x = 8
n = 14 0 0.8687 0.4877 0.2288 0.1028 0.0440 0.0178 0.0068 0.0024 0.0008 0.0002 0.0001 14 1 0.1229 0.3593 0.3559 0.2539 0.1539 0.0832 0.0407 0.0181 0.0073 0.0027 0.0009 13 2 0.0081 0.1229 0.2570 0.2912 0.2501 0.1802 0.1134 0.0634 0.0317 0.0141 0.0056 12 3 0.0003 0.0259 0.1142 0.2056 0.2501 0.2402 0.1943 0.1366 0.0845 0.0462 0.0222 11 4 0.0037 0.0349 0.0998 0.1720 0.2202 0.2290 0.2022 0.1549 0.1040 0.0611 10 5 0.0004 0.0078 0.0352 0.0860 0.1468 0.1963 0.2178 0.2066 0.1701 0.1222 9 6 0.0013 0.0093 0.0322 0.0734 0.1262 0.1759 0.2066 0.2088 0.1833 8 7 0.0002 0.0019 0.0092 0.0280 0.0618 0.1082 0.1574 0.1952 0.2095 7 8 0.0003 0.0020 0.0082 0.0232 0.0510 0.0918 0.1398 0.1833 6 9 0.0003 0.0018 0.0066 0.0183 0.0408 0.0762 0.1222 5
10 0.0003 0.0014 0.0049 0.0136 0.0312 0.0611 4
11 0.0002 0.0010 0.0033 0.0093 0.0222 3
12 0.0001 0.0005 0.0019 0.0056 2
13 0.0001 0.0002 0.0009 1
14 0.0001 0
0.99 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 x p
Write the answer. The probability of 8 successes is about 0.0734.
Fifteen binomial trials are conducted with a probability of success of 0.63. Use a graphics calculator to find the probability of 6 successes.
Solution
Graphics calculators use the probability function, abbreviated to Bpd, binompdf or
pdfbin for the probability of x successes.
Casio CFX-9859GB PLUS
Use the STAT menu.
Press to access DIST, the distributions.
Press again to access BINM, the binomial distribution.
Press to access Bpd, the density function.
Press to access Var, the variable data type.
Now enter the values and press to find the value.
Write the binomial probability distribution values. p = 0.63, n = 15, x = 6
F5
F5
F1
F2
EXE
Example
18
We can use the tables of the binomial distribution to find expected values.
In Example 19, we could have worked out the expected number of spades using the rule for expected frequency (= probability × number of trials).
Texas Instruments TI-83
Press to use the DISTR menu.
Scroll down to 0: binompdf. Press .
Complete the function by entering 15 0.63
6 . Press .
All calculators
Sharp EL-9650
The Sharp instructions are given on the CD-ROM.
Write the answer. The probability of 6 successes is about 0.0407.
2nd VARS
ENTER
,
,
) ENTERCalculator instructions
A card is drawn from a well-shuffled pack and then replaced. The number of spades is noted for each of 8 draws. What is the expected value for the number of spades?
Solution
Use the binomial distribution table on pages 462–65 to find the values of the probability distribution.
Write the characteristics. Success = ‘drawing a spade’
p = = 0.25, n = 8
Calculate the expected value. E(X)= 0 × 0.1001 + 1 × 0.2670 + 2 × 0.3115
+ 3 × 0.2076 + 4 × 0.0865 + 5 × 0.0231
+ 6 × 0.0038 + 7 × 0.0004 + 8 × 0
= 1.9999 ≈ 2
Write the answer. The expected number of spades is 2. 1
4
---x = number of spades P(X = x)
0 0.1001
1 0.2670
2 0.3115
3 0.2076
4 0.0865
5 0.0231
6 0.0038
7 0.0004
8 0
Expected value of a binomial distribution
For a binomial distribution,
E(X) = np
!
Assuming that births of boys and girls are equally likely, how many girls would you expect in a family of 6?
Solution
Write the characteristics. Success = ‘a girl’, p = 0.5, n = 6
Calculate the expected value. E(X)= np
= 6 × 0.5 = 3
Write the answer. We would expect 3 girls.
Example
20
Did you know?
The percentage of boys born is not actually 50%, but is closer to 52%. In general, boys are more likely to die than girls, so that by 5 years old there are as many girls as there are boys. Also, in general, women live longer than men, so there are many more elderly women than elderly men.
Exercise 9.4
Binomial probability distribution
Use the binomial probability table on pages 462–65 or a graphics calculator for this exercise.
1 State whether or not each of the following situations gives rise to a binomial distribution. a Players take it in turns to be dealt a card from a well-shuffled pack. The first player
with a spade wins.
b Players are each dealt 4 cards, and the player with the most spades wins.
c Each player throws a die 5 times. The player with the highest number of sixes wins.
d The numbers of male children in families are counted.
e The numbers of male children in families of 5 children are counted.
2 A die is rolled 4 times and the number of times that a six occurs is noted. a What is n?
b Find p and q.
c Use tables to find the approximate probability that 1 six is rolled.
d Use tables to find the approximate probability that 2 sixes are rolled.
3 The Bernoulli family has 4 children. It is known that the genetic makeup of the parents
is such that there is a 30% chance that a child born will have blue eyes.
a What are the values of n, p and q for this situation?
b What is the probability that all of the Bernoulli children have blue eyes? c What is the probability that 2 of the Bernoulli children have blue eyes?
d What is the probability that at least 1 of the Bernoulli children has blue eyes?
Graphics calculator Additional
4 A fair die is rolled 6 times. Find the probability of rolling:
a 4 sixes b 2 sixes c no sixes d all sixes.
5 A coin is tossed 12 times. Find the probability of tossing:
a 2 heads b 3 tails c 7 heads.
Modelling and problem solving
6 In a particular street, there are 3 sets of traffic lights. There is an even chance that, at any
given time, the lights will show green. (You should stop when the light facing you is orange
or red.) When a car travels down the street, what is the probability that: a it will have a clear run (all green lights)?
b it will have to stop twice? c it will have to stop at least once?
7 A poll found that 65% of people interviewed across Australia were in favour of a proposal
regarding the powers of police. What is the probability that, if 5 people are selected at random:
a exactly 3 will be in favour of the proposal? b at least 4 will be in favour?
8 A manufacturer of toothbrushes finds that 1 in every 20 produced has a fault of some
description. The toothbrushes are marketed in packs of 10. What is the probability that a pack will have 2 faulty toothbrushes?
9 A manufacturer of a new type of tap washer finds that, on average, 1 in every 10 washers
is faulty because it is either oversized or undersized. A machine automatically places 8 washers in the packets that pass under the packaging machine. What is the probability that a pack of 8 washers will contain:
a no more than 1 defective washer? b at least 1 defective washer?
10 School records over a 10-year period show that there is a 15% chance that a student is
left-handed. If a class has 15 students, find the probability that:
a none is left-handed b 4 are left-handed
c at least 2 are left-handed d no more than 3 are left-handed.
11 It is known that fatigue causes 20% of all industrial accidents.
a Last week, industrial inspectors discovered 10 industrial accidents. What is the
probability that at least 3 of them were caused by fatigue?
b The Feldon Chemical Company reported that, of the 8 industrial accidents reported at
its factory in July, it was ‘highly unlikely’ that any more than 1 of them resulted from fatigue. Comment on this statement.
12 It is found that 5% of all electronic components manufactured are rejected by the quality
assurance section because of faulty wiring. If an instrument contains 10 of these components, find the probability that:
a the instrument works (that is, has no faulty components) b the instrument contains at least 1 faulty component c the instrument contains fewer than 3 faulty components.
Chapter summary
Chapter
Review
Communication and justification
1 What is meant by the multiplication principle?
2 Explain the addition rule for probabilities.
3 What is a Venn diagram?
4 If you know that the probability of event A occurring is explain how to find the probability that event A does not occur.
5 Explain how fair odds are calculated from a probability.
6 What is a discrete random variable?
7 Describe the properties of a discrete probability function.
8 Describe a uniform probability distribution.
9 What is meant by the term ‘expected value’?
10 What are meant by Bernoulli trials?
11 Describe the conditions of a binomial probability situation.
Knowledge and procedures
12 If 3 normal coins are tossed together, the probability of obtaining exactly 2 heads is:
A B C D E
13 What is the probability of being dealt 4 fours from a well-shuffled 52-card standard
pack of cards?
14 A coin is weighted so that the probability of heads is 0.7. What is the probability of
getting at least 2 heads from 3 tosses of the coin?
15 A carton of eggs contains three 55 g eggs, five 59 g eggs and two 52 g eggs. The
probability of choosing two 55 g eggs at random is:
A 0.09 B 0.067 C 0.05 D 0.1 E 0.033
16 What are the fair odds for getting exactly 2 heads from 3 tosses of a coin?
17 A discrete variable:
A can have only integer values B can have only integer values or halves C can have only rational values D can have only specific values
E can have only real values.
18 Which of the following could represent a probability distribution?
a x 0 1 2 3 b x 0 1 2 3
p(x) 0.5 0.3 0.2 0 p(x) 0.1 0.3 0.4 0.1
c x 0 1 2 3 d x 0 1 2 3
p(x) 0.1 0.2 0.3 0.4 p(x) −0.4 0.3 0.6 0.7
e x 0 1 2 3
p(x) 2 3 4 5
Ex 9.1
Ex 9.1
Ex 9.1
Ex 9.1 1
3 ---,
Ex 9.1
Ex 9.2
Ex 9.2
Ex 9.2
Ex 9.3
Ex 9.4
Ex 9.4
Ex 9.1 3
10
--- 3
8
--- 5
8
--- 7
8
--- 2
3
---Ex 9.1
Ex 9.1
Ex 9.1
Ex 9.1
Ex 9.2
19 A pair of fair dice are rolled and the difference between the upper faces is noted. a Construct the probability function for D, the random variable of the difference. b Draw a histogram of the distribution.
c Find the expected value of D.
20 What is the expected value from throwing a single die?
21 Find the expected value of the following probability distribution.
22 Find the following binomial probabilities using tables or a graphics calculator. a P(X = 5) when n = 15 and p = 0.3 b P(X = 7) when n = 19 and p = 0.45
23 A binomial probability distribution has 20 trials and the probability of success is 0.7.
The expected value of the distribution is:
A 7 B 17 C 14 D 12.5 E 10
24 An Olympic archer can hit the bullseye an average of 9 times out of 10. The probability
of the archer scoring 18 or more bullseyes from 20 shots is:
A 0.677 B 0.190 C 0.285 D 0.270 E 0.900
Modelling and problem solving
25 A euchre pack of cards consists of the ace, 7–10, jack, queen and king of the four suits,
together with a joker.
a What is the probability of dealing 2 kings in the first 2 cards?
b What is the probability of dealing 2 jacks of the same colour and the ace of hearts
in the first 3 cards?
26 A card game for two players proceeds as follows: Each player is dealt a card. If either
card is a spade, that player wins. If neither card is a spade or both are spades, the cards are replaced and reshuffled, then the cards are dealt again. Play continues for up to 7 deals. If neither player has won by spades on the seventh deal, then the player with the higher card wins; if they are of equal rank, then the player with the higher suit wins. Suits are ranked ♠, ♦, ♥, ♣(highest to lowest). The random variable D is the number of deals in a game. Construct the probability function for the random variable D and find the expected value of D.
27 A target shooter knows from experience that she hits a target 4 times out of each 5 shots.
If she is allowed 10 shots in a competition, what is the probability that she gets:
a 10 hits? b 9 hits? c at least 7 hits?
28 During a flu epidemic, the probability that children will be absent from school is 15%.
What is the probability that from a class of 20 students:
a 7 are away? b at least 7 are away? c at least 5 are away?
29 A new vaccine is claimed to be 85% effective in immunising young children against a
particular childhood disease. In a sample of 15 children who are vaccinated, what is the probability that:
a all become immune to the disease? b at least 2 do not become immune? c fewer than 4 do not become immune?
x 2 3 4 5 6 7 8
P(X = x) 0.1 0.2 0.3 0.15 0.1 0.1 0.05
Ex 9.2, 3
Ex 9.3
Ex 9.3
Ex 9.4
Ex 9.4
Ex 9.4
Ex 9.1
Ex 9.2, 3
Ex 9.4
Ex 9.4