Making Predictions with Data

From understanding the odds of winning a lottery, predicting weather conditions or evaluating risks like health issues, probability provides a framework for assessing the likelihood of various outcomes. It also prevents common misconceptions, such as assuming random events are predictable. Probability empowers you to critically analyse statistics presented in the media, advertisements or health studies, ensuring you make informed choices. By grasping the basics of probability, you can approach situations with a clearer, more logical perspective. This section details the application of addition and multiplication laws and axioms of probability as well as the investigation of these axioms as a build-up of what you learnt in year one.

KEY IDEAS

• If the events are not mutually exclusive, the probability of either event A or event B occurring is the sum of their individual probabilities minus the intersection probability: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

• If two events A and B are dependent, the multiplication rule incorporates the conditional probability of event B occurring, given that event A has already occurred. That is, P(A ∩ B) = P(A) ⋅ P(B∣A)

• If two events A and B are independent the probability of both events happening is the product of their individual probabilities: P(A ∩ B) = P(A) ⋅ P(B)

• If two events A and B are mutually exclusive the probability of either event A or event B occurring is the sum of their individual probabilities:

P(A ∪ B)=P(A) + P(B)

Figure 13.1: A pile of playing cards

The concept of probability assists in quantifying uncertainty, helping us evaluate the likelihood of events in a systematic way. Among its fundamental principles are the addition law and the multiplication law, which are essential for understanding and solving problems involving combinations of events. We shall explore how the addition and multiplication laws help in solving problems faster and simpler.

Addition Law of Probability

The addition law for probability is a fundamental principle in probability theory that finds the possibility of the occurrence of at least one of two, or more, mutually exclusive or not mutually exclusive events. It is also known as the “OR” rule in probability.

Remember that if events are mutually exclusive, they cannot happen at the same time.

Activity 13.1: Establishing the Addition Law of Probability

In small groups, work through the following activity to establish the addition law. Discuss each stage as you go, ensuring you understand what is happening.

1. State the De Morgan’s Law, (A ∪ B)ᶜ= Aᶜ∩ Bᶜ2. Remember that, for example, Aᶜ, which can also be written as A′, means the complement of A, or the probability that A does not occur. P(Aᶜ) = 1 − P(A)

3. State the complement rule P(A ∪ B) = 1 − P((A ∪ B)ᶜ)

4. Substitute the De Morgan’s result into the complement P(A ∪ B) = 1 − P(Aᶜ∩ Bᶜ)

5. Express P(Aᶜ∩ Bᶜ) in terms of probabilities of A and B.

From the complement rule, P)Aᶜ( = 1 − P)A(, P)Bᶜ( = 1 − P)B(

6. Connect P(Aᶜ∩ Bᶜ) to P(A ∩ B).

7. Consider Aᶜand Bᶜas independent then, P(Aᶜ∩ Bᶜ) = P)Aᶜ( ∙ P(Bᶜ( (As we can recall that if events are independent then P(A ∩ B) = P(A) ⋅ P(B))

8. Substitute P)Aᶜ( = 1 – P(A) and P(Bᶜ( = 1 − P)B( into P)Aᶜ( ∙ P(Bᶜ(.

P(Aᶜ∩ Bᶜ) =)1 – P)A( (∙ ( 1 − P)B()

9. Substitute the result of P(Aᶜ∩ Bᶜ) into P(A ∪ B).

P(A ∪ B) = 1− [)1 – P)A( (∙ )1 − P)B((]

10. Simplify the terms on the RHS of the equation P(A ∪ B) = P)A( + P)B(− P)A( ∙ P)B( if events A and B are independent P(A ∪ B) = P)A( + P)B(− P)A ∩ B( We also have P(A ∪ B) = P)A( + P)B( if events A and B are mutually exclusive as, in this case, P(A ∩ B) = 0, as by definition there is no overlap as the two events cannot occur simultaneously.

The addition law can also be represented in a Venn Diagram:

Figure 12.2: Venn diagram

Example 13.1

In a local market in Accra, a vendor sells 30 fruits: 18 are oranges and 12 are bananas. If 5 of the oranges and 3 of the bananas are ripe, what is the probability of randomly selecting either a ripe orange or a ripe banana?

Figure 12.3: A basket of fruits

Solution

Step 1: Extract the totals Total fruits = 30 Total oranges = 18 Ripe oranges (R_(O)) = 5 Total Bananas = 12 Ripe bananas (R_(B)) = 3

Step 2: Determine the probability of randomly picking a ripe orange P( R_(O)) = 5/30

Step 3: Determine the probability of randomly picking a ripe banana P( R_(B)) = 3/30

Step 4: Analyse if a fruit can be both a ripe orange and a ripe banana at the same time A fruit cannot be both a ripe banana and ripe orange so the intersection is zero as the events are mutually exclusive.

Step 5: Substitute the known probabilities in the addition law and simplify.

P(R_(O) ∪ R_(B)) = P(R_(O)) + P(R_(B)) − P(R_(O) ∩ R_(B)) = 5_ 30+ 3_ 30 − 0 = 8_ 30 = 4_ 15 = 0.26˙ Therefore, the probability of randomly selecting either a ripe orange or a ripe banana is 0.26˙.

Example 13.2

In East Legon residential area, the likelihood of a teenager owning a skateboard is 0.37 while the chance of owning a bicycle is 0.81.

If the probability that a teenager owns both a skateboard and a bicycle is 0.36, determine the likelihood that a teenager owns either a skateboard or a bicycle.

Figure 12.4: Skateboard

Solution

Step 1: Extract each probabilities given P(skateboard) = 0.37, P(bicycle) = 0.81, P(skateboard and bicycle)= 0.36

Step 2: Substitute the known probabilities in the addition law.

P(skateboard or bicycle) = P(skateboard) + P(bicycle)− P(skateboard and bicycle) P(skateboard or bicycle) = 0.37 + 0.81 − 0.36 = 0.82 Therefore, the probability of a teenager owning either a skateboard or a bicycle is 0.82.

Example 13.3

There are 200 first year students in a school in Tamale, 120 boys and 80 girls.

Among the students, 30 boys and 20 girls are members of the science club.

If a prefect needs to be selected from the class, what is the probability of the prefect being either a boy or a student from the science club?

Solution

Step 1: Extract the needed totals Total number of Boys = 120 Total number of boys in the science club = 30 Total number of science club members = 50 Total number of students = 200

Step 2: Determine the probability of picking a boy P(B) = 120/200

Step 3: Determine the probability of being a science club member P( S_(M)) = 50/200

Step 4: Determine the probability of being a boy and a science club member P(B ∩ S_(M)) = 30/200

Step 5: Substitute the known probabilities in the addition law and simplify.

P(B ∪ S_(M)) = P(B) + P(S_(M)) − P)B ∩ S_(M)( = 120_ 200+ 50_ 200 − 30_ 200 = 140/200 = 0.7 Therefore, the probability of the prefect being either a boy or a student from the science club is 0.7.

Example 13.4

100 tickets were sold at a national lottery point in Ghana.

20 of these tickets win a cash prize, whilst 15 tickets win a gift prize and 10 of these winning tickets win both.

What is the probability of winning any prize?

Solution

Step 1: Extract the needed totals Number of cash prize tickets = 20 Number of gift prize tickets = 15 Number of these which are cash and gift tickets = 10

Step 2: Determine the probability of picking a cash prize ticket.

P(C ) = 20/100

Step 3: Determine the probability of picking a gift ticket.

P(G) = 15/100

Step 4: Determine the probability of picking a cash and gift prize ticket.

P(C ∩ G ) = 10/100

Step 5: Substitute the known probabilities in the addition law and simplify.

P)C ∪ G( = P)C( + P)G(− P)C ∩ G( = 20_ 100+ 15_ 100 − 10_ 100 = 25_ 100 = 1_ 4 = 0.25 Therefore, the probability of winning any prize is 0.25.

Multiplication Laws of Probability

The multiplication rule of probability is essential for calculating the probability of two or more events occurring together. It can be applied to both independent and dependent events.

For independent events, where the outcome of one event has no effect on the other event happening, the probability of both events occurring is the product of their individual probabilities: P(A ∩ B) = P(A) × P(B) For dependent events, where the outcome of one event affects the other’s outcome, the formula adjusts to: P(A ∩ B) = P(A) × P(B|A) where P(B|A) is the probability of event B occurring given that A has occurred.

Let us go through some examples to make this clear.

Example 13.5

Suppose you are playing a game where you roll two six-sided dice.

What is the probability that both dice show a number greater than 3?

Solution

Step 1: Find the probability of rolling a number greater than 3 on one die The numbers greater than 3 are 4, 5, and 6.

P(x > 3) = 3/6 = 1/2

Step 2: Determine whether the rolls are independent or not They are independent as rolling a number greater than 3 on one die has no impact on whether the other die roll is greater than 3.

Step 3: Apply the independent events law of multiplication P(D₁ ∩ D₂ > 3) = 1/2 × 1/2 = 1/4 The probability that both dice show a number greater than 3 is 1/4.

Figure 12.5: Two six-sided dice

Example 13.6

In Sunyani, the probability of rain on any given day is 0.6. If it rains, there is a 0.8 chance that sales at a local market will increase.

What is the probability that it rains and sales increase?

Solution

Step 1: Restate the probability that it rains(R) P(R) = 0.6

Step 2: Restate the probability that sales at a local market will increase if it rains P(S_(I)|ᴿ) = 0.8

Step 3: Determine whether rains and sales increase are independent or not The events are dependent, as it depends on the rain to increase sales at the market.

Step 4: Apply the dependent events law of multiplication P(R ∩ S_(I)( = P(R) × P(S_(I)|ᴿ) P( R ∩ S_(I)( = 0.6 × 0.8 = 0.48

An axiom is a statement that is accepted as true without proof, serving as a basis for further reasoning or arguments. Axioms are crucial in various fields, particularly in mathematics and logic. They form the starting points from which theorems and other statements are derived. One common axiom is the axiom of equality in mathematics which states that “a number is always equal to itself”. This is based on the idea that “things which are equal to the same thing are also equal to each other”. This underpins many mathematical operations.

Let us look at the axioms of probability.

The Axioms of probability The axioms of probability are principles that define the mathematical framework for probability theory. The axioms of focus in this section are:

1. Unit Measure Axiom:

The probability of the entire sample space (S) is equal to one, P(S)=1.

This axiom asserts that when considering all possible outcomes of a random experiment, at least one outcome must occur, making the total probability of all outcomes equal to 1, or 100%.

2. The probability of an empty set is zero. Thus, P(∅) = 0.

3. Non-negativity Axiom:

For any event B, the probability of B is a non-negative real number, P(B) ≥ 0 This means that probabilities cannot be negative and they range from zero to one.

Hence 0 ≤ P(B) ≤ 1 where B is a proper subset of S.

4. Complement Rule:

P(B) = 1 − P(B′)

Example 13.7

The probabilities of Kwame and Ama solving a physics problem correctly are 0.6 and 0.9 respectively.

Find the probability that:

a. Only Ama solves it correctly.

b. Both Kwame and Ama fail to solve it correctly.

c. Only one of them solves it correctly.

d. What is the probability of at least one of them solves the problem correctly?

Solution

Let P(A) = P(Ama solving correctly) Let P(K) = P(Kwame solving correctly) P(A) = 0.9, P)A′( = 0.1 P(K) = 0.6, P)K′( = 0.4

a. P(Only Ama solving it correctly) = P(A) × P()K′( = 0.9 × 0.4 = 0.36

b. P(both Ama and Kwame fail to solve it correctly) = P()A′( × P))K′( = 0.1 × 0.4 = 0.04

c. P(only one of them solves it correctly) = P(A) × P)K′( + P(K) × P)A′( = 0.9(0.4)+ 0.6(0.1) = 0.42

d. P(at least one of them solves it correctly) = P(A) × P()K′( + P(K) × P)A′( + P(A)P(K) = 0.9(0.4)+ 0.6(0.1)+ 0.9 (0.6) = 0.96 Alternatively:

P(at least one of them solves it correctly) = 1 – P(Neither of them solve it correctly) = 1 − P()A′( × P)K′( = 1 − 0.1)0.4( = 0.96

Example 13.8

A box contains 40 bulbs of which 3 are defective. Dzifa has agreed to buy the whole box of bulbs if, when she selects 3 at random, her 3 selected contains at most 1 which is defective.

Find the probability that she does not buy the box of bulbs.

Solution

Represent defective bulbs with D Represent non-defective bulbs with D′ Calculate the probability that the selection contains 2 or more defective bulbs:

P(D ≥ 2) = ( 3/40 × 2/39 × 37/38 ) +( 3/40 × 37/39 × 2/38)+ (37/40 × 3/39 × 2/38) + ( 3/40 × 2/39 × 1/38 ) = 14/1235 = 0.011 (to 2sf)

1. Accra has 9 coffee shops: 4 Vida e caffe´, 2 Cuppa Cappuccino and 3 Cafe Accra.

If a tourist selects one shop at random to buy a cup of coffee, find the probability that it is either a Cuppa Cappuccino or a Vida e caffe´.

2. The research and development centres for three local companies have the following number of employees:

Noguchi Memorial Institute for Medical Research 200

CSIR Ghana 700

Kintampo Health Research Centre (KHRC) 150

If a research employee is selected at random, find the probability that the employee is employed by Noguchi or KHRC.

3. A single card is drawn at random from an ordinary deck of cards. Find the probability that it is either an ace or a black card.

4. The probability that Roland selects a pizza with mushrooms and/or beef is 0.55 and the probability that the he selects only mushrooms is 0.32.

If the probability that he selects only beef is 0.17, find the probability of him selecting both items.

5. The probability that Stacy wins a 200m race is 7/10 and the chances that she wins a 1500m race is 1/2.

Assuming that these events are independent, what is the probability that she wins:

a. both races

b. only one race?

6. Whenever, I go to the Mandela market, I bump into Kojo 3 days out of 10 and Efua 4 days out of 10.

Assuming that these events are independent, find the probability that, on a particular market day, I shall meet:

a. Both Kojo and Efua

b. Neither of them

c. At least one of them.

7. In a game of darts, the likelihood that Danful hits the target is 3/5 and the chances that Alex hits the target is 2/3. If they both throw the darts together, what is the probability that:

a. neither of them hits the target

b. at least one of them hits the target

c. Exactly one of them hits the target

d. both hit the target?

8. Two dice are rolled.

Find the probability of getting:

a. a sum greater than 9 or less than 4

b. a sum of 7

c. a sum of 8, 9 or 10

9. Research on 300 patients revealed that out of 120 hypertensive patients (those with high blood pressure), 67 had high cholesterol levels and of 180 non-hypertensive patients, 54 had high cholesterol levels.

If a patient is selected at random, find the probability that the patient is;

a. Non-hypertensive

b. Hypertensive with high cholesterol levels

c. Non-hypertensive with low cholesterol levels

10. One box contains 2 pink balls and 1 blue ball. A second box contains 1 pink ball and 2 blue balls.

A fair coin is tossed and if it falls heads up, the first box is selected and a ball is randomly drawn.

If the coin falls tails up, the second box is selected and a ball is randomly drawn.

Find the likelihood of selecting a pink ball.

11. Akwasi is taking a mathematics test which has two papers.

The probability that Akwasi passes both papers is 0.3.

The probability that he passes the first paper is 0.6 and that he passes the second paper is 0.5.

Knowing this information can you tell if passing the two papers are independent?

12. A box contains 5 red balls, 3 blue balls and 2 green balls.

Two balls are drawn randomly one after the other without replacement.

a. What is the probability that at least one of the balls drawn is red?

b. What is the probability that both balls drawn are of the same colour?

c. What is the probability that the first ball is blue, and the second ball is not green?

In this section, we will explore permutations and combinations. Understanding these concepts will help you develop a strong foundation in counting principles, to enable you to solve complex problems related to arrangements and probability with confidence. Embracing these concepts will empower you to make informed decisions in event planning, sports, research and more. Let us embark on this journey together and unlock the power of counting!

KEY IDEAS

• Combinations involve the selection of items where the order does not matter. For example, choosing 2 colours from {red, blue, white} results in 3 combinations: {red, blue}, {red, white} and {blue, white}.

• Key Differences are that in permutations order matters, but in combinations order does not matter

• Key Formulas:

Permutations: n_(P)_(Γ) = n !______ (n − r)!

Combinations: n_(C)_(Γ) = n !_______ (n − r)!r !

• Permutations are the arrangements of items where the order matters.

For example, for the arrangement of the letters in “CAT” there are 6 permutations: CAT, CTA, ACT, ATC, TCA and TAC.

Activity 14.1: Revision of Fundamental Counting Rules

Working in pairs discuss how to go about solving this problem.

A school’s NSMQ main team consists of two males, A and B, and three Females, E, F and G. The school is to choose only two contestants to sit on stage. As a coordinator, show all the possible pairings and indicate how many ways you can do this.

The principle applied above is the multiplication principle, commonly known as the fundamental concept of counting, which is a straightforward method for calculating the number of possible outcomes.

This is how it works:

If you have one event that can occur in m different ways and another event that can occur in n different ways, multiply the two numbers together to determine the total number of ways both events can occur consecutively. So, the total number of ways = m × n.

Example 14.1

You have four pairs of socks and three pairs of shoes.

How many different ways can you combine your socks and shoes?

Solution

4 × 3 = 12 ways This principle is widely used in probability, statistics and various fields of mathematics to solve counting problems efficiently.

Example 14.2

Two coins are flipped and a die is rolled.

Find the number of outcomes for the sequence of events.

Solution

Outcome when two coins are flipped: {HH, HT, TH, TT} Outcome when a die is rolled: {1, 2, 3, 4, 5, 6} 1 2 3 4 5 6 HH HH1 HH2 HH3 HH4 HH5 HH6 HT HT1 HT2 HT3 HT4 HT5 HT6 TH TH1 TH2 TH3 TH4 TH5 TH6 TT TT1 TT2 TT3 TT4 TT5 TT6 The total outcome is 24 Alternatively, The outcome when two coins are rolled = 4 The outcome when a die is rolled = 6 4 × 6 = 24 ways

Example 14.3

A painter wishes to paint a building with different paints. The categories include Colour: red, green, yellow, white, blue, brown, black, violet Type: latex, oil Texture: flat, semi-gloss, high gloss Use: indoors, outdoors How many different combinations of paint are there?

Solution

There are 8 items in the colours category, 2 in the type category, 3 in the texture category and 2 in the use category.

8 × 2 × 3 × 2 = 96 ways

Example 14.4

There are 4 blood groups, A, B, AB and O. Blood can also be Rh+ and Rh-. Finally, a blood donor can be classified as either male or female. How many different ways can a donor have his or her blood labelled?

Solution

4 × 2 × 2 = 16 ways

If you remember back to year 1, you will recall that permutations relate to the act of arranging all the members of a set into some specific sequence or order. The order in which the items are arranged is important.

If we are asked to find the number of ways to arrange r objects from n objects, we us this formula:

nPᵣ = n !_ (n − r)!

Therefore, if we had n objects and we were arranging all of them, we have:

nPₙ = n !

The number of permutations of n different objects taken r at a time with repetitions allowed is given by:

nʳLet us use this knowledge in some examples.

Example 14.5

How many 3 letter words with or without meaning can be formed from the letters in the word DUST when repetitions are allowed?

Solution

Step 1: Identify the Available Letters

The letters in “DUST” are: D, U, S, T. This gives us a total of 4 distinct letters.

Step 2: Calculate the Number of Possible Arrangements

Since repetitions are allowed, for each of the 3 positions in the word, we can choose any of the 4 letters.

For the first letter: 4 choices (D, U, S, T) For the second letter: 4 choices (D, U, S, T) For the third letter: 4 choices (D, U, S, T)

Step 3: Multiply the Choices

The total number of 3-letter combinations can be calculated as follows:

Total combinations = 4 × 4 × 4 = 4³= 64 Alternatively, we can use nʳsince the repetitions are allowed n = 4, r = 3 4³= 64

Example 14.6

A password can be made up of any four-digit combination.

a. How many different passwords are possible?

b. How many are possible if all the digits are odd?

Solution

a. Since the password can be made from any of the four digits, it means repetitions are allowed:

10 × 10 × 10 × 10 = 10⁴= 10 000

b. If all the digits must be odd we are limited to only 5 digits each time, so we have: 5 × 5 × 5 × 5 = 5⁴= 625

Example 14.7

A school wants to award prizes for 1st, 2nd, 3rd and 4th in a class of 20.

How many possible ways can the prizes be awarded, assuming no two students tie?

Solution

This is choosing 4 students from 20 and the order of those 4 matters, so we have:

20_(P)₄ = 116 280 The number of ways is 116 280.

Example 14.8

We are going to use the letters {a, b, c, d, e, f, g, h} to form a 5-character “password” with no repeated characters.

How many different passwords are possible?

Solution

We are choosing 5 letters from 8 and order matters in a password, so we have:

8_(P)₅ = 8 × 7 × 6 × 5 × 4 = 6 720 ways

Example 14.9

Kofi, Adzo, Afiba, Haruna, Ayitey, Ghartey and Mercy form the Executives of the Reading Club. They are to choose from amongst themselves a Chairperson, Secretary and Treasurer. No person can hold more than one position. How many different outcomes are possible?

Solution

We are choosing 3 people from 7 and order matters as to the position, so we have:

7_(P)₃ = 210 ways

Example 14.10

a. 6 prefects are given 5 special desks at which to work. How many ways can the desks be allocated?

b. In how many ways can the letters of the word ‘RECTANGLE’ be arranged?

c. How many four-digit numbers can be formed with digits 5, 7, 8 and 9 with no digit repeated?

Solution

a. We have 6 prefects and 5 desks and which desk they are given matters, so we have: 6_(P)₅ = 6 !______ )6 − 5(! = 6 × 5 × 4 × 3 × 2 = 720 ways

b. There are 9 letters in the word RECTANGLE but the this includes 2 × E’s which are indistinguishable from each other, so the number of arrangements is:

9 !__ 2 ! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/2 × 1 = 362880/2 = 181 440 ways

c. 4 × 3 × 2 × 1 = 24 ways

Activity 14.2: Revision on Combinations

In pairs or small groups, investigate the difference between combinations and permutations and how and why the formulas differ.

Combinations allow us to choose items from a group where the order of selection is not important. Unlike permutations, where the arrangement matters, in combinations, it does not matter how we arrange the items we select.

Let’s solve more problems with the use of combinations.

Example 14.11

15 students vied for 3 slots in a quiz team.

In how many ways can the three slots be filled?

Solution

In this example, order does not matter and there are 15 students from which three will be chosen:

15_(C)₃ = 15 !________ (15 − 3)!3 ! = 455 ways

Example 14.12

There are 9 men and 11 women in a farming cooperative group.

How many committees of 5 men and 7 women can be formed?

Solution

Order does not matter.

There are 9 men, choosing 5 men = 9_(C)₅ = 126 There are 11 women, choosing 7 women = 11_(C)₇ =330 9_(C)₅ × 11_(C)₇ = 126 × 330 = 41 580 ways

Example 14.13

Nhyira decides to form a band. She needs a bass player, 2 guitarists, a keyboard player and a drummer. She invites applications and gets 7 bass players, 6 guitarists, 5 keyboard players and 4 drummers. Assuming each person applies only once, in how many ways can Nhyira put the band together?

Solution

There are 7 bass players, selecting 1 person = 7_(C)₁ There are 6 guitarists selecting 2-person = 6_(C)₂ There are 5 keyboard players selecting 1 person = 5_(C)₁ There are 4 drummers selecting 1 person = 4_(C)₁ 7_(C)₁ × 6_(C)₂ × 5_(C)₁ × 4_(C)₁ = 7 × 15 × 5 × 4 = 2 100 ways

Example 14.14

How many ways can you form a 3-person committee from 5 men and 8 women:

a. with no restrictions

b. the committee must have 1 man and 2 women

c. the committee must have only 1 woman?

Solution

a. With no restrictions:

Total number of people = 5 + 8 = 13 3 are to be selected to form the committee = 13_(C)₃ = 286 ways

b. The committee must have 1 man and 2 women Ways of selecting 1 man from 5 men = 5_(C)₁ =5 Ways of selecting 2 women from 8 women = 8_(C)₂ = 28 Total number of ways = 5 × 28 = 140 ways

c. The committee has only 1 woman If the committee has only 1 woman, then there must be two men.

Ways of selecting 1 woman from 8 women = 8_(C)₁ = 8 Ways of selecting 2 men from 5 men = 5_(C)₂ = 10 Total number of ways = 8 × 10 = 80 ways

Example 14.15

Out of 4 mathematicians and 8 statisticians, a committee consisting of 2 mathematicians and 4 statisticians is to be formed.

In how many ways can this be done if:

a. any mathematician and statistician can be chosen.

b. one particular statistician must be chosen.

Solution

a. Number of ways of choosing 2 mathematicians from 4 mathematicians.

4_(C)₂ = 6 Number of ways of choosing 4 Statisticians from 8 Statisticians 8_(C)₄ = 70 Number of ways of choosing the committee = 6 × 70 = 420 ways

b. If one particular statistician must be on the committee, the person can be selected by 1_(C)₁ = 1 way (no surprise there!)

We are then left with 7 statisticians to choose 3 = 7_(C)₃ = 35 Number of ways of selecting 2 mathematicians from 4 mathematicians 4_(C)₂ = 6 Total number of ways = 1 × 35 × 6 = 210 ways.

Example 14.16

6 people are to be chosen for a new committee from 8 males and 8 females. How many different ways can the committee be chosen if:

a. there are no restrictions on who is chosen

b. there must be equal males and females on the committee

c. the current chairperson must be re-elected to the committee, but no other restrictions

d. there must be at least 4 females on the committee.

Solution

a. Total members = 8 + 8 = 16 6 persons are to be selected to form the committee = 16_(C)₆ = 8 008 ways

b. Equal males and females on the committee Ways of selecting 3 males from 8males = 8_(C)₃ = 56 Ways of selecting 3 females from 8 females = 8_(C)₃ = 56 Total number of ways = 56 × 56 = 3 136 ways

c. If the current chairperson must be re-elected, the person can be selected by 1_(C)₁ = 1 We can then select 5 from the 15 people left 15_(C)₅ = 3 003 Total number of ways = 1 × 3003 = 3 003 ways

d. At least 4 females must be on the committee means, the number of females can be, 4, 5, 6 Females Males Number of ways 4 2 8_(C)₄ × 8_(C)₂ = 70 × 28 = 1960 5 1 8_(C)₅ × 8_(C)₁ = 56 × 8 = 448 6 0 8_(C)₆ × 8_(C)₀ = 28 × 1 = 28 The total number of ways is the sum of the ways obtained in the table:

1960 + 448 + 28 = 2436 ways

Example 14.17

A group consists of 4 girls and 7 boys.

In how many ways can a team of 5 members be chosen if the team has:

a. no girls

b. at least one boy and one girl

c. at least three girls?

Solution

a. No girl means the number of girls = 0 and number of boys = 5 4_(C)₀ × 7_(C)₅ = 1 x 21 = 21 ways

b. at least one boy and one girl, means:

Boys Girls Number of ways 1 4 7_(C)₁ × 4_(C)₄ = 7 x 1 = 7 2 3 7_(C)₂ × 4_(C)₃ = 21 x 4 = 84 3 2 7_(C)₃ × 4_(C)₂ = 35 x 6 = 210 4 1 7_(C)₄ × 4_(C)₁ = 35 x 4 = 140 The total number of ways is the sum of the ways obtained in the table:

7 + 84 + 210 + 140 = 441 ways

c. at least three girls, means:

Girls Boys Number of ways 3 2 4_(C)₃ × 7_(C)₂ = 4 x 21 = 84 4 1 4_(C)₄ × 7_(C)₁ = 1 x 7 = 7 The total number of ways is the sum of the ways obtained in the table:

7 + 84 = 91 ways

Example 14.18

A mathematics paper has 13 questions from which a candidate has to answer any 10 questions.

How many different sets of questions can be chosen?

Solution

13_(C)₁₀ = 286 ways

1. A password consists of three digits, 0 through 9, followed by three letters from an alphabet having 26 letters. If repetition of the digits is allowed, but repetition of the letters is not allowed, determine the number of different passwords that can be made.

2. How many three different course meals can be served from a menu that has 5 choices for drinks, 7 choices for vegetables, and 4 choices for desserts?

3. 4 medical doctors are to be selected from a group of 8 to undertake medical outreach. In how many ways can this be done if:

a. any of the medical doctors can be selected

a. two particular doctors must be part of the team

4. In how many ways can 3 boys and 5 girls be chosen from 20 boys and 15 girls to represent a school for a competition?

5. A committee of 5 is to be formed from 7 men and 8 women.

Determine the number of ways if:

a. only 1 man will be on the committee

a. 2 women will be on the committee

b. at least 4 men will be on the committee

c. at most 3 men will be on the committee

6. A bag contains 5 red, 4 black and 3 yellow balls.

In how many ways can the following be selected?

a. 3 red balls

a. 3 black and 1 red

b. 2 red and 2 yellow

7. When three dice are rolled with two coins, how many outcomes are possible?

8. Kwame has 5 shirts, 4 pairs of shoes and 8 pairs of trousers. How many outfits can Kwame choose from a shirt, pair of shoes and pair of trousers.

9. The digits 1, 2, 3, 4 and 5 are to be used in 3-digit ID Cards. How many different cards are possible if repetitions are permitted?

10. There are 25 people in a class. Any 5 of these can be chosen for a quiz team. How many ways can they choose this team?

11. An examination paper is divided into two sections, A and B. Section A contains 5 questions and section B contains 8 questions. The candidate is to attempt all the questions in section A and choose 5 from section B. In how many ways can the candidate choose his questions?

12. A committee of 6 is to be formed from 15 board members. How many different ways this can be done if it must:

a. include 2 particular board members

b. exclude the chairperson, the secretary and the treasurer?

13. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 8 and 9 which are divisible by 5 and none of the digits is repeated?

14. There are three places A, B and C such that 4 roads connect A and B and 5 roads connect B and C. In how many ways can one travel from A to B?

15. The Mathematics Club will select a president, a vice president and a treasurer for the club.

If there are 20 members in the club, how many different selections of a president, a vice president and a treasurer are possible if each club member can be selected to only one position?

16. A candidate taking the WASSCE Additional Mathematics exam must answer 9 out of 15 questions.

a. How many choices are available to the candidate if there are no restrictions on the questions chosen?

b. How many choices are available to the candidate if they must answer the first 5 questions?

c. How many choices are available to the candidate if they must answer at least 4 of the first five questions?