Statistics and Probability Practice Questions — JAMB UTME — Page 2

Question 21Question →

The grouped frequency distribution table below shows the marks scored by $100$ candidates in a Mathematics screening test:

Class Interval	Frequency
$10 - 19$	15
$20 - 29$	25
$30 - 39$	40
$40 - 49$	20

Match each data representation component on the left with its correct calculated numerical value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The sector angle matches

144^\circ

, frequency density matches

4.0

, preceding upper class boundary matches

29.5

, and median class cumulative frequency matches

80

.

The items match based on direct statistical computations: the modal sector angle is

144^\circ

, frequency density is

4.0

, upper class boundary of the preceding interval is

29.5

, and cumulative frequency at the upper boundary of the median class is

80

.

Step-by-Step Solution

1

Determine the modal class and total frequency

N

.

The highest frequency is

40

, so the modal class is

30 - 39

. Total frequency

N = 15 + 25 + 40 + 20 = 100

.

Modal class identification is essential for pie chart sector, frequency density, and class boundary calculations.

2

Calculate the sector angle for the modal class in a pie chart.

Sector Angle

= \frac{40}{100} \times 360^\circ = 144^\circ

.

The sector angle represents the class frequency as a fraction of total frequency multiplied by

360^\circ

.

3

Calculate the frequency density of the modal class.

Class boundary range for

30 - 39

is

29.5 - 39.5

, so width

= 10

. Frequency density

= \frac{40}{10} = 4.0

.

Frequency density is defined as class frequency divided by class interval width.

4

Determine the preceding upper class boundary and the cumulative frequency for the median class.

Preceding interval is

20 - 29

, so its upper boundary is

29.5

. Median is at position

50

(in interval

30 - 39

). Cumulative frequency up to boundary

39.5

is

15 + 25 + 40 = 80

.

Class boundaries are midpoints between adjacent class limits, and cumulative frequency sums all preceding frequencies up to the upper boundary.

Key Concept

Interpretation and calculation of pie chart sector angles, histogram frequency densities, real class boundaries, and cumulative frequencies from grouped data.

Question 22Question →

In a probability experiment, two fair six-sided dice were rolled $180$ times, yielding an experimental probability of $\frac{5}{18}$ for obtaining a sum divisible by $3$ . If $m$ additional consecutive rolls were conducted and every single one resulted in a sum divisible by $3$ , the updated overall experimental probability equaled the theoretical probability that the absolute difference between the numbers shown on two fair six-sided dice is at most $1$ . Calculate the value of $m$ .

Show answer & explanation

Answer: 54

Answer

54

The initial number of successful trials is

180 \times \frac{5}{18} = 50

. The theoretical probability of rolling two dice with an absolute difference of at most

1

is calculated by counting

6

outcomes with difference

0

and

10

outcomes with difference

1

, giving

\frac{16}{36} = \frac{4}{9}

. Equating the updated experimental probability

\frac{50 + m}{180 + m}

to

\frac{4}{9}

yields

9(50 + m) = 4(180 + m)

, which simplifies to

5m = 270

or

m = 54

.

Step-by-Step Solution

1

Calculate the initial number of successful trials from the given experimental probability.

Initial successful outcomes =

180 \times \frac{5}{18} = 50

.

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

2

Calculate the theoretical probability that the absolute difference between two rolled six-sided dice is at most 1.

Favorable outcomes = 16, so

P(\text{theoretical}) = \frac{16}{36} = \frac{4}{9}

.

Outcomes with difference 0:

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

(6 outcomes). Outcomes with difference 1:

(1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5)

(10 outcomes). Total sample space

= 6 \times 6 = 36

.

3

Formulate the algebraic equation relating the updated experimental probability to the theoretical probability.

\frac{50 + m}{180 + m} = \frac{4}{9}

.

Adding

m

consecutive successful rolls increases both the number of successful outcomes (to

50 + m

) and the total number of trials (to

180 + m

).

4

Solve the equation for

m

.

9(50 + m) = 4(180 + m) \implies 450 + 9m = 720 + 4m \implies 5m = 270 \implies m = 54

.

Cross-multiplication converts the rational expression into a linear equation.

Key Concept

Experimental and Theoretical Probability Synthesis

Estimated Time:3m 0s

Question 23Question →

A box contains $50$ marbles of three different colors: red, blue, and green. The theoretical probability of selecting a red marble at random is $\frac{3}{10}$ . In a probability experiment, a marble was drawn with replacement $250$ times, resulting in a red marble being drawn $85$ times. What is the positive difference between the experimental probability and the theoretical probability of selecting a red marble?

\frac{1}{25}

\frac{7}{25}

\frac{1}{5}

\frac{17}{50}

Show answer & explanation

Answer: $\frac{1}{25}$

Answer

The positive difference between the experimental probability and the theoretical probability of selecting a red marble is

\frac{1}{25}

.

The experimental probability is the ratio of observed successful trials to total trials, which is

\frac{85}{250} = \frac{17}{50}

. The theoretical probability is given as

\frac{3}{10} = \frac{15}{50}

. Subtracting the theoretical probability from the experimental probability gives

\frac{17}{50} - \frac{15}{50} = \frac{2}{50} = \frac{1}{25}

.

Step-by-Step Solution

1

Calculate the experimental probability (relative frequency) of drawing a red marble.

\text{Experimental Probability} = \frac{\text{Number of successful outcomes}}{\text{Total number of trials}} = \frac{85}{250} = \frac{17}{50} = 0.34

Experimental probability is calculated using empirical trial outcomes.

2

Identify the given theoretical probability of drawing a red marble.

\text{Theoretical Probability} = \frac{3}{10} = \frac{15}{50} = 0.30

The theoretical probability is provided directly as

\frac{3}{10}

.

3

Compute the positive difference between experimental and theoretical probabilities.

\text{Difference} = \frac{17}{50} - \frac{15}{50} = \frac{2}{50} = \frac{1}{25} = 0.04

Subtracting theoretical probability from experimental probability gives the required positive difference.

Key Concept

Experimental vs Theoretical Probability

Estimated Time:1m 30s

Question 24Question →

The frequency table below shows the distribution of daily petrol consumption (in litres) recorded by $40$ commercial minibus drivers:

Daily Petrol Consumption (litres)	Frequency ( $f$ )
$10 - 14$	$6$
$15 - 19$	$10$
$20 - 24$	$12$
$25 - 29$	$8$
$30 - 34$	$4$

What is the mean daily petrol consumption of the minibus drivers?

19.25\text{ litres}

22.00\text{ litres}

21.25\text{ litres}

23.25\text{ litres}

Show answer & explanation

Answer: $21.25\text{ litres}$

Answer

The mean daily petrol consumption is

21.25\text{ litres}

.

The mean of a grouped frequency distribution is computed by multiplying the midpoint of each class interval by its frequency, summing these products, and dividing by the total frequency. For the given distribution,

\sum fx = 850

and

\sum f = 40

, giving a mean of

21.25\text{ litres}

.

Step-by-Step Solution

1

Calculate the class mark (midpoint)

x

for each class interval.

Midpoints are:

12

for

10 - 14

,

17

for

15 - 19

,

22

for

20 - 24

,

27

for

25 - 29

, and

32

for

30 - 34

.

For grouped data, each class interval is represented by its midpoint.

2

Multiply each midpoint

x

by its corresponding frequency

f

to find

fx

.

6 \times 12 = 72

,

10 \times 17 = 170

,

12 \times 22 = 264

,

8 \times 27 = 216

,

4 \times 32 = 128

.

This yields the total sum contribution of each class.

3

Find total frequency

\sum f

and total sum of products

\sum fx

.

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

and

\sum fx = 72 + 170 + 264 + 216 + 128 = 850

.

Required components for the grouped mean formula.

4

Calculate the mean

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

.

\bar{x} = \frac{850}{40} = 21.25\text{ litres}.

Applying the formula for the mean of grouped frequency distribution.

Key Concept

Grouped Mean Calculation using Midpoints

Estimated Time:1m 30s

Question 25Question →

The table below shows the distribution of marks obtained by $40$ candidates in a computer-based recruitment test:

Mark Interval	Frequency ( $f$ )
$1 - 10$	$4$
$11 - 20$	$8$
$21 - 30$	$12$
$31 - 40$	$6$
$41 - 50$	$10$

What is the estimated mean mark of the candidates?

23.5

25.5

28.0

32.5

Show answer & explanation

Answer: $28.0$

Answer

The estimated mean mark of the candidates is

28.0

.

To find the mean of grouped frequency data, each class interval must be represented by its midpoint (

x

). Finding the midpoints (

5.5, 15.5, 25.5, 35.5, 45.5

) and multiplying by their frequencies gives total products summing to

1120

. Dividing by the total number of candidates (

40

) yields

28.0

.

Step-by-Step Solution

1

Calculate the midpoint (

x

) for each class interval

Class midpoints are:

5.5

,

15.5

,

25.5

,

35.5

, and

45.5

.

The class mark or midpoint is the representative value for data within a grouped interval.

2

Multiply each class midpoint (

x

) by its corresponding frequency (

f

) to find

f \cdot x

4 \times 5.5 = 22

,

8 \times 15.5 = 124

,

12 \times 25.5 = 306

,

6 \times 35.5 = 213

,

10 \times 45.5 = 455

.

Determines the total weighted value contribution of each class.

3

Sum all

f \cdot x

values and calculate the total frequency

\sum f

\sum f \cdot x = 22 + 124 + 306 + 213 + 455 = 1120

, and

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

.

Provides the overall sum of data estimates and total sample size needed for the grouped mean formula.

4

Apply the grouped mean formula

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

\bar{x} = \frac{1120}{40} = 28.0.

Computes the estimated mean for grouped frequency data.

Key Concept

Grouped Mean Calculation using Class Midpoints

Estimated Time:1m 30s

Question 26Question →

In an agricultural experiment, two crop varieties, $X$ and $Y$ , are tested independently for germination under drought conditions. The probability that variety $X$ germinates is $0.65$ , and the probability that at least one of the two varieties germinates is $0.86$ . What is the probability that variety $Y$ germinates?

Show answer & explanation

Answer: 0.6

Answer

The probability that variety Y germinates is 0.6

Using the law of addition for independent events

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

, substituting

P(X) = 0.65

and

P(X \cup Y) = 0.86

gives

0.86 = 0.65 + 0.35 P(Y)

, which yields

P(Y) = 0.6

.

Step-by-Step Solution

1

Apply the general addition law for two probability events

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

The addition law relates the union, individual probabilities, and intersection of compound events.

2

Express the intersection using the law of multiplication for independent events

P(X \cap Y) = P(X) \times P(Y) = 0.65 \times P(Y)

Because germination of variety X and variety Y are independent events.

3

Substitute given values into the combined probability formula and solve for

P(Y)

0.86 = 0.65 + P(Y) - 0.65 P(Y) \implies 0.21 = 0.35 P(Y) \implies P(Y) = 0.6

Isolating the unknown probability

P(Y)

yields the correct decimal value.

Key Concept

Compound Probability Laws and Independent Events

Question 27Question →

The table below shows the speed distribution (in km/h) recorded for $50$ vehicles passing a police checkpoint on a highway:

Speed Interval (km/h)	Frequency ( $f$ )
$40 - 49$	$6$
$50 - 59$	$12$
$60 - 69$	$18$
$70 - 79$	$10$
$80 - 89$	$4$

Calculate the mean speed of the vehicles.

Show answer & explanation

Answer: 63.3

Answer

The mean speed of the vehicles is

63.3\text{ km/h}

.

The mean of a grouped frequency distribution is computed using the formula

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

. The midpoints (

x

) for the intervals are

44.5, 54.5, 64.5, 74.5,

and

84.5

. Multiplying these midpoints by their respective frequencies yields products totaling

3165

. Dividing this total by the total number of vehicles (

50

) gives a mean speed of

63.3\text{ km/h}

.

Step-by-Step Solution

1

Find the class midpoints (

x

) for each speed interval

Midpoints are

x_1 = 44.5

,

x_2 = 54.5

,

x_3 = 64.5

,

x_4 = 74.5

, and

x_5 = 84.5

.

The midpoint of a grouped class interval is calculated as

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

.

2

Calculate the product

fx

for each class

6 \times 44.5 = 267

,

12 \times 54.5 = 654

,

18 \times 64.5 = 1161

,

10 \times 74.5 = 745

, and

4 \times 84.5 = 338

.

Multiplying class midpoint by frequency estimates the total contribution of all items within that interval.

3

Find total frequency

\sum f

and sum of products

\sum fx

\sum f = 6 + 12 + 18 + 10 + 4 = 50

and

\sum fx = 267 + 654 + 1161 + 745 + 338 = 3165

.

Summing frequencies gives total sample size, and summing

fx

gives estimated grand total.

4

Compute the estimated mean speed

\bar{x} = \frac{\sum fx}{\sum f} = \frac{3165}{50} = 63.3\text{ km/h}

.

The standard formula for the mean of grouped frequency data is

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

.

Key Concept

Grouped Mean Calculation

Question 28Question →

The table below shows the frequency distribution of marks obtained by a group of students in a mathematics test:

Mark ( $x$ )	2	4	6	8	10
Frequency ( $f$ )	2	1	4	1	2

Find the mean deviation of the distribution.

Show answer & explanation

Answer: 2

Answer

The mean deviation of the distribution is 2.

To find the mean deviation, first calculate the mean

\bar{x} = \frac{\sum fx}{\sum f} = \frac{60}{10} = 6

. Next, sum the absolute deviations multiplied by their frequencies:

\sum f|x - \bar{x}| = 2(4) + 1(2) + 4(0) + 1(2) + 2(4) = 20

. Dividing this total by the sum of frequencies

10

yields a mean deviation of

2

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the distribution

\bar{x} = \frac{\sum f x}{\sum f} = \frac{(2 \times 2) + (1 \times 4) + (4 \times 6) + (1 \times 8) + (2 \times 10)}{2 + 1 + 4 + 1 + 2} = \frac{60}{10} = 6

The mean is required as the central benchmark from which individual deviations are measured.

2

Calculate the sum of absolute deviations weighted by frequency

\sum f |x - \bar{x}| = 2|2 - 6| + 1|4 - 6| + 4|6 - 6| + 1|8 - 6| + 2|10 - 6| = 8 + 2 + 0 + 2 + 8 = 20

Each absolute difference from the mean must be multiplied by its frequency to account for the total deviation.

3

Divide the total absolute deviation by the total frequency

\text{Mean Deviation} = \frac{\sum f |x - \bar{x}|}{\sum f} = \frac{20}{10} = 2

The mean deviation represents the average distance of all observations from the arithmetic mean.

Key Concept

Mean Deviation of a Frequency Distribution

Estimated Time:1m 30s

Question 29Question →

The distribution of daily solar energy generation (in kWh) recorded at an agricultural research station over a period of days is summarized in the table below:

Daily Generation (kWh)	Frequency ( $f$ )
$10 - 19$	$3$
$20 - 29$	$x$
$30 - 39$	$8$
$40 - 49$	$5$
$50 - 59$	$x + 2$

If the mean daily solar energy generation for the recorded period is $37.5\text{ kWh}$ , what is the value of $x$ ?

1

3

6

11

Show answer & explanation

Answer: $6$

Answer

6

The exact class marks are

14.5, 24.5, 34.5, 44.5,

and

54.5

. The total frequency is

\sum f = 18 + 2x

and the sum of products is

\sum f x = 651 + 79x

. Substituting these into

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

with

\bar{x} = 37.5

gives

675 + 75x = 651 + 79x

, which solves to

4x = 24

, giving the value

6

.

Step-by-Step Solution

1

Determine the class mark (midpoint

x_i

) for each class interval

The midpoints are

14.5, 24.5, 34.5, 44.5,

and

54.5

.

The class mark is calculated as

\frac{\text{Lower Limit} + \text{Upper Limit}}{2}

.

2

Calculate total frequency

\sum f

and total product sum

\sum f x_i

\sum f = 3 + x + 8 + 5 + (x + 2) = 18 + 2x

.

\sum f x_i = 3(14.5) + x(24.5) + 8(34.5) + 5(44.5) + (x+2)(54.5) = 43.5 + 24.5x + 276 + 222.5 + 54.5x + 109 = 651 + 79x

To compute the mean of grouped data, sum the products of each frequency and its corresponding class mark.

3

Set up the mean formula equation and solve for

x

\bar{x} = \frac{\sum f x_i}{\sum f} \implies 37.5 = \frac{651 + 79x}{18 + 2x}

37.5(18 + 2x) = 651 + 79x

675 + 75x = 651 + 79x

79x - 75x = 675 - 651

4x = 24 \implies x = 6

Equating the expression for mean to the given mean value

37.5\text{ kWh}

yields a linear equation in

x

.

Key Concept

Measures of Central Tendency for Grouped Data - Finding Unknown Frequencies from Grouped Mean

Question 30Question →

The test scores of five students in a mathematics quiz are $5, 8, 11, 12,$ and $14$ . What is the variance of these test scores?

10

\sqrt{10}

50

2.8

Show answer & explanation

Answer: $10$

Answer

The variance of the test scores is

10

.

The mean of the scores is

10

. Subtracting

10

from each score gives deviations of

-5, -2, 1, 2,

and

4

. Squaring these gives

25, 4, 1, 4,

and

16

, which sum to

50

. Dividing this sum by the number of data points (

5

) yields a variance of

10

.

Step-by-Step Solution

1

Calculate the mean (

\bar{x}

) of the data set

\bar{x} = \frac{5 + 8 + 11 + 12 + 14}{5} = \frac{50}{5} = 10

The mean is required to find the deviation of each score from the central value.

2

Compute the squared deviations from the mean

(x_i - \bar{x})^2

(5-10)^2 = 25

,

(8-10)^2 = 4

,

(11-10)^2 = 1

,

(12-10)^2 = 4

,

(14-10)^2 = 16

Squaring deviations ensures all negative differences become positive.

3

Find the sum of all squared deviations

\sum (x_i - \bar{x})^2 = 25 + 4 + 1 + 4 + 16 = 50

Summing the squared deviations measures total variation around the mean.

4

Divide the total sum of squared deviations by the number of data values (

N = 5

)

\text{Variance } (\sigma^2) = \frac{50}{5} = 10

Variance is defined as the average of the squared deviations.

Key Concept

Variance of Ungrouped Data

Estimated Time:1m 30s

Question 31Question →

A small business recorded the number of customer inquiries received per day over six consecutive days as follows: $4$ , $7$ , $8$ , $11$ , $13$ , and $17$ . What is the variance of the daily customer inquiries?

Show answer & explanation

Answer: 18

Answer

The variance of the daily customer inquiries is

18

.

To find the variance of the data set

\{4, 7, 8, 11, 13, 17\}

, first calculate the mean:

\bar{x} = \frac{4+7+8+11+13+17}{6} = 10

. Next, compute the squared deviation of each data point from the mean:

(4-10)^2 = 36

,

(7-10)^2 = 9

,

(8-10)^2 = 4

,

(11-10)^2 = 1

,

(13-10)^2 = 9

, and

(17-10)^2 = 49

. Summing these squared deviations gives

108

. Dividing this total by the number of observations (

6

) yields a variance of

18

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the given data set.

\bar{x} = 10

The mean is required as the central point from which deviations are calculated.

2

Determine the squared deviation of each data value from the mean.

(-6)^2 = 36

,

(-3)^2 = 9

,

(-2)^2 = 4

,

1^2 = 1

,

3^2 = 9

,

7^2 = 49

Variance measures the average squared distance of data points from the mean.

3

Sum all calculated squared deviations.

\sum (x - \bar{x})^2 = 36 + 9 + 4 + 1 + 9 + 49 = 108

This provides the total sum of squares for the data set.

4

Divide the total sum of squares by the number of observations (

n = 6

).

\sigma^2 = \frac{108}{6} = 18

Population variance formula is

\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}

.

Key Concept

Variance of Ungrouped Data

Estimated Time:1m 30s

Question 32Question →

Five daily rainfall measurements (in mm) recorded in a city are $3, 6, 7, 9,$ and $15$ . What is the standard deviation of these rainfall measurements?

3.2

4

16

20

Show answer & explanation

Answer: $4$

Answer

The standard deviation of the rainfall measurements is

4

.

First compute the mean of the data set:

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8

. Next, sum the squared deviations from the mean:

(3-8)^2 + (6-8)^2 + (7-8)^2 + (9-8)^2 + (15-8)^2 = 25 + 4 + 1 + 1 + 49 = 80

. Dividing by the number of observations (

5

) yields the variance

\sigma^2 = \frac{80}{5} = 16

. Taking the square root gives the standard deviation

\sigma = \sqrt{16} = 4

.

Step-by-Step Solution

1

Calculate the arithmetic mean (

\bar{x}

) of the dataset

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = \frac{40}{5} = 8

The mean is needed to find deviations for each data point.

2

Find the deviation of each number from the mean and square it

(3-8)^2 = 25

,

(6-8)^2 = 4

,

(7-8)^2 = 1

,

(9-8)^2 = 1

,

(15-8)^2 = 49

Squaring ensures all deviation values are non-negative.

3

Sum the squared deviations and divide by the total number of items (

n = 5

) to find the variance

\text{Variance } (\sigma^2) = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16

Variance measures the average squared distance from the mean.

4

Take the square root of the variance to obtain standard deviation

\text{Standard Deviation } (\sigma) = \sqrt{16} = 4

Standard deviation returns the dispersion measure back to the original units.

Key Concept

Standard Deviation of Ungrouped Data

Question 33Question →

Find the positive integer value of $n$ such that $^{n}P_4 = 42 \times {^{n}P_2}$ .

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)

allows dividing out

n(n-1)

, leading to

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

. Expanding and factoring gives

n^2 - 5n - 36 = 0

, which yields

n = 9

as the only valid positive integer.

Step-by-Step Solution

1

Apply the permutation formula

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

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

and

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

By definition of permutations, selecting

r

items from

n

distinct items without replacement.

2

Substitute the expansions into the given relation

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

Direct substitution into

^{n}P_4 = 42 \times {^{n}P_2}

.

3

Simplify by dividing out common non-zero terms

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

Since

n \ge 4

,

n(n-1) \neq 0

and can be safely divided from both sides.

4

Form and solve the quadratic equation

n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0

Expanding terms and factoring the resulting quadratic expression.

5

Determine the valid root

n = 9

Permutation total items

n

must satisfy

n \ge r \ge 0

, rejecting the negative root

n = -4

.

Key Concept

Algebraic equations involving permutations

Estimated Time:1m 15s

Question 34Question →

Six members of a board of directors are to be seated around a circular conference table. If two specific members refuse to sit next to each other, how many different seating arrangements are possible?

72

48

120

240

Show answer & explanation

Answer: 72

Answer

72

To find the number of circular seating arrangements where two specific members do not sit next to each other, use the complement rule. First, the total unrestricted circular arrangements of 6 members is

(6 - 1)! = 5! = 120

. Next, calculate the arrangements where the two members sit together by treating them as 1 unit (giving 5 units in total). The circular arrangements of these 5 units is

(5 - 1)! = 4! = 24

, and the 2 members can swap seats in

2! = 2

ways, yielding

24 \times 2 = 48

arrangements together. Subtracting this from the total yields

120 - 48 = 72

.

Step-by-Step Solution

1

Calculate total circular arrangements without restrictions

Total arrangements =

(6 - 1)! = 5! = 120

The number of ways to arrange

n

distinct items in a circle is

(n - 1)!

.

2

Calculate arrangements where the two specific members sit together

Restricted arrangements =

(5 - 1)! \times 2! = 4! \times 2 = 24 \times 2 = 48

Treat the two members as a single block, giving 5 units to arrange around a circle in

(5-1)!

ways, and multiply by

2!

for internal ordering of the pair.

3

Subtract the together arrangements from total arrangements

Ways apart =

120 - 48 = 72

Complementary counting gives the number of arrangements where the two members do not sit next to each other.

Key Concept

Circular Permutations with Restrictions

Estimated Time:1m 30s

Question 35Question →

A school committee of $4$ members is to be selected from $6$ male teachers and $4$ female teachers. If the committee must contain exactly $2$ male teachers and $2$ female teachers, how many different committees can be formed?

Show answer & explanation

Answer: 90

Answer

The total number of different committees that can be formed is

90

.

Selecting 2 female teachers from 4 yields

\binom{4}{2} = 6

ways. Selecting 2 male teachers from 6 yields

\binom{6}{2} = 15

ways. By the multiplication principle, the total number of ways to form the committee is

6 \times 15 = 90

.

Step-by-Step Solution

1

Find the number of ways to select 2 female teachers out of 4

$\binom{4}{2} = 6$

Selection order does not matter within the committee, so combination formula $nCr$ applies.

2

Find the number of ways to select 2 male teachers out of 6

$\binom{6}{2} = 15$

Selection order does not matter within the committee, so combination formula $nCr$ applies.

3

Multiply the possibilities for selecting male and female members

6 \times 15 = 90

According to the fundamental counting principle, independent group selections are multiplied.

Key Concept

Combinations with restricted subset selections (Product Rule of Counting)

Question 36Question →

$10$ distinct points are marked on the circumference of a circle. How many different triangles can be formed by connecting any $3$ of these points as vertices?

45

120

360

720

Show answer & explanation

Answer: $120$

Answer

The number of distinct triangles that can be formed is

120

.

To form a triangle, any

3

points must be chosen from the

10

available points. Because no three points on a circle are collinear, every choice of

3

points forms a unique triangle. Since the order of choosing vertices does not alter the triangle, we use combinations:

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

.

Step-by-Step Solution

1

Identify the total number of items

n

and the subset size

r

.

n = 10

points on the circle, and

r = 3

points required to form a triangle.

Any set of

3

non-collinear points uniquely determines a triangle. Since all points lie on a circle, no three points are collinear.

2

Determine whether order matters.

Order does not matter because choosing points

A, B, C

produces the same triangle as choosing

B, C, A

.

Selection of vertices for a geometric shape is a combination problem, not a permutation problem.

3

Apply the combination formula

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

.

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120

Evaluating the combination gives the total number of distinct triangles.

Key Concept

Combinations for Geometric Formations

Estimated Time:1m 30s

Question 37Question →

A box contains $30$ tickets numbered $1$ to $30$ . In a probability experiment, a ticket is drawn at random from the box and its number recorded before being replaced. This trial is performed $150$ times, and a ticket with a number that is a multiple of $4$ is recorded $45$ times. What is the absolute difference between the experimental probability and the theoretical probability of selecting a ticket bearing a multiple of $4$ ?

\frac{1}{15}

\frac{3}{10}

\frac{7}{30}

\frac{8}{15}

Show answer & explanation

Answer: $\frac{1}{15}$

Answer

The absolute difference between the experimental probability and the theoretical probability is

\frac{1}{15}

.

The theoretical probability of drawing a multiple of 4 is

\frac{7}{30}

since there are 7 favorable tickets (

4, 8, 12, 16, 20, 24, 28

) out of 30 total tickets. The experimental probability from 150 draws is

\frac{45}{150} = \frac{9}{30}

. Taking the absolute difference gives

\frac{9}{30} - \frac{7}{30} = \frac{2}{30} = \frac{1}{15}

.

Step-by-Step Solution

1

Determine the theoretical probability

Theoretical probability

P(T) = \frac{7}{30}

The multiples of

4

between

1

and

30

are

4, 8, 12, 16, 20, 24, 28

, giving

7

favorable outcomes out of

30

possible outcomes.

2

Determine the experimental probability

Experimental probability

P(E) = \frac{45}{150} = \frac{3}{10}

The event occurred

45

times out of

150

experimental trials.

3

Calculate the absolute difference between

P(E)

and

P(T)

|P(E) - P(T)| = |\frac{9}{30} - \frac{7}{30}| = \frac{2}{30} = \frac{1}{15}

Express both fractions with a common denominator of

30

and subtract the smaller probability from the larger.

Key Concept

Experimental probability is calculated from trial data (favorable trials divided by total trials), whereas theoretical probability is calculated from expected sample space outcomes (favorable outcomes divided by total sample space size).

Estimated Time:1m 30s

Question 38Question →

In a mathematics examination, a student is required to answer $5$ questions out of $8$ available questions. If the first $2$ questions are compulsory, in how many different ways can the student select the questions to answer?

Show answer & explanation

Answer: 20

Answer

The student can select the questions in 20 different ways.

With 2 compulsory questions, the student only has to select 3 more questions from the remaining 6 questions. The number of ways to select 3 items from 6 without regard to order is given by ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.

Step-by-Step Solution

1

Deduct compulsory questions from both the required total and available total

The student must choose 3 additional questions from the remaining 6 questions.

Compulsory questions are fixed and provide only 1 selection choice.

2

Apply the combination formula ^6C_3 to calculate selection ways

^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

The order in which the student chooses the examination questions does not alter the group of questions selected.

Key Concept

Combinations with restricted/compulsory choices

Question 39Question →

A container holds a large number of red, blue, and yellow counters. The theoretical probability of selecting a red counter at random is $\frac{2}{5}$ , and the theoretical probability of selecting a blue counter is $\frac{1}{3}$ . In an experiment where a counter is drawn and replaced $300$ times, a yellow counter is selected $72$ times. What is the absolute difference between the theoretical expected number of yellow counters and the experimental frequency of yellow counters obtained?

8

148

28

80

Show answer & explanation

Answer: $8$

Answer

The absolute difference between the theoretical expected number of yellow counters and the experimental frequency is

8

.

The theoretical probability of drawing a yellow counter is

1 - (\frac{2}{5} + \frac{1}{3}) = \frac{4}{15}

. For

300

trials, the theoretical expected count of yellow counters is

\frac{4}{15} \times 300 = 80

. Subtracting the experimental observed count of

72

from

80

gives an absolute difference of

8

.

Step-by-Step Solution

1

Calculate the theoretical probability of drawing a yellow counter.

P(\text{Yellow}) = 1 - \left(P(\text{Red}) + P(\text{Blue})\right) = 1 - \left(\frac{2}{5} + \frac{1}{3}\right) = 1 - \frac{11}{15} = \frac{4}{15}

The sum of probabilities of all mutually exclusive outcomes in the sample space must equal

1

.

2

Calculate the theoretical expected frequency of yellow counters in

300

trials.

\text{Expected frequency} = P(\text{Yellow}) \times \text{Total trials} = \frac{4}{15} \times 300 = 80

The expected number of occurrences of an outcome is the product of its theoretical probability and the number of trials.

3

Find the absolute difference between expected frequency and experimental frequency.

|80 - 72| = 8

The question asks for the difference between the theoretical expected value (

80

) and the actual experimental result (

72

).

Key Concept

Experimental and Theoretical Probability

Estimated Time:1m 30s

Question 40Question →

Two marksmen, Kemi and Chidi, independently shoot at a target once. The probability that Kemi hits the target is $0.7$ , and the probability that Chidi hits the target is $0.6$ . What is the probability that at least one of them hits the target?

Show answer & explanation

Answer: 0.88

Answer

The probability that at least one marksman hits the target is 0.88.

Since the two shooting attempts are independent, the probability that both marksmen miss is

(1 - 0.7) \times (1 - 0.6) = 0.3 \times 0.4 = 0.12

. Subtracting this probability from 1 yields

1 - 0.12 = 0.88

, which represents the probability that at least one marksman hits the target.

Step-by-Step Solution

1

Find the probability of each event not occurring (missing the target).

P(\text{Kemi misses}) = 0.3

and

P(\text{Chidi misses}) = 0.4

The sum of an event's probability and its complement is always 1.

2

Determine the probability of both events failing simultaneously.

P(\text{both miss}) = 0.3 \times 0.4 = 0.12

For independent events

A

and

B

,

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

.

3

Use the complement rule to find the probability of at least one success.

P(\text{at least one hits}) = 1 - 0.12 = 0.88

The event 'at least one hits' is the exact complement of 'neither hits'.

Key Concept

Compound probability laws for independent events and complement of combined events

Statistics and Probability

Answer

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