Statistics and Probability Practice Questions — JAMB UTME — Page 6

Question 101Question →

A sports delegation of $6$ athletes is to be selected from a pool of $7$ sprinters and $5$ distance runners. If the delegation must contain at least $4$ sprinters, in how many different ways can the delegation be formed?

Show answer & explanation

Answer: 462

Answer

The total number of different ways to form the delegation is 462.

To select at least 4 sprinters in a delegation of 6 athletes, we must evaluate three mutually exclusive cases: selecting 4 sprinters and 2 distance runners (

{}^7\text{C}_4 \times {}^5\text{C}_2 = 350

), selecting 5 sprinters and 1 distance runner (

{}^7\text{C}_5 \times {}^5\text{C}_1 = 105

), and selecting 6 sprinters and 0 distance runners (

{}^7\text{C}_6 \times {}^5\text{C}_0 = 7

). Adding these yields

350 + 105 + 7 = 462

total ways.

Step-by-Step Solution

1

Determine all valid combinations of sprinters and distance runners satisfying the condition of having at least 4 sprinters in a group of 6.

Three valid cases: (4 sprinters, 2 distance runners), (5 sprinters, 1 distance runner), and (6 sprinters, 0 distance runners).

The delegation requires 6 members and at least 4 sprinters.

2

Calculate the combinations for each case using

{}^n\text{C}_r = \frac{n!}{r!(n-r)!}

.

Case 1:

{}^7\text{C}_4 \times {}^5\text{C}_2 = 350

; Case 2:

{}^7\text{C}_5 \times {}^5\text{C}_1 = 105

; Case 3:

{}^7\text{C}_6 \times {}^5\text{C}_0 = 7

.

Apply the product rule of counting for selecting sprinters and distance runners independently within each case.

3

Sum the results of the mutually exclusive cases.

350 + 105 + 7 = 462.

Apply the addition principle of counting for mutually exclusive scenarios.

Key Concept

Combinations with restrictions and the addition principle of counting

Estimated Time:2m 0s

Question 102Question →

A continuous grouped frequency distribution consists of four class intervals: $10 - 14$ , $15 - 24$ , $25 - 29$ , and $30 - 44$ . The total frequency of the distribution is $160$ , and the frequency of the class interval $25 - 29$ is $22$ .

In a histogram representing this data, the height of the rectangle for the interval $15 - 24$ corresponds to a frequency density of $6$ . In a pie chart representing the same distribution, the sector angle for the class interval $30 - 44$ is $108^\circ$ .

What is the frequency density of the class interval $10 - 14$ ?

Show answer & explanation

Answer: 6

Answer

The frequency density of the class interval

10 - 14

is

6

.

By converting the pie chart sector angle of

108^\circ

into a frequency of

48

out of

160

, and using the frequency density of

6

with class width

10

to find a frequency of

60

for

15 - 24

, the remaining frequency for

10 - 14

is

30

. Dividing this by the true class width of

5

(from boundaries

9.5

to

14.5

) gives a frequency density of

6

.

Step-by-Step Solution

1

Calculate the frequency of the class interval

30 - 44

from the pie chart sector angle.

Frequency

f_{30-44} = \frac{108^\circ}{360^\circ} \times 160 = 48

.

The sector angle in a pie chart is directly proportional to the frequency relative to the

360^\circ

total.

2

Determine the class width and frequency of the class interval

15 - 24

.

Class boundaries are

14.5

and

24.5

, so width

w = 10

. Frequency

f_{15-24} = 6 \times 10 = 60

.

Frequency density is defined as frequency divided by class width, so frequency equals frequency density multiplied by class width.

3

Determine the frequency of the class interval

10 - 14

.

Frequency

f_{10-14} = 160 - (60 + 22 + 48) = 30

.

The sum of all class frequencies must equal the total frequency of

160

.

4

Calculate the class width and frequency density of

10 - 14

.

Class width

w_{10-14} = 14.5 - 9.5 = 5

. Frequency density

= \frac{30}{5} = 6

.

Dividing the frequency of the class (

30

) by its exact class boundary width (

5

) yields the frequency density.

Key Concept

Integration of Frequency Density and Pie Chart Sector Angles

Question 103Question →

A container contains $6$ red counters, $4$ blue counters, and $3$ green counters. In how many different ways can a selection of $5$ counters be made if the selection must contain at least $2$ red counters, at least $1$ blue counter, and at most $1$ green counter?

Show answer & explanation

Answer: 750

Answer

The total number of different ways to make the selection is 750.

Breaking the problem into disjoint cases based on the number of green counters (0 or 1) and applying the combination formula for red, blue, and green counters in each valid configuration yields

240 + 510 = 750

total ways.

Step-by-Step Solution

1

Determine the allowable counts for Green (

G

), Red (

R

), and Blue (

B

) counters

Green counters can be 0 or 1. If

G=0

,

R+B=5

with

R \geq 2, B \geq 1

. If

G=1

,

R+B=4

with

R \geq 2, B \geq 1

.

The constraints state

G \leq 1

,

R \geq 2

, and

B \geq 1

for a total of 5 counters.

2

Calculate combinations for Case 1 (

G = 0

)

Ways for

(2R, 3B, 0G) = 60

;

(3R, 2B, 0G) = 120

;

(4R, 1B, 0G) = 60

. Total for Case 1 = 240.

Apply the combinations formula

\binom{n}{r} = \frac{n!}{r!(n-r)!}

to each sub-case.

3

Calculate combinations for Case 2 (

G = 1

)

Ways for

(2R, 2B, 1G) = 270

;

(3R, 1B, 1G) = 240

. Total for Case 2 = 510.

Apply combinations to choose 1 Green counter along with the valid Red and Blue combinations.

4

Sum all mutually exclusive cases

Total selection ways =

240 + 510 = 750

.

According to the addition principle of counting, the totals of mutually exclusive cases are added together.

Key Concept

Combinations with Multiple Conditional Constraints

Question 104Question →

The cumulative frequency distribution of the operational lifespans (in hours) for a batch of $100$ precision LED modules tested in a laboratory is summarized below:

Lifespan Interval (hours)	Class Boundaries	Cumulative Frequency
$100 - 119$	$99.5 - 119.5$	$10$
$120 - 139$	$119.5 - 139.5$	$25$
$140 - 159$	$139.5 - 159.5$	$60$
$160 - 179$	$159.5 - 179.5$	$85$
$180 - 199$	$179.5 - 199.5$	$100$

Using linear interpolation for cumulative frequency distributions, calculate the 75th percentile ( $P_{75}$ ) of the lifespan of these modules in hours.

Show answer & explanation

Answer: 171.5

Answer

The 75th percentile of operational lifespan is 171.5 hours.

The 75th percentile rank position is

0.75 \times 100 = 75

. The percentile falls within the class boundary

159.5 - 179.5

. Substituting lower class boundary

L = 159.5

, preceding cumulative frequency

c.f. = 60

, class frequency

f = 25

, and class width

c = 20

into

P_{75} = L + \left(\frac{75 - c.f.}{f}\right) \times c

gives

159.5 + \left(\frac{15}{25}\right) \times 20 = 171.5

hours.

Step-by-Step Solution

1

Determine the rank position of the 75th percentile.

Rank position = 75th value out of 100.

The 75th percentile corresponds to 75% of the total frequency N = 100.

2

Identify the percentile class interval and extract relevant parameters.

Class interval is 159.5 - 179.5, with L = 159.5, c.f. = 60, f = 25, and c = 20.

The cumulative frequency increases from 60 to 85 across the boundary 159.5 to 179.5, which contains the 75th value.

3

Compute the percentile value using ogive linear interpolation.

P_75 = 159.5 + [(75 - 60) / 25] * 20 = 171.5 hours.

Applying the cumulative frequency interpolation formula yields the exact value.

Key Concept

Calculating Percentiles from Cumulative Frequency / Ogives

Question 105Question →

A box contains $6$ red balls and $4$ blue balls. A ball is drawn at random, its color is recorded, and it is then returned to the box. A second ball is subsequently drawn at random. What is the probability of selecting a red ball on the first draw and a blue ball on the second draw?

\frac{6}{25}

\frac{4}{15}

\frac{1}{10}

1

Show answer & explanation

Answer: $\frac{6}{25}$

Answer

\frac{6}{25}

Because the first ball is replaced after being recorded, the two draws are independent compound events. The probability of drawing a red ball first is

\frac{6}{10} = \frac{3}{5}

, and the probability of drawing a blue ball second is

\frac{4}{10} = \frac{2}{5}

. Multiplying these probabilities yields

P(\text{Red and Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

.

Step-by-Step Solution

1

Calculate the probability of drawing a red ball on the first draw

P(\text{Red}) = \frac{6}{6 + 4} = \frac{6}{10} = \frac{3}{5}

There are

6

red balls out of a total of

10

balls.

2

Determine independence and calculate the probability of drawing a blue ball on the second draw

P(\text{Blue}) = \frac{4}{10} = \frac{2}{5}

Because the first ball is replaced before the second draw, the total number of balls and their composition remain unchanged.

3

Apply the multiplication law for independent compound events

P(\text{Red and then Blue}) = P(\text{Red}) \times P(\text{Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

For independent events

A

and

B

,

P(A \cap B) = P(A) \times P(B)

.

Key Concept

Multiplication Law of Probability for Independent Events

Estimated Time:45s

Question 106Question →

Two independent security systems, $X$ and $Y$ , operate in a facility. The probability that system $Y$ detects an intrusion is $0.20$ greater than the probability that system $X$ detects it. If the probability that at least one of the systems detects an intrusion is $0.92$ , what is the probability that exactly one of the systems detects the intrusion?

0.44

0.48

0.52

0.32

Show answer & explanation

Answer: $0.44$

Answer

The probability that exactly one of the systems detects the intrusion is

0.44

.

The correct answer is

0.44

. Setting

P(X) = p

and

P(Y) = p + 0.20

, we use the independence rule

P(X \cap Y) = P(X)P(Y)

in the addition law

P(X \cup Y) = P(X) + P(Y) - P(X)P(Y) = 0.92

. Solving

p^2 - 1.80p + 0.72 = 0

gives

P(X) = 0.60

and

P(Y) = 0.80

. The probability of both detecting the intrusion is

0.60 \times 0.80 = 0.48

. Subtracting the probability of both from the probability of at least one (

0.92 - 0.48

) gives

0.44

for exactly one system detecting the intrusion.

Step-by-Step Solution

1

Define variables for the individual probabilities

Let

P(X) = p

. Then

P(Y) = p + 0.20

.

System

Y

's probability is given as

0.20

greater than System

X

's probability.

2

Apply the addition law for independent events

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + (p + 0.20) - p(p + 0.20) = 0.92

Since

X

and

Y

are independent,

P(X \cap Y) = P(X)P(Y)

.

3

Solve the quadratic equation for

p

2p + 0.20 - p^2 - 0.20p = 0.92 \implies -p^2 + 1.80p - 0.72 = 0 \implies p^2 - 1.80p + 0.72 = 0

. Factoring yields

(p - 0.60)(p - 1.20) = 0

. Since

p \le 1

,

p = 0.60

.

Probability values cannot exceed

1

, so

p = 0.60

is the valid root.

4

Calculate individual probabilities and the probability of both occurring

P(X) = 0.60

,

P(Y) = 0.80

, and

P(X \cap Y) = 0.60 \times 0.80 = 0.48

.

These are needed to evaluate the compound probability of exactly one event occurring.

5

Calculate the probability that exactly one system detects the intrusion

P(\text{exactly one}) = P(X \cup Y) - P(X \cap Y) = 0.92 - 0.48 = 0.44

.

The probability of exactly one event occurring is the probability of at least one minus the probability of both.

Key Concept

Probability laws for independent and compound events

Question 107Question →

A fair spinner is divided into $8$ equal sectors numbered $1$ to $8$ . In an experiment, the spinner is spun $120$ times, and it lands on a multiple of $3$ a total of $36$ times. What is the positive difference between the experimental probability and the theoretical probability of landing on a multiple of $3$ ?

\frac{1}{20}

\frac{3}{10}

\frac{1}{4}

\frac{1}{10}

Show answer & explanation

Answer: $\frac{1}{20}$

Answer

The positive difference between the experimental probability and theoretical probability is

\frac{1}{20}

.

The correct option is

\frac{1}{20}

. The theoretical probability of hitting a multiple of

3

on an

8

-sector spinner is

\frac{2}{8} = \frac{1}{4}

. The experimental probability over

120

spins with

36

successes is

\frac{36}{120} = \frac{3}{10}

. Subtracting theoretical probability from experimental probability gives

\frac{3}{10} - \frac{1}{4} = \frac{6-5}{20} = \frac{1}{20}

.

Step-by-Step Solution

1

Calculate the theoretical probability of landing on a multiple of

3

.

The sample space is

\{1, 2, 3, 4, 5, 6, 7, 8\}

(total

8

outcomes). Multiples of

3

in this set are

\{3, 6\}

(

2

outcomes). Thus,

P(\text{Theoretical}) = \frac{2}{8} = \frac{1}{4}

.

Theoretical probability is the ratio of favorable outcomes to total possible outcomes in a fair trial.

2

Calculate the experimental probability (relative frequency) from the trial results.

P(\text{Experimental}) = \frac{36}{120} = \frac{3}{10}

.

Experimental probability is defined as the number of observed successful events divided by the total number of trials conducted.

3

Find the positive difference between the experimental and theoretical probabilities.

\text{Difference} = \frac{3}{10} - \frac{1}{4} = \frac{6}{20} - \frac{5}{20} = \frac{1}{20}

.

Subtracting the smaller theoretical probability from the larger experimental probability yields the positive difference.

Key Concept

Experimental versus Theoretical Probability

Estimated Time:1m 30s

Question 108Question →

A bag contains $5$ red balls, $3$ blue balls, and $n$ green balls. The theoretical probability of drawing a blue ball at random from the bag is $\frac{3}{20}$ . In a probability experiment, a ball is drawn at random from the bag and replaced $500$ times. If a green ball is observed $340$ times, calculate the absolute difference between the experimental probability and the theoretical probability of drawing a green ball.

Show answer & explanation

Answer: 0.08

Answer

The absolute difference between the experimental probability and the theoretical probability of drawing a green ball is

0.08

(or

\frac{2}{25}

).

The theoretical probability of drawing a blue ball establishes that the bag contains

20

total balls, meaning there are

12

green balls. The theoretical probability of picking a green ball is therefore

\frac{12}{20} = 0.60

. The experimental probability is

\frac{340}{500} = 0.68

. Taking the positive difference gives

|0.68 - 0.60| = 0.08

.

Step-by-Step Solution

1

Find the total number of balls and the value of

n

n = 12

, Total balls =

20

The theoretical probability of picking a blue ball is

\frac{3}{8+n} = \frac{3}{20}

, giving

8 + n = 20

.

2

Find the theoretical probability of drawing a green ball

P(\text{Green})_{\text{theo}} = 0.60

There are

12

green balls out of

20

total balls, so

\frac{12}{20} = 0.60

.

3

Find the experimental probability of drawing a green ball

P(\text{Green})_{\text{exp}} = 0.68

In

500

trials, a green ball was observed

340

times, so

\frac{340}{500} = 0.68

.

4

Calculate the absolute difference

|0.68 - 0.60| = 0.08

Subtract the theoretical probability from the experimental probability and take the absolute value.

Key Concept

Comparison of theoretical probability based on outcome sample space and experimental probability based on observed trial relative frequencies.

Question 109Question →

The mean of four positive integers is $9$ . If the mode of the set of numbers is $12$ and their median is $10$ , what is the value of the smallest integer?

4

2

6

5

Show answer & explanation

Answer: $4$

Answer

The smallest integer is 4.

Let the four positive integers in ascending order be

a \le b \le c \le d

. The sum of the four numbers is

4 \times 9 = 36

. For

12

to be the mode, it must appear at least twice, so

c = 12

and

d = 12

. The median of four numbers is

\frac{b + c}{2} = 10

, which gives

\frac{b + 12}{2} = 10

, so

b = 8

. Substituting these into the sum equation

a + 8 + 12 + 12 = 36

gives

a = 4

.

Step-by-Step Solution

1

Calculate the sum of all four integers from the given mean.

Sum

= 4 \times 9 = 36

.

The mean of

n

numbers is the sum divided by

n

.

2

Determine the two largest numbers using the mode.

Let the ordered numbers be

a \le b \le c \le d

. Since the mode is

12

and the set contains

4

numbers with a single mode,

c = 12

and

d = 12

.

For

12

to be the mode in a 4-element set without tie, it must appear at least twice.

3

Use the median to find the second integer

b

.

Median

= \frac{b + c}{2} = \frac{b + 12}{2} = 10 \implies b + 12 = 20 \implies b = 8

.

The median of an even number of ordered elements is the arithmetic mean of the two middle elements.

4

Solve for the smallest integer

a

.

a + b + c + d = 36 \implies a + 8 + 12 + 12 = 36 \implies a + 32 = 36 \implies a = 4

.

Subtracting the sum of the known three numbers from the total sum gives the smallest integer.

Key Concept

Using mean, median, and mode definitions simultaneously to deduce unknown values in an ungrouped dataset

Question 110Question →

Two independent weather forecasting stations, $A$ and $B$ , operate in a region. The probability that station $A$ makes an accurate forecast on any given day is $0.80$ , and the probability that station $B$ makes an accurate forecast is $0.75$ . What is the probability that at least one of the two stations makes an accurate forecast on a given day?

Show answer & explanation

Answer: 0.95

Answer

The probability that at least one of the two stations makes an accurate forecast is

0.95

.

The probability of at least one event occurring is given by

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Because events

A

and

B

are independent,

P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60

. Substituting into the addition rule yields

0.80 + 0.75 - 0.60 = 0.95

. Alternatively, using the complement rule:

1 - P(A')P(B') = 1 - (1 - 0.80)(1 - 0.75) = 1 - (0.20 \times 0.25) = 1 - 0.05 = 0.95

.

Step-by-Step Solution

1

Calculate the probability of both events occurring simultaneously using the multiplication law for independent events.

P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60

Since the two forecasting stations operate independently, the joint probability is the product of their individual probabilities.

2

Apply the general addition law of probability to calculate the probability of at least one station making an accurate forecast.

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.80 + 0.75 - 0.60 = 0.95

The probability of compound event 'at least one' corresponds to the union of the two events.

Key Concept

Addition and Multiplication Laws of Probability for Independent Events

Question 111Question →

The table below presents the cumulative frequency distribution of delay times (in minutes) for $50$ regional flights recorded at an airport:

Delay Time (minutes)	Cumulative Frequency
$10 - 19$	$8$
$20 - 29$	$20$
$30 - 39$	$40$
$40 - 49$	$50$

What is the $70\text{th}$ percentile of the delay times?

37.0

minutes

37.5

minutes

32.0

minutes

35.0

minutes

Show answer & explanation

Answer: $37.0$ minutes

Answer

The

70\text{th}

percentile of the delay times is

37.0

minutes.

The

70\text{th}

percentile position is found by taking

70\%

of

50

, which yields

35

. Looking at the cumulative frequency table, the 35th item falls in the

30 - 39

interval (class boundaries

29.5 - 39.5

). Applying the percentile interpolation formula:

P_{70} = 29.5 + \left(\frac{35 - 20}{20}\right) \times 10 = 29.5 + 7.5 = 37.0\text{ minutes}

.

Step-by-Step Solution

1

Determine the position of the 70th percentile rank.

Position rank

= \frac{70}{100} \times 50 = 35\text{th value}

.

The 70th percentile corresponds to 70% of the total frequency

N = 50

.

2

Identify the 70th percentile class interval and its parameters.

Interval:

30 - 39

; Lower class boundary

L = 29.5

, cumulative frequency preceding class

F = 20

, frequency of percentile class

f = 20

, class width

c = 10

.

The cumulative frequency reaches 20 at

29.5

and 40 at

39.5

, so the 35th value lies in the

30 - 39

class interval.

3

Apply the percentile interpolation formula

P_{70} = L + \left(\frac{\frac{70N}{100} - F}{f}\right) \times c

.

P_{70} = 29.5 + \left(\frac{35 - 20}{20}\right) \times 10 = 29.5 + \left(\frac{15}{20}\right) \times 10 = 29.5 + 7.5 = 37.0\text{ minutes}

.

Linear interpolation distributes the cumulative frequencies evenly across the class interval.

Key Concept

Calculating percentiles from grouped cumulative frequency data using class boundaries and linear interpolation.

Question 112Question →

A research committee of $5$ members is to be formed from $6$ male scientists and $5$ female scientists. The committee must contain strictly more male scientists than female scientists. If one particular male scientist and one particular female scientist refuse to serve on the same committee together, in how many different ways can the committee be selected?

231

281

1,281

256

Show answer & explanation

Answer: 231

Answer

The total number of valid ways to select the committee is 231.

The correct response of 231 is found by calculating the total committees having strictly more men than women (281) and subtracting the subset of those committees where both restricted individuals are chosen together (50).

Step-by-Step Solution

1

Calculate the unconstrained ways to form a 5-member committee with strictly more males than females.

Three possible gender distributions exist:
- 5 Males, 0 Females:

\binom{6}{5} \times \binom{5}{0} = 6 \times 1 = 6

- 4 Males, 1 Female:

\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75

- 3 Males, 2 Females:

\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200

Total unconstrained ways =

6 + 75 + 200 = 281

.

Establishing the total pool of valid gender-ratio committees before applying the exclusion condition.

2

Calculate the number of forbidden committee selections where both the restricted male scientist and restricted female scientist are included together.

Since 1 male and 1 female are already chosen, 3 remaining members must be selected from the remaining 5 males and 4 females:
- For 4 Males, 1 Female total (need 3 more males, 0 more females):

\binom{5}{3} \times \binom{4}{0} = 10 \times 1 = 10

- For 3 Males, 2 Females total (need 2 more males, 1 more female):

\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40

Total forbidden ways =

10 + 40 = 50

.

To enforce the restriction that both specific members cannot serve together, calculate all committee configurations containing both.

3

Subtract the forbidden selections from the unconstrained selections.

Valid ways =

281 - 50 = 231

.

Applying the subtraction principle of combinations to yield the final allowed arrangements.

Key Concept

Combinations with gender ratio constraints and mutual exclusion restrictions

Estimated Time:2m 30s

Question 113Question →

A committee of $6$ delegates is to be seated around a circular table for a conference. If two specific delegates must always sit next to each other, in how many distinct ways can the delegates be arranged?

48

24

240

15

Show answer & explanation

Answer: 48

Answer

48

The correct answer is 48 because treating the 2 restricted delegates as a single unit leaves 5 items to arrange around a circle, which yields

(5-1)! = 24

arrangements. Since the 2 delegates can arrange themselves in

2! = 2

ways within their block, the total number of arrangements is

24 \times 2 = 48

.

Step-by-Step Solution

1

Group the restricted delegates into a single block

2 specified delegates are treated as 1 unit, leaving 4 remaining delegates, making a total of 5 items to arrange.

Since the two delegates must sit next to each other, treating them as a single entity ensures they remain together in all seating positions.

2

Calculate the circular arrangements of the 5 items

(5 - 1)! = 4! = 24

ways.

The number of ways to arrange

n

distinct items around a circular table is given by

(n - 1)!

.

3

Account for internal arrangements of the paired delegates

2! = 2

ways.

The two delegates within the single block can swap positions between themselves.

4

Multiply the circular arrangements by the internal arrangements

24 \times 2 = 48

total distinct arrangements.

By the fundamental counting principle, total arrangements equal the product of the independent steps.

Key Concept

Circular Permutations with Restrictions

Estimated Time:1m 15s

Question 114Question →

A pie chart illustrates the annual allocation of funds for an agricultural research station across four sectors: Crop Research, Livestock, Irrigation, and Equipment Maintenance. The sector angles for Crop Research, Livestock, and Irrigation are $135^\circ$ , $90^\circ$ , and $60^\circ$ respectively. If the total annual budget is $\text{N}7,200,000$ , what is the amount allocated to Equipment Maintenance in Naira?

Show answer & explanation

Answer: 1500000

Answer

The amount allocated to Equipment Maintenance is 1,500,000 Naira.

A complete pie chart has a total angle of

360^\circ

, representing the full amount of

\text{N}7,200,000

. The angle corresponding to Equipment Maintenance is

360^\circ - (135^\circ + 90^\circ + 60^\circ) = 75^\circ

. The dollar/naira allocation is given by

\frac{75^\circ}{360^\circ} \times 7,200,000 = 1,500,000

.

Step-by-Step Solution

1

Calculate the sum of the given sector angles

Sum of known angles = 285 degrees

Determining the total angular measure already accounted for by the three known sectors.

2

Subtract the sum of known angles from 360 degrees

Sector angle for Equipment Maintenance = 75 degrees

The sum of all sector angles in a pie chart is always 360 degrees.

3

Multiply the fraction of the pie chart by the total funds

1,500,000 Naira

Converting the sector angle representation to its corresponding quantitative value.

Key Concept

Pie chart sector angle computation and value conversion

Question 115Question →

A plane contains $12$ distinct points such that exactly $5$ of them lie on the same straight line, while no other subset of three points is collinear. How many distinct straight lines can be formed by joining pairs of these points?

Show answer & explanation

Answer: 57

Answer

The total number of distinct straight lines that can be formed is 57.

The total number of distinct straight lines is obtained by taking all possible pairs of points from 12, which is

^{12}C_2 = 66

, subtracting the pairs formed among the 5 collinear points (

^5C_2 = 10

), and adding 1 back to account for the single straight line on which those 5 points lie. This gives

66 - 10 + 1 = 57

.

Step-by-Step Solution

1

Calculate the total possible lines by selecting any 2 points out of 12.

^{12}C_2 = \frac{12 \times 11}{2} = 66

Any pair of non-collinear points uniquely determines a straight line.

2

Calculate the combinations of 2 points selected from the 5 collinear points.

^5C_2 = \frac{5 \times 4}{2} = 10

These 10 point pairs all lie on the exact same single straight line instead of forming 10 distinct lines.

3

Adjust the count to find the net number of distinct straight lines.

66 - 10 + 1 = 57

Subtracting the 10 overcounted combinations and adding back 1 for the single line passing through all 5 collinear points gives the correct total.

Key Concept

Combinations with Collinear Constraints

Question 116Question →

A biased four-sided die numbered $1$ , $2$ , $3$ , and $4$ is rolled $200$ times in a probability experiment. The frequency of each outcome is recorded in the table below:

Outcome	Frequency
$1$	$45$
$2$	$60$
$3$	$55$
$4$	$40$

What is the experimental probability of rolling an odd number?

Show answer & explanation

Answer: 0.5

Answer

0.5

The correct experimental probability is calculated by dividing the total frequency of observed odd outcomes (

45 + 55 = 100

) by the total number of trials (

200

), which gives

\frac{100}{200} = 0.5

.

Step-by-Step Solution

1

Identify the favorable outcomes for rolling an odd number

The odd outcomes on the die are

1

and

3

.

Odd numbers are integers that are not divisible by

2

.

2

Calculate the total frequency of the favorable outcomes

Frequency of

1

is

45

and frequency of

3

is

55

. Total frequency

= 45 + 55 = 100

.

The total number of times an odd number was rolled is the sum of the frequencies of outcome

1

and outcome

3

.

3

Compute the experimental probability

\text{Experimental Probability} = \frac{\text{Frequency of odd outcomes}}{\text{Total number of trials}} = \frac{100}{200} = 0.5

.

Experimental probability measures the relative frequency of an event occurring in an actual experiment.

Key Concept

Experimental Probability

Question 117Question →

The frequency distribution table below shows the masses (in $\text{kg}$ ) of $40$ cassava tubers harvested from an agricultural test plot:

Mass ( $\text{kg}$ )	Frequency ( $f$ )
$10 - 14$	$4$
$15 - 19$	$8$
$20 - 24$	$12$
$25 - 29$	$10$
$30 - 34$	$6$

What is the estimated mean mass of the harvested cassava tubers?

20.75\text{ kg}

22.40\text{ kg}

22.75\text{ kg}

24.75\text{ kg}

Show answer & explanation

Answer: $22.75\text{ kg}$

Answer

22.75\text{ kg}

The value

22.75\text{ kg}

is correct because the mean for grouped data is obtained by multiplying each class midpoint by its frequency, summing these products to get

910

, and dividing by the total frequency of

40

.

Step-by-Step Solution

1

Find the class midpoint (

x

) for each class interval using

x = \frac{\text{Lower limit} + \text{Upper limit}}{2}

.

Midpoints are:

12

for

10-14

,

17

for

15-19

,

22

for

20-24

,

27

for

25-29

, and

32

for

30-34

.

Grouped data calculations require representative central values for each class interval.

2

Multiply each class midpoint (

x

) by its corresponding frequency (

f

) to get

fx

, then sum all

fx

values.

\sum fx = (4 \times 12) + (8 \times 17) + (12 \times 22) + (10 \times 27) + (6 \times 32) = 48 + 136 + 264 + 270 + 192 = 910

.

This calculates the total estimated mass of all harvested cassava tubers.

3

Sum the frequencies to get the total number of items

\sum f

.

\sum f = 4 + 8 + 12 + 10 + 6 = 40.

The mean formula divides the total sum by the total frequency.

4

Compute the grouped mean using

\bar{x} = \frac{\sum fx}{\sum f}

.

\bar{x} = \frac{910}{40} = 22.75\text{ kg}.

Dividing the sum of products by the total frequency yields the estimated mean.

Key Concept

Grouped Mean Calculation

Question 118Question →

During a quality control inspection at a manufacturing plant, two independent automated scanners, $S_1$ and $S_2$ , inspect items along a production line. The probability that scanner $S_1$ detects a surface flaw on a defective item is $0.90$ , and the probability that scanner $S_2$ detects the flaw is $0.80$ . What is the probability that a surface flaw is detected by exactly one of the two scanners?

Show answer & explanation

Answer: 0.26

Answer

The probability that a surface flaw is detected by exactly one of the two scanners is 0.26.

The probability of a surface flaw being detected by exactly one scanner is calculated by combining two mutually exclusive compound events: scanner

S_1

detects it while scanner

S_2

fails (

0.90 \times 0.20 = 0.18

), and scanner

S_1

fails while scanner

S_2

detects it (

0.10 \times 0.80 = 0.08

). Adding these two values yields

0.18 + 0.08 = 0.26

.

Step-by-Step Solution

1

Determine the probabilities of non-detection for each scanner.

P(S_1') = 0.10, P(S_2') = 0.20

The probability of an event's complement is 1 minus the probability of the event.

2

Calculate the probability that scanner S_1 detects the flaw while scanner S_2 misses it.

P(S_1 \cap S_2') = 0.90 \times 0.20 = 0.18

Because the scanners operate independently, joint probability is the product of their individual probabilities.

3

Calculate the probability that scanner S_1 misses the flaw while scanner S_2 detects it.

P(S_1' \cap S_2) = 0.10 \times 0.80 = 0.08

Independence allows multiplication of individual probabilities.

4

Sum the probabilities of the two mutually exclusive outcomes.

P(\text{exactly one}) = 0.18 + 0.08 = 0.26

The addition law applies because detecting the flaw with only S_1 and only S_2 are mutually exclusive events.

Key Concept

Compound probability laws for independent and mutually exclusive events

Question 119Question →

Given the permutation equation ${^{n}P_4} = 42 \cdot {^{n}P_2}$ , what is the positive integer value of $n$ ?

Show answer & explanation

Answer: 9

Answer

The positive integer value of

n

is

9

.

Expanding

{^{n}P_4}

as

n(n-1)(n-2)(n-3)

and

{^{n}P_2}

as

n(n-1)

transforms the equation to

n(n-1)(n-2)(n-3) = 42n(n-1)

. Dividing both sides by

n(n-1)

gives

(n-2)(n-3) = 42

, which simplifies to

n^2 - 5n - 36 = 0

. Factoring this quadratic gives

(n-9)(n+4) = 0

. Since

n

must be a positive integer greater than or equal to

4

, the correct answer is

9

.

Step-by-Step Solution

1

Write out the expanded expressions for

{^{n}P_4}

and

{^{n}P_2}

.

{^{n}P_4} = n(n-1)(n-2)(n-3)

and

{^{n}P_2} = n(n-1)

.

By definition,

{^{n}P_r} = \frac{n!}{(n-r)!} = n(n-1)\dots(n-r+1)

.

2

Set up the algebraic equation based on the given problem statement.

n(n-1)(n-2)(n-3) = 42n(n-1)

.

Substitute the expanded permutation formulas into the given identity.

3

Simplify the equation by dividing both sides by the non-zero common product

n(n-1)

.

(n-2)(n-3) = 42

.

Since

n \ge 4

,

n(n-1) \neq 0

, so we can cancel these terms from both sides.

4

Expand and rearrange the resulting expression into a standard quadratic equation.

n^2 - 5n - 36 = 0

.

Expanding

(n-2)(n-3)

gives

n^2 - 5n + 6

; subtracting

42

yields

n^2 - 5n - 36 = 0

.

5

Solve the quadratic equation for

n

and discard non-physical roots.

n = 9

.

Factoring yields

(n-9)(n+4) = 0

, so

n = 9

or

n = -4

. A permutation requires

n

to be a positive integer

\ge 4

, so

n = 9

.

Key Concept

Solving algebraic equations involving permutations

{^{n}P_r} = \frac{n!}{(n-r)!}

Question 120Question →

Two independent events $A$ and $B$ in a sample space satisfy $P(A') = \frac{2}{5}$ and $P(A \cup B) = \frac{7}{10}$ , where $A'$ denotes the complement of event $A$ . What is the probability that exactly one of the two events occurs?

\frac{11}{20}

\frac{7}{10}

\frac{17}{20}

\frac{3}{20}

Show answer & explanation

Answer: $\frac{11}{20}$

Answer

\frac{11}{20}

First, find

P(A) = 1 - \frac{2}{5} = \frac{3}{5}

. Since

A

and

B

are independent,

P(A \cap B) = P(A)P(B)

. Using the addition law

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, we substitute the values to get

\frac{7}{10} = \frac{3}{5} + P(B)\left(1 - \frac{3}{5}\right)

, which gives

P(B) = \frac{1}{4}

. The probability of exactly one event occurring is

P(A)P(B') + P(A')P(B) = \frac{3}{5} \times \frac{3}{4} + \frac{2}{5} \times \frac{1}{4} = \frac{11}{20}

.

Step-by-Step Solution

1

Calculate the probability of event A

P(A) = 1 - P(A') = 1 - \frac{2}{5} = \frac{3}{5}

The sum of the probabilities of an event and its complement equals 1.

2

Apply the addition law for independent events to find P(B)

P(A \cup B) = P(A) + P(B) - P(A)P(B) \implies \frac{7}{10} = \frac{3}{5} + P(B)\left(1 - \frac{3}{5}\right) \implies P(B) = \frac{1}{4}

For independent events, the intersection probability is

P(A \cap B) = P(A)P(B)

.

3

Calculate the complement of event B

P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}

The complement of event B represents the event that B does not occur.

4

Compute the probability that exactly one event occurs

P(\text{exactly one}) = P(A \cap B') + P(A' \cap B) = P(A)P(B') + P(A')P(B) = \left(\frac{3}{5} \times \frac{3}{4}\right) + \left(\frac{2}{5} \times \frac{1}{4}\right) = \frac{9}{20} + \frac{2}{20} = \frac{11}{20}

The occurrence of exactly one event means either A occurs and B does not, or A does not occur and B occurs.

Key Concept

Compound Probability and Probability Laws for Independent Events

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