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Probability (3.1 – 3.3)
3.1 Random experiments, equally likely outcomes; probability of a given result.
3.2 Sum and product of results.
3.3
Experiments involving
successive outcomes: tree diagrams.” (syllabus)
For a more detailed description of the requirements for this topic, see the mathematics syllabus on the Board of Studies website click here
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PROBABILITY
1.
One card is drawn at random from a normal pack of 52 cards.
What is the probability of drawing
(i)
A red
card?
(ii)
A black
card?
(iii)
A
heart?
(iv)
A
picture card?
(v)
The
ace of spades?
2.
A coin
is tossed 3 times. What is the probability of tossing
(i)
Three
heads?
(ii)
Two
heads and a tail?
(iii)
At
least one head?
(iv)
No
heads.
3.
A
drawer contains 10 black socks and 10 brown socks. If Josh goes to the drawer
in the dark and chooses two socks at random, what is the probability of him
choosing
(i)
Two
black socks?
(ii)
A
matching pair?
4.
The
drawer now contains 12 black socks and 8 brown socks. If Josh again goes to the
drawer in the dark and chooses two socks at random, what is the probability of
him choosing
(i)
Two
black socks?
(ii)
A
matching pair?
5.
Two
dice are thrown and the score is taken as the sum of the dots on the two
uppermost faces.
(i)
Draw a
grid showing all of the possible outcomes. Use your grid to determine
(ii)
The
probability of throwing 12.
(iii)
The
probability of throwing an even total.
(iv)
The
probability of throwing a total greater than 7.
(v)
The
probability of throwing a 1 on at least one of the dice?
Answers:
1. (i)
(ii)
(iii)
(iv)
(v) ![]()
2. (i)
(ii)
(iii)
(iv) ![]()
3. (i)
(ii) ![]()
4. (i)
(ii) ![]()
5. (i)
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1 |
2 |
3 |
4 |
5 |
6 |
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1 |
2 |
3 |
4 |
5 |
6 |
7 |
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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 |
(ii)
(iii)
(iv)
(v) ![]()
MORE PROBABILITY
(i) Even?
(ii) Divisible by 4?
(iii) Less than 120?
(iv) Greater than 230?
(i) First prize
(ii) First and second prize
(iii) Second prize
(iv) No prize
(v) Explain why answers (i) + (ii) + (iii) + (iv) =1
(i) All 4 are girls?
(ii) There are 2 girls?
(iii) At least one is a girl?
(iv) There are no girls?
(i) They both play tennis?
(ii) One plays tennis and one plays soccer?
(iii) Neither plays tennis?
(i) The first marble drawn is blue?
(ii) The first marble drawn is white?
(iii) Both the first and second marbles drawn are white?
(iv) Both marbles are red?
(v) Neither marble is white?
Answers:
1. (i)
(ii)
(iii) 0 (iv) ![]()
2. (i)
(ii)
(iii)
(iv)
(v) Answers (i) to (iv) represent all
possible outcomes.
All
possible outcomes add to 1.
3. (i)
(ii)
(iii)
(iv) ![]()
4. (i)
(ii)
(iii) ![]()
5. (i)
(ii)
(iii)
(iv)
(v) ![]()
EVEN MORE PROBABILITY
(i) Lauren’s lucky dip contains a whistle?
(ii) Emma’s lucky dip contains a comic?
(iii) Both Lauren’s and Emma’s lucky dip contain a ring?
(iv) Neither Lauren’s nor Emma’s lucky dip contain a badge?
(v) Lauren’s lucky dip contains a ring and Emma’s lucky dip contains a badge?
(i) Even?
(ii) Odd?
(iii) Divisible by 5?
(iv) Greater than 600?
(v) Less than 600?
(vi) Greater than 750?
(i) How many two digit numbers could the teacher form?
(ii) If he allocated each student in the class of 30, a different two digit number, what is the probability of Sarah, a member of the class, being allocated the number 63?
(iii) What is the probability that the first number that the teacher rolls has not been allocated to any student?
(iv) What is the probability that the first two numbers that the teacher rolled were the same?
(v) What is the probability that the three numbers that the teacher rolls have not been allocated so that the teacher does not check anybody’s homework?
(i) He could spell his own name with the letters chosen?
(ii) He could spell the name “Tom” with the letters chosen?
Answers: 1. (i)
, (ii) 1, (iii)
, (iv)
, (v)
2. (i)
, (ii)
, (iii)
, (iv)
, (v)
, (vi) ![]()
3.
(i) 36, (ii)
, (iii)
, (iv)
, (v)
4. (i)
, (ii) ![]()
PROBABILITY PROBLEMS
1. A letter is chosen at random from the letters of the word “MATHEMATICS”. What is the probability that it is
(i) an “A”?
(ii) a vowel?
(iii) a consonant?
2. A coin is tossed 5 times. What is the probability of tossing
(i) 5 heads?
(ii) no heads?
(iii) 4 heads?
(iv) at least 3 heads?
3. A card is drawn at random from a normal deck of playing cards.
What is the probability that the card drawn will be
(i) an ace?
(ii) a spade?
(iii) an ace and a spade?
(iv) either an ace or a spade?
(v) not a spade?
4. In a group of 25 students, 17 study physics, 13 study chemistry and 6 study no
science subjects at all. If a student is chosen at random, what is the probability that he or she studies
(i) no science?
(ii) physics?
(iii) physics but not chemistry?
(iv) physics, chemistry or both?
(v) both physics and chemistry?
5. The “El- Cheapo Calculator Company” manufactures calculators but 10% of them are faulty. Zaf bought 3 El-Cheapo calculators. What is the probability that
(i) the first one he used was faulty?
(ii) all 3 were faulty?
(iii) at least one was faulty?
(iv) all 3 were good?
Answers:
1. (i)
(ii)
(iii) ![]()
2. (i)
(ii)
(iii)
(iv) ![]()
3. (i)
(ii)
(iii)
(iv)
( don’t count the ace of spades twice) (v) ![]()
4. (i)
(ii)
(iii)
(iv)
(v) ![]()
5. (i)
(ii)
(iii)
(iv) ![]()
PROBABILITY: THE PRODUCT RULE
The number of ways that two mutually exclusive events can occur is equal to the product of the number of ways that the first can occur and the number of ways that the second can occur.
The probability of two mutually exclusive events occurring is equal to the product of the probability of the first event occurring and the probability of the second event occurring.
PAB = PA x PB
Example: Matthew has a choice of a beef burger, chicken burger, fish burger or vegie burger and also of a cola drink or orange drink.
(i) How many combinations of a burger and a drink are there?
(ii) If Matthew chooses a burger and a drink, what is the probability of him choosing a vegie burger and orange drink if all burger choices and all drink choices are equally probable?
Answer: (i) There are four choices of burger and two choices of drink, so the number of combinations is 4 x 2 = 8.
(ii) Probability
of choosing a vegie burger is
and probability of
choosing an orange drink is
.
Probability of vegie burger and orange drink is
x
= ![]()
Exercise:
1. Students at
What is the probability that a student will choose biology and history?
2. Hickstown, Squaresville and Crudvale are three towns. There are 4 roads from Hickstown to Squaresville and 6 roads from Squaresville to Crudvale. In how many ways may a motorist travel from Hickstown to Crudvale via Squaresville?
3. A school committee consisting of 2 boys and 2 girls is to be chosen at random. Bert, Charlie, Fred and George have applied for the boys’ positions and Mary, Sue and Sally have applied for the girls’ positions. What is the probability that both Charlie and Sue are on the committee?
4. A coffee shop offers a choice of cappuccino, flat white or late`, with normal milk, skim milk or soy milk and sugar or saccharine. How many combinations of coffee, milk and sweetener are possible?
5. A code for a person’s registration consists of a letter of the alphabet followed by a single digit, e.g. B5.
(i) How many combinations are possible?
(ii) What is the probability of a person being allocated the code Z4?
(iii) What is the probability of a particular person being allocated a code ending in 5?
Answers: 1.
2. 24 3.
4. 18 5. (i) 260
(ii)
(iii) ![]()
PROBABILITY: PERMUTATIONS
Permutations refer to the number of ways a given set can be arranged. For instance the letters abc can be arranged 6 ways: abc, acb, bac, bca, cab, cba.
Example: In how many ways can 5 people queue in single file for movie tickets?
Answer: The first position in the queue may be filled in one of 5 ways. There are now 4 people left so the second position may be filled in one of 4 ways. Similarly the third position can be filled in 3 ways, the fourth position in 2 ways and the fifth position in 1 way.
Hence the number of arrangements for 5 people in a queue is 5 x 4 x 3 x 2 x 1 = 120.
The number of arrangements for n items is: n x (n-1) x (n-2) x …………x 2 x 1
The product of an integer n and all the positive integers below it is called “factorial n” and is written n!
Exercise 1:
1. In how many ways may the letters of the word “MONDAY” be arranged?
2. Cathy received 5 DVDs for her birthday. In how many ways can she arrange the order of viewing her 5 DVDs?
3. Thirty students enter a classroom in single file. In how many ways can the order of them entering the classroom be arranged?
Sometimes not all of the positions have to be allocated.
Consider a raffle where there are 3 prizes and 100 tickets.
The first prize may be allocated in one of 100 ways. Once the first prize has been drawn there are 99 tickets left so the second prize may be allocated in one of 99 ways. There are 98 tickets left for the third prize so the third prize may be allocated 98 ways.
Thus the number of ways that the first three prizes may be distributed is:
100 x 99 x 98 = 970200
Another way of writing this is
i.e. ![]()
The number of ways r objects can be
arranged from n different objects is: ![]()
This is written nPr and can be determined directly by most calculators.
Exercise 2:
(ii) What is the probability of picking the first 5 places in correct order from a field of 18 horses of equal ability?
3. A librarian has 80 equal size books but only has room for 10 of them on the shelf. In how many ways can she arrange the ten books on the shelf?
Answers : Ex.1. 1. 6! = 720 2. 5! = 120 3. 30! = 2.65 x 1032
Ex.2.
1. 2193360 2. (i)
(ii)
3. 5.97 x 1018
PROBABILITY: COMBINATIONS
The order in which elements of the set are selected is often not important. For instance if you are playing cards, the cards in your hand are important while the order in which you were dealt those cards is not.
Example: In how many ways can a committee of three be chosen from a group of 10?
Answer:
If order is important then the number of ways is 10P3 = 720 ways.
Suppose Tom, Dick and Harry are the three
representatives. The chance of them being chosen in that order is
. But there are 3! i.e.6 ways in which Tom, Dick & Harry
can be chosen. (TDH, THD, DTH, DHT, HTD, HDT) so the probability of Tom Dick
and Harry being on the committee becomes
= ![]()
And the number of possible arrangements is 120.
This is known as the number of combinations and is written nCr and may be calculated by :
nCr
=
This can be
determined directly by most calculators.
Exercise 1:
Sometimes we combine the product rule with combinations.
Example: In how many ways can a committee of 3 boys and 3 girls be selected from a class of 16 boys and 14 girls?
Answer: The boys may be chosen 16C3 ways and the girls 14C3 ways.
Boys 16C3 = 560 ways Girls 14C3 = 364 ways
Total = 560 x 364 = 203840 ways.
Exercise 2:
1. How many ways may a Parliamentary Committee be formed if it is to consist of 4 government members and 3 opposition members if 8 government and 10 opposition members apply?
2. A bookshelf contains 20 mathematics books and 14 science books. In how many ways can a person choose 2 mathematics books and two science books?
3. A class consists of 17 boys and 13 girls. In how many ways is it possible to choose a group of 3 boys and 2 girls?
Answers:
Ex.1. 1.
2598960 2. 3268760 3.
=
4. 3.54 x 1018
Ex.2. 1. 8400 2. 17290 3. 53040
taken;
(i) 2 at a time?
(ii) 3 at a time?
(iii) 4 at a time?
(i) 2 at a time?
(ii) 3 at a time?
(iii) 4 at a time?
(i) 2 As?
(ii) an A and an I?
(iii) one vowel only?
(iv) no vowels?
(i) in correct order?
(ii) in any order?
(i) both are the letter “A”?
(ii) both are the same?
Answers:
1.
x
x
x
x
x
x
x
x
= ![]()
2. (i) 22 (ii) 42 (iii) 56
3. (i) 43 (ii) 229 (iii) 1044
4. (i)
(ii)
(iii)
(iv) ![]()
5. (i)
(ii) ![]()
6. Yes. Both have 56 combinations.
7. 24
8. (i)
(ii) ![]()