Making Predictions with Data

Probability and counting are fundamental concepts that play a crucial role in solving real- world problems. Understanding these concepts will equip us with the tools to analyse situations involving uncertainty, make predictions, and make informed decisions based on data. We will explore several key topics, including conditional probability, equally likely events, permutations and combinations and the binomial distribution. Understanding these concepts is not just theoretical; they have practical applications in everyday life, including:

1. School Leadership Selection: When choosing a student council, schools can use combinations to determine how many different groups can be formed from a larger pool of candidates.
2. Business Planning: Companies can apply binomial probability to assess risks and make informed decisions about investments or product launches.
3. Industrial Processes: In manufacturing, understanding permutations can help in optimising the arrangement of products on assembly lines.

KEY IDEAS

• Binomial probability deals with experiments that have two possible outcomes, such as success and failure. • Combinations refer to the selection of objects where the order does not matter. • Conditional probability is the probability of an event occurring given that another event has already occurred. • Events are considered equally likely if they have the same chance of occurring. • Permutations refer to the different ways to arrange a set of objects where the order matters. For instance, arranging books on a shelf involves permutations.

Activity 8.1 Probability Concepts Activity

This activity is designed to help us engage with key probability concepts. We will work individually or in small groups to apply what we have learned about experiments, the multiplication rule, mutually exclusive events, the law of total probability, and dependent events. Materials Needed • Coins (at least 2 per group) • Dice (1 per group) • A bag containing coloured balls (e.g., 2 red, 3 blue) • Paper and pens/pencils for calculations Part 1: Experiments and Outcomes Task: Conduct a simple experiment and observe the outcomes.

1. Coin Flip Experiment • Flip a coin 3 times. Record the outcomes (heads or tails). • Calculate the experimental probability of getting all heads. • Discuss how this relates to the theoretical probability of $\frac{1}{8}$ .
2. Dice Roll Experiment • Roll a die. Record the outcomes. • Calculate the experimental probability of rolling a 3. • Compare your results with the theoretical probability of $\frac{1}{6}$ . Part 2: Multiplication Rule of Probability Task: Solve problems using the multiplication rule.
3. Dependent Events • Using the bag of coloured balls, randomly draw one ball without replacement, record its colour, then draw a second ball. • Calculate the probability of drawing a red ball first and then a blue ball second. • Use the formula: $P(A \cap B) = P(A) \cdot P(B \mid A)$
4. Independent Events • Flip a coin and roll a die at the same time. Record the outcomes.

• Calculate the probability of getting heads on the coin and a 4 on the die. • Use the formula: $P(A \cap B) = P(A) \times P(B)$ Part 3: Mutually Exclusive Events Task: Identify mutually exclusive events (two events that cannot occur at the same time). Group Discussion • Think of examples of mutually exclusive events in daily life (e.g., weather conditions: sunny vs. rainy). • Create a list of at least 3 pairs of mutually exclusive events and explain why they cannot occur simultaneously. Part 4: Law of Total Probability Task: Apply the law of total probability.

1. Fruit Basket Scenario • Imagine you have a basket with 3 apples, 2 bananas, and 5 oranges. Calculate the probability of picking a banana using the law of total probability. • Use the formula: $P(B) = P(B \mid A_1) \cdot P(A_1) + P(B \mid A_2) \cdot P(A2) + P(B \mid A_3) \cdot P(A_3)$ • Discuss how partitioning the sample space helps in calculating probabilities. Part 5: Dependent Events (Sampling Without Replacement) Task: Explore dependent events through selection. Group Leader and Secretary Selection • From a group of 5 learners, randomly select a group leader and then a secretary. • Discuss how the selection of the secretary is dependent on who was chosen as the leader. • Calculate the probability of selecting a specific learner as the leader and a different specific learner as the secretary.

How conditional probabilities arise

In probability theory, we often encounter situations involving multiple events, particularly when we need to determine the likelihood of an event occurring based on the occurrence of another event. This is where the concept of conditional probability comes into play. Conditional probability measures the probability of an event happening given that we know another event has already occurred. For example, if we want to find the probability of drawing a red card from a deck of cards after knowing that the first card drawn was a heart, we recognise that the first event

influences the outcome of the second. Understanding this relationship helps us make more informed predictions and decisions based on prior knowledge. To illustrate, consider a scenario where a student takes a mathematics test and a science test. If we know that the student scored well on the mathematics test, we might want to determine the probability that they will also score well on the science test. This situation exemplifies conditional probability because the outcome of the mathematics test can provide insight into the performance on the science test. By analysing how these events are related, we gain a deeper understanding of the underlying patterns and dependencies in various situations, which is crucial for making accurate assessments in real-life scenarios.

Example 8.1

A box contains 12 identical balls, 5 of which are black and the rest are green. Two balls are drawn at random from the bag without replacement. What is the probability of getting one black ball and one green ball?

Solution

To find the probability of drawing one black ball and one green ball from a box containing 12 identical balls (5 black and 7 green), we can follow these steps: Step 1: Determine the Total Number of Balls Total Balls: 12 Black Balls: 5 Green Balls: $12 - 5 = 7$ Step 2: Calculate the Total Ways to Draw Two Balls The total number of ways to choose 2 balls from 12 is given by the combination formula: Total Ways = ${}^{12}C_2$ = 66 Step 3: Calculate the Ways to Draw One Black and One Green Ball To find the number of ways to draw one black ball and one green ball, we can choose:

1. One Black Ball: The number of ways to choose 1 black ball from 5 = ${}^{5}C_1 = 5$
2. One Green Ball: The number of ways to choose 1 green ball from 7 = ${}^{7}C_1 = 7$ Step 4: Calculate the Total Ways to Draw One Black and One Green The total ways to draw one black ball and one green ball is the product of the two combinations: Ways to choose 1 black and 1 green $= 5 \times 7 = 35$ Step 5: Calculate the Probability Now, the probability of drawing one black ball and one green ball is given by the ratio of the favourable outcomes to the total outcomes: P(1 black and 1 green) = $\frac{\text{Ways to choose 1 black and 1 green}}{\text{Total Ways}} = \frac{35}{66}$

Example 8.2

A bag contains 8 counters: 2 pink, 1 green, and 5 yellow. A counter is drawn at random from the bag and not replaced. A second counter is then drawn. Find the probability that both counters drawn are yellow.

Solution

To find the probability that both counters drawn are yellow, we can follow these steps:

1. Total Ways to Choose 2 Counters from 8: Total Ways = ${}^{8}C_2 = 28$
2. Ways to Choose 2 Yellow Counters: Since there are 5 yellow counters, the number of ways to choose 2 yellow counters is ${}^{5}C_2 = 10$
3. Probability of Both Counters Being Yellow P(Both Yellow) = $\frac{\text{Ways to choose 2 yellow}}{\text{Total Ways}} = \frac{{}^{5}C_2}{{}^{8}C_2}$ P(Both Yellow) = $\frac{10}{28} = \frac{5}{14}$

Example 8.3

Two marbles are drawn randomly from a bag containing 5 red, 7 white, and 8 blue marbles, one after the other, without replacement. Find the probability that: a) both are red b) one is white and the other is blue

Solution

Red marbles = 5 White marbles = 7 Blue marbles = 8 Total number of marbles = 5 + 7 + 8 = 20 Total ways of selecting 2 marbles out of 20 $= {}^{20}C_2 = 190$ a) Ways of selecting two red marbles from 5 marbles is ${}^{5}C_2 = 10$ P (both red balls) $= \frac{{}^{5}C_2}{{}^{20}C_2} = \frac{10}{190} = \frac{1}{19}$ b) Ways of selecting 1 white marble from 7 is ${}^{7}C_1 = 7$ Ways of selecting 1blue marble from 8 marbles is ${}^{8}C_1 = 8$ P (1 white and 1 blue marbles) $= \frac{{}^{7}C_1 \times {}^{8}C_1}{{}^{20}C_2} = \frac{7 \times 8}{190} = \frac{28}{95}$

Conditional probability $\mathrm{P(B|A)}$ is the probability of event B happening given the information that A has already taken place. P $(B|A) = \frac{P(B \cap A)}{P(A)}$ , $P(A) > 0$ . What will happen to the value of $\mathrm{P(B|A)}$ if the two events are mutually exclusive? In this special case, $\mathrm{P(B|A)} = 0$ .

Example 8.4

A fair die is rolled once. Let A be the event that the outcome is 2. Let B be the event that the outcome is even. Find P $(A|B)$

Solution

$S = \{1, 2, 3, 4, 5, 6\}$ $A = \{2\}, \quad B = \{2, 4, 6\}, \quad (A \cap B) = \{2\}$ $P(A) = \frac{1}{6}$ , $P(B) = \frac{1}{2}, P(A \cap B) = \frac{1}{6}$ P $(A|B) = \frac{P(A \cap B)}{P(B)}$ P $(A|B) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$

Example 8.5

A box has 3 red marbles and 2 black ones. 2 marbles are randomly selected one after the other without replacement. What is the probability that the second marble is red, given that the first one is red?

Solution

Let R 2 be the event that the second marble is red and R 1 the event that the first one is red. Total marbles, $3 + 2 = 5$ $P(R_2) = \frac{3}{5}$ $P((R_1 \cap R_2)) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ P $(R_1|R_2) = \frac{P(R_1 \cap R_2)}{P(R_2)}$

P $(R_1|R_2) = \frac{\frac{3}{10}}{\frac{3}{5}} = \frac{1}{2}$ Alternatively: Selecting 2 from 5, $5_{C_2} = 10$ Selecting 2 red marbles from 3 red marbles, $3_{C_2} = 3$ Selecting 0 marbles from 2 black marbles, $2_{C_0} = 1$ P (selecting two red) = $\frac{3_{C_2} \times 2_{C_0}}{5_{C_2}} = \frac{3 \times 1}{10} = \frac{3}{10}$ $P(R_2) = \frac{3}{5}$ P $(R_1|R_2) = \frac{\frac{3}{10}}{\frac{3}{5}} = \frac{1}{2}$

Example 8.6

In a survey among high school learners, $60\%$ use Facebook, $40\%$ use WhatsApp and $20\%$ use both. A person is chosen at random and it is given that they already use WhatsApp, find the probability that they also use Facebook.

Solution

Let F and W represent those who use Facebook and WhatsApp $P(F) = \frac{60}{100}$ $P(F) = 0.6$ $P(W) = \frac{40}{100}$ $P(F) = 0.4$ $P(F \cap W)$ = $\frac{20}{100}$ $P(F \cap W)$ $= 0.2$ P $(F|W) = \frac{0.2}{0.4}$ P $(F|W) = 0.5$

Conditional probabilities can be applied to solve real-life problems. We will now work through the following examples to understand how conditional probabilities can be applied to real life. We know that: $P(B|A) = \frac{P(B \cap A)}{P(A)}$ , $P(A) > 0.$

Example 8.7

In a school with 50 students, 11 females study French while 12 females study Arabic. In addition, 7 males study French and 20 males study Arabic. a) Organise the information in tabular form. b) If a student is selected at random, what is the probability that they are : i) studying Arabic, given that they are female. ii) male.

Solution

a) French (Fr) Arabic (A) Total Female (Fe) 11 12 23 Male (M) 7 20 27 Total 18 22 50 b) i) $P(A|Fe)$ = $\frac{12}{23}$ ii) P(M) = $\frac{27}{50}$

Example 8.8

The probability that a student is absent on a Friday is $0.06$ . The probability that today is Friday is $0.2$ . Find the probability that a student is absent given that today is Friday.

Solution

$P(\text{absent}|\text{Friday})$ $=$ $\frac{P(\text{absent} \cap \text{Friday})}{P(\text{Friday})}$ $P(\text{absent}|\text{Friday})$ $=$ $\frac{0.06}{0.2}$ $P(\text{absent}|\text{Friday})$ $= 0.30$

Example 8.9

In a community, 50% of families own a car. Of those families, 90% have a garage. Among families that do not own a car, only 20% have a garage. If a family is selected at random and found to have a garage, what is the probability that they own a car?

Solution

Let: $C$ : Event that a family owns a car $\mathrm{C}'$ : Event that a family does not own a car $\mathrm{G}$ : Event that a family has a garage. Now, $P(C|G) = \frac{P(C \cap G)}{P(G)}$ Also, $P(C) = 0.50$ , $P(C') = 1 - 0.50 = 0.50$ $P(G|C) = 0.90$ $P(G|C') = 0.20$ We determine the probability that a family owns a car and has a garage $P(C \cap G) = P(C) \times P(G|C)$ $= 0.50 \times 0.90$ $= 0.45$ For the probability that a family does not own a car and has a garage, $P(C' \cap G)$ $P(C' \cap G) = P(C') \times P(G|C')$ $= 0.50 \times 0.20$ $= 0.10$ The probability that a family owns a garage, $\mathrm{P(G)}$ $P(G) = P(C \cap G) + P(C' \cap G)$ $= 0.45 + 0.10$ $= 0.55$ Now, we calculate the probability that a family owns a car given that the family has a garage. $P(G|C) = \frac{P(C \cap G)}{P(G)}$ $= \frac{0.45}{0.55}$ $= 0.8182$ Thus, the probability that a family owns a car given that it has a garage is $0.83$ to 2 d. p.

Example 8.10

In a class of 70 students, 25 use tablets for their studies. Among the tablet users, 12 are on MTN. Of the students who do not use tablets, 7 are on the MTN network. If a student is selected at random and is found to be on MTN, what is the probability that this student uses a tablet for his/her studies?

Solution

Use Tablet Does not Total use a tablet Uses MTN 12 7 19 Does not use MTN 13 38 51 Total 25 45 70 Let T = students who use tablets for their studies. and M = students on MTN. This means that: $P(T) = \frac{25}{70} = \frac{5}{14}$ $P(M) = \frac{19}{70}$ Probability of selecting a student who is on MTN, given that the student uses a tablet $P(M|T) = \frac{12}{25}$ $P(T|M) = \frac{P(M|T) \times P(T)}{P(M)}$ $P(T|M) = \frac{\left(\frac{12}{25}\right) \times \left(\frac{5}{14}\right)}{\frac{19}{70}}$ $P(T|M) = \frac{12}{19}$ Therefore, the probability that a student uses a tablet given that the student is on MTN is $\frac{12}{19}$

In many real-life situations, we encounter events that can have only two possible outcomes. These outcomes can be categorised as success or failure. The study of these scenarios falls under the umbrella of binomial probability.

Examples of Binomial Events

1. Winning an Election: A candidate can either win (success) or lose (failure) an election.

2. Gender of a new-born child: When a baby is born, it can be either male or female. one gender can be described as “success” and the other, “failure”.

3. Conducting a Malaria Test: A malaria test can yield a positive result (success) or a negative result (failure). The probability of success (event of interest) is usually denoted by the letter p and the probability of failure by the letter q. Recall, $P(A) + P(A') = 1$ $P(A) = 1 - P(A')$ This means that $p + q = 1$ and $q = 1 - p$ To find the probability of obtaining exactly $x$ successes in $n$ trials, we use the binomial probability formula : $\mathrm{P}(X = x) = {}^{n}C_x p^x q^{n-x}$ Understanding binomial probability is crucial for analysing events with two possible outcomes. It helps us make informed predictions and decisions in various fields, such as healthcare, finance, and social sciences.

Example 8.11

A farmer observed that the probability of a coconut seed germinating is 0.40. She plants 12 coconut seeds. What is the probability that exactly 5 of them will germinate?

Solution

Let $\text{P(a seed germinating)} = p$ Let $\text{P(a seed not germinating)} = q$ $p = 0.4$ $q = 1 - 0.4 = 0.6$ N = total number seeds planted= 12 x = number of germinated seeds (success)=5 $\mathrm{P}(X = x) = {}^{n}C_x p^x q^{n-x}$ $P(X = 5) = {}^{12}C_5(0.4)^5(0.6)^{12-5}$ $P(X = 5) = {}^{12}C_5(0.4)^5(0.6)^7$ $P(X = 5) = 0.227$ There is about a $22.7\%$ chance that exactly $5$ out of the $12$ coconut seeds will germinate.

Example 8.12

A factory produces bulbs, and the probability that a bulb is defective is 0.35. A quality inspector randomly checks 15 bulbs. Find the probability that:

a) None of the bulbs is defective. b) Exactly 7 bulbs are defective. c) At least 10 bulbs are defective. d) More than 4 bulbs are defective.

Solution

Let P(a bulb is defective) $= p = 0.35$ Let P(a bulb is not defective) $= q = 1 - 0.35 = 0.65$ $n =$ total number of bulbs inspected=15 $x =$ the number of defective bulbs among those inspected. a) $x = 0$ (none of the inspected bulbs were defective) $P(X = x) = {}^{n}C_x p^x q^{n-x}$ $P(X = 0) = {}^{15}C_0(0.35)^0(0.65)^{15-0}$ $P(X = 0) = {}^{15}C_0(0.35)^0(0.65)^{15}$ $P(X = 0) = 0.00152069$ b) Exactly 7 bulbs are defective. Let $x = 7$ $P(X = x) = {}^{n}C_x p^x q^{n-x}$ $P(X = 7) = {}^{15}C_7(0.35)^7(0.65)^{15-7}$ $P(X = 7) = {}^{15}C_7(0.35)^7(0.65)^8$ $P(X = 7) = 0.0525$ c) At least 10 bulbs are defective P(At least 10 bulbs are defective) $= P(x \geq 10)$ $P(x \geq 10) = P(x = 10) + P(x = 11) + P(x = 12) + P(x = 13) + P(x = 14) + P(x = 15)$ We need to calculate the probability of each and sum them. $P(X = 10) = {}^{15}C_{10}(0.35)^{10}(0.65)^{15-10}$ $= 0.00961175$ $P(X = 11) = {}^{15}C_{11}(0.35)^{11}(0.65)^{15-11}$ $= 0.00235253$ $P(X = 12) = {}^{15}C_{12}(0.35)^{12}(0.65)^{15-12}$ $= 0.00042225$

$P(X = 13) = {}^{15}C_{13}(0.35)^{13}(0.65)^{15-13}$ $= 0.00005247$ $P(X = 14) = {}^{15}C_{14}(0.35)^{14}(0.65)^{15-14}$ $= 0.000004036$ $P(X = 15) = {}^{15}C_{15}(0.35)^{15}(0.65)^{15-15}$ $= 0.000000148$ Finding the sum: $0.00961175 + 0.00235253 + 0.00042225 + 0.00005247 + 0.000004036 + 0.000000148$ $P(x \geq 10) = 0.00124$ d) More than 4 bulbs are defective P(More than 4 bulbs are defective) $= P(x > 4)$ $= P(x = 5) + P(x = 6) +$ … $+P(x = 15)$ This is the same as finding $1 - P(x \leq 4)$ $P(X = 0) = {}^{15}C_0(0.35)^0(0.65)^{15-0}$ $= 0.001562$ $P(X = 1) = {}^{15}C_1(0.35)^1(0.65)^{15-1}$ $= 0.01262$ $P(X = 2) = {}^{15}C_2(0.35)^2(0.65)^{15-2}$ $= 0.04755$ $P(X = 3) = {}^{15}C_3(0.35)^3(0.65)^{15-3}$ $= 0.11096$ $P(X = 4) = {}^{15}C_4(0.35)^4(0.65)^{15-4}$ $= 0.17923$ $0.001562 + 0.01262 + 0.04755 + 0.11096 + 0.17923 = 0.35192$ P(More than 4 bulbs are defective) $= 1 - 0.35192$ P(More than 4 bulbs are defective) $= 0.64808 \approx 0.65\ ($ 2 decimal places)

Example 8.13

A fair coin is tossed 9 times. Find the probability of obtaining exactly 4 heads

Solution

Probability of success, tossing to get a head, $p = \frac{1}{2}$ $q = 1 - \frac{1}{2}$ $q = \frac{1}{2}$ a) exactly 4 heads $x = 4$ $P(X = x) = {}^{n}C_x p^x q^{n-x}$ $P(X = 4) = {}^{9}C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^{9-4}$ $P(X = 4) = {}^{9}C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^5$ $P(X = 4) = \frac{63}{256}$

Example 8.14

The probability that a customer orders a particular model of a new car in silver is 0.2. In the next 10 random orders, find the probability that a) Exactly 2 cars are ordered in silver. b) At most 3 cars are ordered in silver.

Solution

a) Exactly 2 cars are ordered in silver. Let $x = 2$ $P(X = x) = {}^{n}C_x p^x q^{n-x}$ $P(X = 2) = {}^{10}C_2(0.2)^2(0.8)^{10-2}$ $P(X = 2) = {}^{10}C_2(0.2)^2(0.8)^8$ $P(X = 2) = 0.3020$

b) At most 3 cars are ordered in silver. We can find $x = 0, 1, 2, 3$ and sum the results. $P(X = 0) = {}^{10}C_0(0.2)^0(0.8)^{10-0}$ $= 0.107374$ $P(X = 1) = {}^{10}C_1(0.2)^1(0.8)^{10-1}$ $= 0.268435$ $P(X = 2) = {}^{10}C_2(0.2)^2(0.8)^{10-2}$ $= 0.301990$ $P(X = 3) = {}^{10}C_3(0.2)^3(0.8)^{10-3}$ $= 0.201327$ $P(x \leq 3) = 0.107374 + 0.268435 + 0.301990 + 0.201327$ $= 0.879126$ Thus, the probability that at most 3 cars are ordered in silver is $0.87912$ .

Example 8.15

The proportions of people with blood groups $A, B, AB,$ and $O$ in the village are in the ratio. $20 : 15 : 10 : 35$ , respectively. Determine the probability that a random sample of 8 people from the village contains: (a) Exactly 5 with blood group O (b) at most 2 with blood group A

Solution

Total ratio $= 20 + 15 + 10 + 35 = 80$ Let $A =$ event that a randomly selected individual has blood group $A$ Let $B =$ event that a randomly selected individual has blood group $B$ Let $AB =$ event that a randomly selected individual has blood group $AB$ Let $O =$ event that a randomly selected individual has blood group $O$ This means that; $P(A) = \frac{20}{80} = \frac{1}{4}$ $P(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}$ $P(B) = \frac{15}{80} = \frac{3}{16}$ $P(\bar{B}) = 1 - \frac{3}{16} = \frac{13}{16}$ $P(AB) = \frac{10}{80} = \frac{1}{8}$ $P(\overline{AB}) = 1 - \frac{1}{8} = \frac{7}{8}$ $P(O) = \frac{35}{80} = \frac{7}{16}$ $P(\bar{O}) = 1 - \frac{7}{16} = \frac{9}{16}$

Let Sample $(n) = 8$ people Let $x =$ number of people out of the sample of 8 with a particular blood group (a) P(Exactly 5 with blood group $O$ ) $= {}^{8}C_5\left(\frac{7}{16}\right)^5\left(\frac{9}{16}\right)^3$ $= 56 \times 0.0160284 \times 0.17797852$ $= 0.15975$ (b) P(at most 2 with blood group $A$ ) $= P(x = 0) + P(x = 1) + P(x = 2)$ $P(x = 0) + P(x = 1) + P(x = 2) = {}^{8}C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^8 + {}^{8}C_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^7 + {}^{8}C_2\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^6$ $= \frac{6561}{65536} + \frac{8 \times 2187}{65536} + \frac{28 \times 729}{65536}$ $= \frac{6561}{65536} + \frac{17496}{65536} + \frac{20412}{65536}$ $= \frac{44469}{65536}$ $= 0.6785$

Fundamental Counting Principle

It states that: If one event can occur in m ways and a second event can occur in n ways, then the two events together can occur in $m \times n$ ways. This can be extended to more events: If event 1 can occur in $a$ ways, event 2 in $b$ ways, and event 3 in $c$ ways, then all three events together can occur in $a \times b \times c$ ways

Example 8.16

A dress? student has 5 shirts and 3 shorts. How many ways can the student [Figure]

Solution

Number of ways to choose a shirt = 5 Number of ways to choose shorts = 3

Total number of outfits $= 5 \times 3 = 15$ So, the student can dress in 15 different ways.

Example 8.17

A license plate is made up of 2 letters followed by 4 digits. How many number plates are possible?

Solution

There are 26 letters of the English alphabet. i.e., A, B, C, D, …., Z Number of ways to choose 2 letters $= 26 \times 26$ The decimal number system has 10 digits. i.e., 0, 1, 2, 3, …., 9. Number of ways to choose 4 digits $= 10 \times 10 \times 10 \times 10 = 10\,000$ Total $= 26 \times 26 \times 10 \times 10 \times 10 \times 10 = 6\,760\,000$ So, there are 6 760 000 possible license plates.

Example 8.18

A How password many passwords consists of can 3 letters be formed? (A–Z) followed by 2 digits (0–9). [Figure]

Solution

Letters: $26 \times 26 \times 26$ Digits: $10 \times 10$ Total $= 17\,576 \times 100 = 1\,757\,600$ So, 1 757 600 different passwords can be formed.

Example 8.19

If a sports club decides to elect a chairperson and a secretary from 5 members, assuming the same person cannot hold both positions, find the number of ways this can be done.

Solution

Number of people to choose a chairperson from = 5 Number of people to choose a secretary from = 4 Number of ways to choose a chairperson then a secretary $5 \times 4 = 20$ . Therefore, there are 20 ways in which this can be done.

A combination is a selection of items from a set where the order of selection does not matter. The combination formula is: ${}^{n}C_r = \frac{n!}{r!\,(n - r)!}$ Where: • n = total number of items • r = number of items chosen

Situations Where Order of Arrangement is NOT

Important (Combination)

1. Choosing a Committee Example: Selecting 2 prefects from a class of 30. The 2 students will be chosen regardless of the order.

2. Lottery Games Example: Selecting 4 winning numbers from 98.

3. Food Choices Example: Choosing 2 drinks out of 5 options in a shop. Sobolo and Asaana are the same as Asaana and Sobolo.

4. Subject Selection Example: If Naadu selects 4 elective subjects out of a total of 10 subjects in the Arts department, the order in which she selects the subjects is not important.

5. Team Selection Example: Choosing 11 players from 15 to form a football team. The lineup order doesn’t matter at this stage (positions not yet assigned).

6. Clothing Choices Example: Picking 4 shirts from 10 in a wardrobe. Order of picking the shirts does not matter.

7. Test Question Selection Example: Kofi chooses 5 questions out of 8 in an exam. The order of chosen questions is irrelevant.

8. Meal Menu Example: Selecting a set of dishes for a party. Whether fufu is chosen before or after konkonte does not matter.

9. Selecting Students for a Prize Example: Picking any 2 students to receive awards. The order of calling their names is not important.

10. Choosing Books to Borrow Example: Borrowing 3 books from a library. It does not matter in what order the books are selected.

How to Answer a Combination Question

Step 1: Identify if it is a combination Ask yourself: Does order matter? If NO, it’s a combination. Look for words like: choose, select, pick, form, group, committee. Step 2: Recall the formula ${}^{n}C_r = \frac{n!}{r!\,(n - r)!}$ Where n = total number of items and r = number of items being chosen Step 3: Substitute values Put in the given values for n and r. Step 4: Simplify factorials Cancel out terms where possible to make calculations simpler. Step 5: State the final answer Write your result clearly, with units if needed (e.g., “ways”, “groups”, “teams”).

Example 8.20

A committee of 3 people is to be selected from 7 teachers. How many different committees can be formed?

Solution

1. Check order: Committees don’t depend on order → Combination.

2. Formula: ${}^{n}C_r = \frac{n!}{r!(n - r)!}$

3. Substitute: we have n=7 and r=3 ${}^{7}C_3 = \frac{7!}{3!\,(7 - 3)!}$

4. Simplify: ${}^{7}C_3 = \frac{7!}{3! \times 4!} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = 35$

5. Answer: There are 35 different committees.

Example 8.21

A DJ is making a 5-track playlist from 7 afrobeat, 4 highlife, and 3 reggae tracks, but track order does not matter for selection. In how many ways can the DJ select tracks if the playlist contains: (i) exactly 3 afrobeat and 2 highlife? (ii) at least one reggae? (iii) at most 2 highlife?

Solution

This is a combination question since the track order does not matter. (i) selecting 3 afrobeats from 7 can be done in ${}^{7}C_3$ ways = Selecting 2 highlife tracks from 4 can be done in ${}^{4}C_2$ ways = exactly 3 afrobeat and 2 highlife = (exactly 3 afrobeat) $\times$ ( 2 highlife) $=$ ${}^{7}C_3 \times {}^{4}C_2 = 35 \times 6 = 210$ ways So, the number of ways the DJ can select 3 afrobeats and 2 highlights from the list of available tracks is $210$ . (ii) The are a total of 3 reggae and $(7 + 4) = 11$ non-reggae tracks the DJ can choose from. Choosing at least one reggae track $=$ [Choosing 1 reggae and 4 non-reggae tracks or Choosing 2 reggae track and 3 non-reggae tracks or Choosing 3 reggae tracks and 2 non-reggae tracks] Number of ways the DJ can choose 1 reggae and 4 non-reggae tracks $= {}^{3}C_1 \times {}^{11}C_4 = 3 \times 330 = 990$ Number of ways the DJ can choose 2 reggae and 3 non-reggae tracks $= {}^{3}C_2 \times {}^{11}C_3 = 3 \times 165 = 495$ Number of ways the DJ can choose 3 reggae and 2 non-reggae tracks $= {}^{3}C_3 \times {}^{11}C_2 = 1 \times 55 = 55$ At least one reggae track $= 990 + 495 + 55 = 1\,540$ ways.

(iii) There are a total of 4 highlife and $(3 + 7) = 10$ non-highlife tracks the DJ can choose from. Choosing at most 2 highlife tracks $=$ [Choosing 0 or no highlife and 5 non-highlife tracks or Choosing 1 highlife track and 4 non-highlife tracks or Choosing 2 highlife tracks and 3 non-highlife tracks] Number of ways the DJ can choose 0 highlife and 5 non-highlife tracks $= {}^{4}C_0 \times {}^{10}C_5 = 1 \times 252 = 252$ Number of ways the DJ can choose 1 highlife track and 4 non-highlife tracks $= {}^{4}C_1 \times {}^{10}C_4 = 4 \times 210 = 840$ Number of ways the DJ can choose 2 highlife tracks and 3 non-highlife tracks $= {}^{4}C_2 \times {}^{10}C_3 = 6 \times 120 = 720$ At most two highlife tracks $= 252 + 840 + 720 = 1\,812$ ways.

Example 8.22

If ${}^{n}C_{r-1} = 36$ , ${}^{n}C_r = 84$ and ${}^{n}C_{r+1} = 126$ , find the value of ${}^{r}C_2$ .

Solution

We know ${}^{n}C_r = \frac{n!}{r!(n - r)!}$ So, ${}^{n}C_r = 84$ is the same as $\frac{n!}{r!(n - r)!} = 84$ ………………….(1) Likewise, ${}^{n}C_{r-1} = 36$ is same as $\frac{n!}{(r - 1)!(n - r + 1)!} = 36$ …………………… (2) And ${}^{n}C_{r+1} = 126$ is same as $\frac{n!}{(r + 1)!(n - r - 1)!} = 126$ …………………………(3) Divide equation (1) by equation (2) $\frac{n!}{r!\,(n - r)!} \div \frac{n!}{(r - 1)!\,(n - r + 1)!} = 84 \div 36$ Change $\div$ to $\times$ by reciprocating the second fractions on each side of the equation $\frac{\cancel{n!}}{r!\,(n - r)!} \times \frac{(r - 1)!\,(n - r + 1)!}{\cancel{n!}} = 84 \times \frac{1}{36}$ Recall that $r! = r(r - 1)!$ Also, $(n - r + 1)! = (n - r + 1)(n - r)!$ Cancel out $n!$ , and substitute the expansions of $r!$ and $(n - r + 1)!$

$\frac{\cancel{n!}}{r(r - 1)!\,(n - r)!} \times \frac{(r - 1)!\,(n - r + 1)(n - r)!}{\cancel{n!}} = 7 \times \frac{1}{3}$ Cancel out the common factors $\frac{n - r + 1}{r} = \frac{7}{3}$ Cross multiplying, we have $3n - 3r + 3 = 7r$ $3n - 10r = -3$ …………………….(4) Divide equation (1) by equation (3) $\frac{n!}{r!\,(n - r)!} \div \frac{n!}{(r + 1)!\,(n - r - 1)!} = 84 \div 126$ Change $\div$ to $\times$ by reciprocating the second fraction on each side of the equation $\frac{\cancel{n!}}{r!\,(n - r)!} \times \frac{(r + 1)!\,(n - r - 1)!}{\cancel{n!}} = 84 \times \frac{1}{126}$ Recall that $(r + 1)! = (r + 1)r!$ Also, $(n - r)! = (n - r)(n - r - 1)!$ Cancel out $n!$ , and substitute the expansions of $(r + 1)!$ and $(n - r)!$ $\frac{1}{r!\,(n - r)(n - r - 1)!} \times \frac{(r + 1)r!\,(n - r - 1)!}{1} = 2 \times \frac{1}{3}$ Cancel out the common factors $\frac{r + 1}{n - r} = \frac{2}{3}$ Cross-multiplying, we have $2n - 2r = 3r + 3$ $2n - 5r = 3$ …………………(5) Multiply (5) by 2 and subtract (4) from the result $(4n - 10r) - (3n - 10r) = 6 - -3$ $n = 9$ Substitute n into (5) $2(9) - 5r = 3$ $-5r = 3 - 18$ $-5r = -15$ $r = 3$ Therefore, ${}^{r}C_2 = {}^{3}C_2 = 3$

Permutations

A permutation is an arrangement of items where the order of the arrangement is important. The general formula is: ${}^{n}C_r = \frac{n!}{(n - r)!}$ Where: n = total number of objects r = number of objects arranged

Situations Where Order of Arrangement is Important

(Permutations)

1. Password and Security Codes Example: A 3-digit PIN, such as $\text{“457”},$ is not the same as $\text{“574”}$ . The order of the digits matters.
2. Seating Arrangements Arranging students in seats: (Kofi, Naadu, Afi, Oko) sitting in order is different from (Naadu, Afi, Oko, Kofi).
3. Award Positions Selecting 1st, 2nd, and 3rd place winners in a race. Who comes first, second, or third is important.
4. Job Assignments Assigning roles like Manager, Secretary, and Treasurer. The same three people in different roles create different outcomes.
5. Telephone, vehicle, index, account, and other identification numbers A change in the position of just one digit will render it wrong.
6. Arranging Books on a Shelf The order of books (Math, Science, English) is different from (Science, Math, English).
7. Forming words with letters Example: Cork and rock, listen and silent, state and taste, secure and rescue. These pairs of letters use the same letters but have very different meanings.
8. Dance or Performance Line-Up The order of dancers or speakers affects the presentation.
9. Sports Team Formation with Positions In sports, the position of players is important.
10. Queue/Line Order Whether it is a waakye queue or a queue at a bank or a trotro station, the order is important.

How to Answer a Permutation Question

S tep 1: Identify if it’s a permutation Ask yourself: Does order matter? If YES, it’s a permutation. Step 2: Recall the formula ${}^{n}P_r = \frac{n!}{(n - r)!}$ Where $n =$ total number of items and $r =$ number of items being chosen. Step 3: Substitute values and simplify the factorials Put in the given numbers for n and r. Step 4: State the final answer Write your result clearly, with units as necessary.

Example 8.23

A computer password is created by selecting 5 distinct letters from the English alphabet (26 letters). How many possible passwords can be formed?

Solution

Step 1: The order is important since it forms a password. Step 2: We will use the formula ${}^{n}P_r = \frac{n!}{(n - r)!}$ Where $n =$ total number of items $= 26$ and $r =$ number of items being chosen = 5 Step 3: Substitute values and simplify the factorials ${}^{26}P_5 = \frac{26!}{(26 - 5)!} = \frac{26!}{21!} = \frac{26 \times 25 \times 24 \times 23 \times 22 \times 21!}{21!}$ $= 26 \times 25 \times 24 \times 23 \times 22$ $= 7\,893\,600$ Step 4: Answer: $7\,893\,600$ possible passwords.

Example 8.24

(a) In how many different ways can the letters of the word CRATES be arranged?

(b) A group of 10 people is to elect a president, a secretary, and a treasurer. In how many ways can these three officers be chosen if no person can hold more than one position? (c) In how many ways can the letters of the word STATISTICS be arranged if all the vowels must be together?

Solution

(a) CRATES has 6 letters. None is repeated. So, the number of arrangements is ${}^{6}P_6 = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $= 6! = 720$ ways (b) $n = 10, r = 3$ ${}^{10}P_3 = \frac{10!}{(10 - 3)!} = 720$ ways (c) The word is STATISTICS has 10 letters Letter counts: $S = 3$ counts, $T = 3$ counts, $A = C = 1$ count, $I = 2$ counts. Number of vowels = 3 counts (I, I, A) Since all vowels must be together, treat them as a single letter. This will give us 7 consonants and 1 vowel $= 8$ items to arrange. But some of the consonants repeat. S has 3 counts, and $T$ has 3 counts. So, the number of arrangements of the 8 items is $\frac{8!}{3! \times 3!}$ Number of arrangements of the vowels (I, I, A) $= \frac{3!}{2!} = 3$ So, total arrangements $= 3 \times \frac{8!}{3! \times 3!} = 3 \times 1120$ $= 3\,360$ ways

Example 8.25

In a certain country, car licence plates consist of 5 characters. The first two characters are letters, chosen from one of the fixed prefixes: GE, AS, WR, VR, or UE. The last three characters are digits chosen from {0,1,3,4,5,6,7,8,9}. No letter or digit is repeated within a plate. How many different licence plates can be formed?

Solution

Since the first two characters have been chosen already, we will focus on the last three characters. We must choose and arrange 3 distinct digits from 10. Order matters (each digit goes into a specific position), so we use permutations: ${}^{10}P_3 = 10 \times 9 \times 8 = 720$

To find the total number of plates, multiply the number of prefixes (5) by the number of digit arrangements. This gives: $5 \times 720 = 3\,600$ ways

Example 8.26

In a certain town, all telephone numbers have eight digits, the first three always being $058$ or $139$ or $246$ or $732$ . How many distinct telephone numbers can be formed if all eight digits must be different?

Solution

Each telephone number is 8 digits long and must begin with one of the four fixed three-digit prefixes: $058$ or $139$ or $246$ or $732$ . In each prefix, the three starting digits are already fixed and are different from each other. After choosing a prefix (4 choices), the remaining 5 digits (4 th to 8 th positions) must be chosen from the 10 digits $0, 1, \ldots, 9$ , but must be different from each other and from the three digits already used in the prefix. That leaves $10 - 3 = 7$ available digits, and we must arrange 5 distinct digits in order: This arrangement can be done in ${}^{7}P_5 = 2\,520$ To find the total number of telephone numbers that have all eight digits distinct, multiply the number of arrangements by the 4 possible prefixes: $4 \times 2\,520 = 10\,080$ ways Therefore, 10 080 telephone numbers have all eight digits distinct.

Permutations vs. Combinations

Aspect Permutation Combination An arrangement of objects A selection of objects where order Definition where order matters. does not matter. “Does the order/position make a “Does the order/position NOT make a Key Question difference?” difference?” Formula ${}^{n}P_r = \frac{n!}{(n - r)!}$ ${}^{n}C_r = \frac{n!}{r!\,(n - r)!}$ Order Important Not important Passwords & security codes, Committee selection, food/menu Real-Life seating plans, job assignments, choices, subject selection, team Applications race results, sports line-up with formation, lottery, exam question positions selection Key Words to Arrange, Line-up, Order, Select, Choose, Pick, Form, Group, Watch Position, Rank, Schedule Combination

Mixed Questions: Permutations or Combinations

Example 8.27

Read each situation carefully. State whether the problem is a permutation (order matters) or a combination (order does not matter). Justify your answer.

1. A school is selecting 3 prefects (without specific roles) from a group of 12 students.
2. A football coach is arranging 11 players into field positions.
3. A customer chooses 2 drinks from 5 available in a shop.
4. The manager wants to assign Chairperson, Secretary, and Treasurer roles to 3 people from a group of 10 workers.
5. A lottery game requires selecting 6 numbers from 49.
6. A student forms a password using 3 letters from the alphabet without repetition.
7. A teacher selects 4 questions from a test bank of 10 questions to form a mini test.

Solution

1. Combination – order doesn’t matter in team selection.
2. Permutation – field positions make order important.
3. Combination – choosing drinks doesn’t depend on order.
4. Permutation – roles assigned make order important.
5. Combination – lottery numbers are the same regardless of order.
6. Permutation – passwords depend on order.
7. Combination – selection of questions doesn’t depend on order.

Example 8.28

A school choir has 10 students: 6 girls and 4 boys. The music director wants to form a 4-member singing group to perform a special piece. The group must include at least 1 boy. The order in which the singers stand on stage matters (positions are important). a. In how many different ways can the music director select and arrange 4 students to form the singing group under the given condition? b. Explain why your solution involves both combination and permutation, and how the model reflects the conditions of the problem.

Solution

a. Ways of selecting 4 members with at least 1 boy $({}^{6}C_3 \times {}^{4}C_1) + ({}^{6}C_2 \times {}^{4}C_2) + ({}^{6}C_1 \times {}^{4}C_3) + ({}^{6}C_0 \times {}^{4}C_4)$ $(20 \times 4) + (15 \times 6) + (6 \times 4) + (1 \times 1)$ $80 + 90 + 24 + 1 = 195$ ways After selecting the 4 students, the teacher needs to arrange them. The number of ways of arranging 4 students $= 4! = 24$ ways Therefore, the number of ways the teacher can select and arrange 4 students $= 24 \times 195 = 4\,680$ ways b. First, we choose which students form the group. When choosing who will be in the group, we use combinations to calculate the choices since the order is not important. Once the 4 students are chosen, the order in which they stand on stage matters. For each chosen set of 4, there are 4! ways to arrange them; this is the permutation part.

Applications of Permutations and Combinations

Permutations and combinations are mathematical tools used to count, arrange, and select items under different conditions. Beyond the classroom, these concepts have applications in business, commerce, industry, and many aspects of daily life. Understanding these applications helps us appreciate mathematics as a tool for solving real-world problems. The list below gives a few examples of their applications.

Applications in Business and Commerce

1. Marketing Strategies – Companies can use permutations to calculate the number of ways products can be arranged on shelves to attract customers.
2. Menu Planning – Restaurants can use combinations to determine the different possible sets of meals that can be served to customers.
3. Staff Scheduling – Businesses can use combinations to select workers for shifts or teams efficiently.

Example 8.29

A supermarket wants to display 5 different brands of fruit juice on a shelf to attract customers. a) In how many different ways can the 5 brands be arranged in a row?

b) If the manager insists that two particular brands must always be placed next to each other, how many arrangements are possible? c) If the manager decides that one brand must always be placed at the centre of the shelf, how many possible arrangements are there?

Solution

(a) Arranging 5 brands in a row Number of arrangements $= {}^{5}P_5$ $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ So, the five brands can be arranged in 120 ways, (b) Two particular brands are always together. Treat the two brands as one item. So, instead of 5 brands, we now have 4 items. These 4 items can be arranged in ${}^{4}P_4$ $= 4!$ Ways. However, within the block, the 2 brands can be arranged in $2!$ Ways. So, total arrangements $= 4! \times 2! = 24 \times 2 - 48$ ways. Therefore, if two brands are placed next to each other, the total number of arrangements is 48 ways. (c) One brand fixed in the centre If one brand is fixed at the centre, the remaining 4 brands can be arranged in any order without any restrictions. Number of arrangements ${}^{4}P_4$ $= 4!$ $= 24$ So, if one brand is fixed at the centre, the total number of arrangements = 24 ways.

Applications in Industry

1. Password and Security Codes – Permutations are used to calculate the possible arrangements of letters and digits for creating strong passwords and PIN codes.
2. Colour Schemes in Fashion and Manufacturing – Factories apply permutations in arranging colour patterns for clothing, textiles, or packaging designs.
3. Product Design – Combinations are used in deciding features to include in different versions of a product (e.g., car models with different accessories).

Example 8.30

A company requires its employees to create a 4-digit security code using the digits 0–9. Each digit can be used only once in a code. a) How many different security codes can be created? b) If the first digit cannot be 0, how many valid codes can be formed? c) If the code must begin with an odd number and end with an even number, how many possible codes are there?

Solution

(a) Total number of security codes We are choosing and arranging 4 digits out of 10 digits ( $0 - 9$ ). This is a permutation: ${}^{10}P_4$ $= 5\,040$ 5 040 different security codes can be created. (b) The first digit cannot be 0 Choices for the first digit = 9 (since it can be any digit except 0). After choosing the first digit, we have 9 digits left for the second, 8 digits for the third, and 7 digits for the fourth. Number of codes $= 9 \times 9 \times 8 \times 7 = 4\,536$ Thus, 4 536 valid codes can be formed. (c) Code begins with an odd number and ends with an even number Odd digits available = {1, 3, 5, 7, 9}. This gives 5 choices for the first digit. Even digits available = {0, 2, 4, 6, 8}. This also gives 5 choices for the last digit. After choosing the first and last digits, 8 digits remain for the second, and 7 digits remain for the third. Number of codes $= 5 \times 8 \times 7 \times 5 = 1\,400$ Hence, there will be 1 400 possible codes.

Applications in Science and Technology

1. Cryptography and Network Security – Permutations and combinations help in designing secure systems for communication and online banking.
2. Computer Architecture – Combinations are used in memory allocation and coding.
3. Molecular Biology – Scientists use permutations and combinations to study arrangements of DNA sequences and protein structures.
4. Pattern Analysis – Applications are found in language processing, artificial intelligence, and image recognition.

Example 8.31

In molecular biology, DNA sequences are made up of 4 nitrogen bases: Adenine (A), Thymine (T), Cytosine (C), and Guanine (G). a) How many different sequences can be formed if a DNA chain has 4 bases arranged in a specific order (repetition allowed)? b) If no base is repeated, how many possible sequences can be formed using the 4 bases? Proteins are built from 20 amino acids. c) In how many ways can a protein segment of 3 amino acids be arranged if order matters and no amino acid is repeated?

Solution

a) DNA sequence of 4 bases, repetition allowed Each of the 4 positions can be filled with any of the 4 bases. Total arrangements $= 4^4 = 4 \times 4 \times 4 \times 4 = 16 \times 16 = 256$ The number of sequences that can be formed is 256 b) DNA sequence of 4 bases, no repetition We are arranging 4 distinct items (A, T, C, G). Total arrangements $= 4! = 24$ 24 sequences can be formed using the 4 bases. c) Protein segment of 3 amino acids, no repetition We are arranging 3 amino acids out of 20. This is a permutation: ${}^{20}P_3$ $= \frac{20!}{(20 - 3)!} = \frac{20 \times 19 \times 18 \times 17!}{17!}$ $= 20 \times 19 \times 18$ $= 6\,840$ So, given the restrictions, a protein segment of 3 amino acids can be arranged in 6 840 possible ways. This shows how permutations are applied in molecular biology to model DNA and protein arrangements.

Applications in Games and Sports

1. Team Selection – Coaches use combinations to select players from a squad for a match.
2. Tournament Arrangements – Permutations are applied in fixing fixtures and arranging match orders.
3. Card Games and Lotteries – Both permutations and combinations are used in calculating the probability of outcomes.

Example 8.32

A football coach has 15 players available and wants to select a team of 11 players for a match. a) In how many different ways can the coach select the team of 11 players? b) If the coach must include 2 particular players (the captain and the goalkeeper), how many different teams can be formed from the remaining players? c) If the coach wants to select a captain in addition to the 11 players, how many possible selections are there?

Solution

(a) Selecting 11 players out of 15 This is a combination: The number of ways $= {}^{15}C_{11} = 1\,365$ (b) If 2 particular players are always included Then we only need to choose 9 more players from the remaining 13. Number of ways ${}^{13}C_9$ $=$ $715$ different teams (c) Selecting a captain in addition to the 11 players First select 11 players from 15: ${}^{15}C_{11}$ =1 365. Then select a captain from these 11 players: ${}^{11}C_1$ $=$ $11$ . Total selections $= 1\,365 \times 11 = 15\,015$ There are $15\,015$ more possible selections.\

Activity 8.2 Application of Permutations and Combinations

Investigate how permutations and combinations are applied in:

1. Business and commerce (e.g., menu planning, marketing strategies, scheduling)
2. Industry (e.g., product designs, passwords, colour schemes in factories)
3. Games and sports (e.g., team selection, card games, tournaments)
4. Technology and science (e.g., cryptography, communication networks, computer memory, molecular biology)

1. In a class of 40 students, 18 study Physics, 12 study Chemistry, and 8 study both Physics and Chemistry. If a student is selected at random: a) Find the probability that the student studies Physics given that they study Chemistry. b) Find the probability that the student studies Chemistry given that they study Physics.
2. A card is drawn at random from a standard pack of 52 playing cards. a) Find the probability that it is a King given that it is a face card. b) Find the probability that it is red given that it is a heart.
3. In a school, 60% of the students play football, and 30% of those who play football also play basketball. If 25% of all students do not play either game: a) What is the probability that a student plays basketball given that they play football? b) What is the probability that a student plays football given that they play basketball?
4. The probability that it rains on a particular day is 0.4. If it rains, the probability that students bring umbrellas is 0.8. If it does not rain, the probability that students bring umbrellas is 0.1. a) Find the probability that it rains given that a student has brought an umbrella. b) Interpret your result.
5. A box contains 3 red balls and 5 blue balls. Two balls are drawn at random without replacement. a) Find the probability that the second ball is red given that the first ball was red. b) Find the probability that the second ball is blue given that the first ball was red.
6. A fair coin is tossed 6 times. a) Find the probability of getting exactly 3 heads. b) Find the probability of getting at least 4 heads.
7. A box contains red and white balls. The probability of picking a red ball in a single draw is 0.3. If a ball is drawn and replaced 5 times, a) Find the probability of obtaining exactly 2 red balls. b) Find the probability of obtaining more than 3 red balls.
8. A factory produces bulbs, and each bulb has a probability of 0.1 of being defective. If a random sample of 12 bulbs is chosen, a) Find the probability that exactly 2 are defective.

b) Find the probability that at most 3 are defective. 9. In a certain city, the probability of rain on any given day is 0.4. Consider a week (7 days). a) Find the probability that it rains on exactly 4 days in the week. b) Find the probability that it rains on at least 5 days in the week. 10. The probability that a student passes a mathematics test is 0.7. If 10 students take the test, a) Find the probability that exactly 8 students pass. b) Find the probability that fewer than 6 students pass. 11. A basketball player has a free-throw success rate of 0.75. If she takes 8 free throws: a) Find the probability that she scores all 8. b) Find the probability that she scores at least once. 12. A restaurant offers 3 types of rice dishes, 2 types of stew, and 4 types of drinks. How many possible meal combinations can a customer order? 13. A student has 5 pairs of shoes, 3 trousers, and 4 shirts. In how many ways can the student dress up? 14. A computer system generates codes using 2 letters followed by 1 digit. How many codes can be formed? 15. (a) A school prefect is to be chosen from 8 candidates. In how many ways can a prefect and 2 assistants be chosen? (b) A company wants to form a committee of 4 workers from 12 staff members. How many committees can be formed? (c) A basketball coach must select 5 players out of 12. In how many ways can the team be formed? (d) A shop has 7 varieties of fruits. In how many ways can a customer select 3 fruits for a fruit salad? (e) In how many ways can 3 science subjects be selected from the 5 core science subjects? 16. An organisation needs a 4-member panel selected from 6 teachers, 5 parents, and 3 alumni. In how many ways can the panel be chosen if: (i) it must include exactly 2 teachers and 2 parents? (ii) it must include at least one alumnus? (iii) it must include no parents? 17. A designer must select 4 items to feature in a set from 5 shirts, 4 trousers, and 3 jackets. In how many ways can the designer choose the set if: (i) it must include 1 jacket, 2 shirts, and 1 trouser?

(ii) it must include at least two shirts? (iii) the set has no jacket? 18. (a) If ${}^{n}C_{r-1} = 10$ , ${}^{n}C_r = 45$ and ${}^{n}C_{r+1} = 120$ , find ${}^{r}C_2$ (b) If ${}^{n}C_{r-1} = 15$ , ${}^{n}C_r = 35$ and ${}^{n}C_{r+1} = 70$ , find ${}^{r}C_3$ (c) If ${}^{n}C_r = 210$ , ${}^{n}C_{r+1} = 252$ and ${}^{n}C_{r+2} = 210$ , find ${}^{r}C_2$ (d) If ${}^{n}C_{r-2} = 21$ , ${}^{n}C_{r-1} = 56$ and ${}^{n}C_r = 126$ , find ${}^{r}C_4$ 19. In a small village, there are 17 families, of which 11 families have at most 2 children. In a rural development program, 7 families are to be chosen for assistance, of which at least 3 families must have at most 2 children. In how many ways can the choice be made? 20. In an examination, there are three multiple-choice questions, and each question has 4 choices. Find the number of ways a student can fail to get all answers correct. 21. In how many ways can 4 books be arranged on a shelf among 7 different books? 22. How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition is not allowed? 23. In how many ways can the letters of the word “MATH” be arranged? 24. In how many different ways can a school assign 5 prefects to 5 distinct offices? 25. A company has 8 candidates for 3 different job positions (Manager, Secretary, Treasurer). In how many ways can the positions be filled? 26. A security code is created using 4 different digits chosen from $0, 1, 2, 3, 4, 5, 6, 7, 8, 9.$ How many different codes can be formed if digits cannot be repeated? 27. In how many ways can the letters of the word MARKET be arranged? 28. From a committee of 12 members, a chairperson, a vice-chairperson, and a financial secretary are to be chosen. In how many ways can these positions be filled if no one can hold more than one office? 29. In the arrangement of the letters of the word SUCCESS, in how many ways can the letters be arranged if the two C’s are always together? 30. A car registration number consists of 2 letters followed by 3 digits. The first two letters must be chosen from the prefixes: BA, KO, NI, SE, or TU. The digits are chosen from 0−9, but no digit is repeated. How many possible car registration numbers can be formed? 3. A certain mobile network assigns 7-digit phone numbers. The first two digits are fixed as $05, 07$ , or $09$ , while the remaining five digits are distinct and chosen from $0 - 9$ .

31. How many such numbers can be assigned?
32. A school is organising a cultural performance, and 12 students have volunteered to participate; 7 girls and 5 boys. The teacher wants to form a 5-member dance group such that the group must contain at least 2 girls, and the order in which students perform on stage matters. (a) In how many different ways can the teacher select and arrange 5 students to form the dance group under the given condition? (b) Explain how your model accounts for both selection and order, and why a combination or permutation (or both) is appropriate in this context.
33. For each question: State whether it is a Permutation or a Combination and solve the problem. a) A school is selecting 3 prefects from a class of 15 students. How many different groups of prefects can be chosen? b) A password is formed using 4 different letters from the 26 letters of the English alphabet. How many possible passwords can be created? c) A football coach wants to assign the roles of Captain, Assistant Captain, and Goalkeeper from a team of 18 players. In how many ways can this be done? d) From 12 books, a student chooses 5 books to borrow from the library. In how many ways can this selection be made? e) In a race with 10 athletes, how many different ways can the Gold, Silver and Bronze medals be awarded? f) A restaurant offers 6 types of drinks. A customer wants to choose 2 drinks. How many different drink choices are possible? g) A factory wants to arrange 5 different products in a row for display on a shelf. How many possible arrangements are there? h) In a quiz competition, 8 questions are given. A student must answer 3 questions. How many ways can the questions be chosen?