Statistics and Probability Practice Questions — JAMB UTME — Page 8

Question 141Question →

The mean mass of $6$ packages in a delivery van is $20\text{ kg}$ . If one package weighing $35\text{ kg}$ is unloaded from the van, what is the mean mass of the remaining $5$ packages?

17\text{ kg}

23\text{ kg}

31\text{ kg}

14\text{ kg}

Show answer & explanation

Answer: $17\text{ kg}$

Answer

The mean mass of the remaining packages is

17\text{ kg}

.

The total weight of the initial

6

packages is

6 \times 20\text{ kg} = 120\text{ kg}

. Unloading one package of

35\text{ kg}

leaves

120 - 35 = 85\text{ kg}

. Dividing this remaining mass by the remaining

5

packages yields

\frac{85}{5} = 17\text{ kg}

.

Step-by-Step Solution

1

Calculate the initial total mass of the 6 packages

\text{Total mass} = 6 \times 20 = 120\text{ kg}

The sum of a set of numbers is equal to the product of their count and their mean.

2

Calculate the total mass after unloading the 35 kg package

\text{New total mass} = 120 - 35 = 85\text{ kg}

Removing a package decreases the total combined mass by its individual weight.

3

Calculate the new mean mass for the remaining 5 packages

\text{New mean} = \frac{85}{5} = 17\text{ kg}

The mean of ungrouped data is found by dividing the new sum by the updated count of items (

6 - 1 = 5

).

Key Concept

Calculation of the mean of ungrouped data after removing an item

Question 142Question →

The table below shows the distribution of examination scores of $50$ candidates in a selection test:

Score Class	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$9$
$30 - 39$	$16$
$40 - 49$	$14$
$50 - 59$	$6$

Using the cumulative frequency distribution (ogive), what is the estimated $70^{\text{th}}$ percentile score?

33.1

43.1

43.6

38.1

Show answer & explanation

Answer: $43.1$

Answer

The estimated

70^{\text{th}}

percentile score is

43.1

.

The

70^{\text{th}}

percentile corresponds to the score at the

35^{\text{th}}

candidate (

70\%

of

50

). This falls within the

40 - 49

score class, which has a lower boundary of

39.5

, a frequency of

14

, and a class width of

10

. Interpolating linearly gives

39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 43.07 \approx 43.1

.

Step-by-Step Solution

1

Construct the cumulative frequency table to find class boundaries and cumulative frequencies.

Cumulative frequencies:

10-19

(

CF = 5

),

20-29

(

CF = 14

),

30-39

(

CF = 30

),

40-49

(

CF = 44

),

50-59

(

CF = 50

).

Cumulative frequencies are required to locate the position of the desired percentile.

2

Determine the rank position of the

70^{\text{th}}

percentile (

P_{70}

).

Rank position =

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

position.

The

70^{\text{th}}

percentile corresponds to the score below which

70\%

of the total candidates fall.

3

Identify the class interval containing the

35^{\text{th}}

cumulative frequency and state its parameters.

The target class is

40 - 49

. Parameters: Lower class boundary

L = 39.5

, preceding cumulative frequency

CF_{\text{prev}} = 30

, class frequency

f = 14

, class width

c = 10

.

Since

30 < 35 \le 44

, the

35^{\text{th}}

entry falls within the

40 - 49

class.

4

Apply the linear interpolation formula for percentiles from an ogive.

P_{70} = L + \left(\frac{\frac{70N}{100} - CF_{\text{prev}}}{f}\right) \times c = 39.5 + \left(\frac{35 - 30}{14}\right) \times 10 = 39.5 + \frac{50}{14} \approx 43.07 \approx 43.1

.

This calculates the exact score estimate on the cumulative frequency curve.

Key Concept

Estimating Percentiles from Grouped Data and Ogives

Question 143Question →

The table below shows the distribution of weekly overtime hours worked by $10$ technicians in a manufacturing plant:

Overtime Hours ( $x$ )	2	4	6	8
Number of Technicians ( $f$ )	4	3	2	1

What is the variance of the overtime hours?

4

2

1.6

4.4

Show answer & explanation

Answer: 4

Answer

The variance of the overtime hours is

4

.

The correct answer is

4

. First, compute the mean

\bar{x} = \frac{40}{10} = 4

. Then compute the sum of weighted squared deviations

\sum f(x - \bar{x})^2 = 4(2-4)^2 + 3(4-4)^2 + 2(6-4)^2 + 1(8-4)^2 = 16 + 0 + 8 + 16 = 40

. Dividing this by total frequency

\sum f = 10

yields

\sigma^2 = \frac{40}{10} = 4

.

Step-by-Step Solution

1

Calculate the total frequency (

\sum f

) and the sum of the products of values and frequencies (

\sum fx

).

\sum f = 4 + 3 + 2 + 1 = 10

and

\sum fx = (2 \times 4) + (4 \times 3) + (6 \times 2) + (8 \times 1) = 8 + 12 + 12 + 8 = 40

.

These sums are required to compute the mean of the distribution.

2

Determine the mean (

\bar{x}

) of the distribution.

\bar{x} = \frac{\sum fx}{\sum f} = \frac{40}{10} = 4

.

The mean is used as the central reference point to find deviations.

3

Find the squared deviation

(x - \bar{x})^2

for each score and multiply by its respective frequency

f

.

For

x = 2

:

4(2 - 4)^2 = 16

.
For

x = 4

:

3(4 - 4)^2 = 0

.
For

x = 6

:

2(6 - 4)^2 = 8

.
For

x = 8

:

1(8 - 4)^2 = 16

.
Total sum

\sum f(x - \bar{x})^2 = 16 + 0 + 8 + 16 = 40

.

Variance evaluates the average of these weighted squared deviations.

4

Compute the variance (

\sigma^2

) by dividing the sum of weighted squared deviations by the total frequency.

\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{40}{10} = 4

.

This gives the population variance for the given frequency distribution.

Key Concept

Variance for Frequency Distributions

Estimated Time:1m 30s

Question 144Question →

A bag contains red, blue, and yellow marbles. The theoretical probability of selecting a red marble at random from the bag is $\frac{2}{5}$ . In a probability experiment, a marble is drawn from the bag, its color recorded, and then replaced. This procedure is repeated $250$ times, resulting in a red marble being drawn $115$ times. What is the absolute difference between the observed experimental frequency of red marbles and the theoretical expected frequency?

Show answer & explanation

Answer: 15

Answer

15

The theoretical expected frequency of selecting a red marble over

250

trials is found by multiplying the total trials by the theoretical probability:

250 \times \frac{2}{5} = 100

. The experimental frequency observed was

115

. Taking the absolute difference gives

|115 - 100| = 15

.

Step-by-Step Solution

1

Calculate the theoretical expected frequency of red marble outcomes.

Theoretical expected frequency =

250 \times \frac{2}{5} = 100

.

Expected frequency is determined by multiplying the number of trials (

N = 250

) by the theoretical probability (

P = \frac{2}{5}

).

2

Find the absolute difference between the observed experimental frequency and the theoretical expected frequency.

|115 - 100| = 15

.

The observed experimental frequency is

115

and the theoretical expected frequency is

100

, so the positive difference is

15

.

Key Concept

Experimental Frequency vs. Theoretical Expected Frequency

Estimated Time:1m 30s

Question 145Question →

The table below shows the distribution of test scores obtained by a group of students:

Score	1	2	3	4	5
Frequency	3	$x$	4	2	1

If the mean score of the distribution is $2.5$ , what is the median score?

2

2.5

3

6

Show answer & explanation

Answer: $2$

Answer

The median score is

2

.

First, calculate the missing frequency

x

using the mean formula

\bar{x} = \frac{\sum fx}{\sum f} = \frac{28 + 2x}{10 + x} = 2.5

, which yields

x = 6

. Total number of scores is

N = 16

. The median is the mean of the

8\text{th}

and

9\text{th}

values. Looking at cumulative frequencies, score 1 covers positions 1–3 and score 2 covers positions 4–9. Thus, both the

8\text{th}

and

9\text{th}

scores are 2, giving a median of 2.

Step-by-Step Solution

1

Set up the equation for the mean using

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

.

Total frequency

\sum f = 3 + x + 4 + 2 + 1 = 10 + x

, and total score sum

\sum fx = 1(3) + 2(x) + 3(4) + 4(2) + 5(1) = 28 + 2x

.

The mean of an ungrouped frequency distribution is the sum of all score-frequency products divided by the total frequency.

2

Solve for the unknown frequency

x

.

2.5 = \frac{28 + 2x}{10 + x} \implies 2.5(10 + x) = 28 + 2x \implies 25 + 2.5x = 28 + 2x \implies 0.5x = 3 \implies x = 6

.

Equating the mean expression to

2.5

gives the value of

x

.

3

Find total number of observations

N

and determine the median position.

N = 10 + 6 = 16

. Since

N

is even, median position is between the

8\text{th}

and

9\text{th}

values.

For an even number of observations

N

, the median is the average of the

\frac{N}{2}\text{th}

and

(\frac{N}{2} + 1)\text{th}

items.

4

Locate the

8\text{th}

and

9\text{th}

items using cumulative frequencies.

Cumulative frequencies: Score 1 has 3; Score 2 reaches

3 + 6 = 9

. Both the

8\text{th}

and

9\text{th}

values are

2

. Therefore, median

= \frac{2 + 2}{2} = 2

.

Positions 4 through 9 correspond to the score

2

.

Key Concept

Measures of central tendency for ungrouped frequency distributions

Question 146Question →

The frequency distribution table below displays the examination marks of a group of candidates:

Mark Class	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$9$
$30 - 39$	$16$
$40 - 49$	$12$
$50 - 59$	$8$

If an ogive (cumulative frequency curve) is constructed to represent this dataset, which of the following represents the correct coordinate pair for the point corresponding to the modal class?

(39.5, 30)

(29.5, 30)

(34.5, 30)

(39.5, 16)

Show answer & explanation

Answer: $(39.5, 30)$

Answer

The correct coordinate pair is

(39.5, 30)

.

The modal class is

30 - 39

because it has the highest frequency of

16

. The upper class boundary for this interval is

39.5

. Summing frequencies up to this class gives

5 + 9 + 16 = 30

. Therefore, the plotted point on the ogive must be

(39.5, 30)

.

Step-by-Step Solution

1

Identify the modal class interval from the frequency table.

The class with the highest frequency (

16

) is

30 - 39

.

The modal class is defined as the interval with the maximum frequency.

2

Determine the upper class boundary of the modal class.

Upper class boundary

= 39 + 0.5 = 39.5

.

Cumulative frequency (ogive) curves are plotted using the upper class boundaries on the horizontal axis.

3

Calculate the cumulative frequency up to and including the modal class.

Cumulative frequency

= 5 + 9 + 16 = 30

.

An ogive plots accumulated frequencies up to each boundary.

4

Form the coordinate pair

(x, y) = (\text{Upper Class Boundary}, \text{Cumulative Frequency})

.

(39.5, 30)

.

Points on an ogive take the form

(\text{Upper Boundary}, \text{Cumulative Frequency})

.

Key Concept

Plotting Cumulative Frequency Curves (Ogives)

Question 147Question →

In an archery competition, two archers, Kemi and Chidi, attempt to hit a target independently. The probability that Kemi hits the target is $\frac{1}{3}$ and the probability that Chidi hits the target is $\frac{2}{5}$ . What is the probability that at least one of them hits the target?

\frac{3}{5}

\frac{11}{15}

\frac{2}{15}

\frac{13}{15}

Show answer & explanation

Answer: $\frac{3}{5}$

Answer

The probability that at least one archer hits the target is

\frac{3}{5}

.

For independent events, the probability of both events occurring is

P(K \cap C) = P(K) \times P(C) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}

. Applying the addition law of probability

P(K \cup C) = P(K) + P(C) - P(K \cap C)

gives

\frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{9}{15} = \frac{3}{5}

.

Step-by-Step Solution

1

Identify event probabilities and independence

P(\text{Kemi}) = \frac{1}{3}

and

P(\text{Chidi}) = \frac{2}{5}

The problem specifies that the two attempts are independent events.

2

Calculate the joint probability (intersection) using the multiplication law for independent events

P(\text{Kemi} \cap \text{Chidi}) = P(\text{Kemi}) \times P(\text{Chidi}) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}

For independent events, the probability of both occurring together is the product of their individual probabilities.

3

Apply the addition law of probability to find the union

P(\text{Kemi} \cup \text{Chidi}) = \frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{5 + 6 - 2}{15} = \frac{9}{15} = \frac{3}{5}

The addition law states

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

to prevent double counting the intersection.

Key Concept

Compound Events, Multiplication Law for Independent Events, and Addition Law of Probability

Estimated Time:1m 30s

Question 148Question →

In an electrical power station, two independent backup transformers, $T_1$ and $T_2$ , operate simultaneously during power surges. The probability that $T_1$ fails during a surge is $0.15$ , and the probability that $T_2$ fails during the same surge is $0.20$ . What is the probability that at least one transformer functions correctly during a power surge?

Show answer & explanation

Answer: 0.97

Answer

0.97

The failure probabilities are

P(F_1) = 0.15

and

P(F_2) = 0.20

. By the multiplication law for independent events, the probability of both failing is

P(F_1 \cap F_2) = 0.15 \times 0.20 = 0.03

. Using the complement rule, the probability of at least one functioning correctly is

1 - P(F_1 \cap F_2) = 1 - 0.03 = 0.97

.

Step-by-Step Solution

1

Determine the joint probability of both transformers failing.

P(F_1 \cap F_2) = 0.15 \times 0.20 = 0.03

Since the operational failures of the two transformers are independent events, the probability of both failing together is the product of their individual failure probabilities.

2

Calculate the probability that at least one transformer functions correctly.

P(\text{at least one functions}) = 1 - 0.03 = 0.97

The event that at least one transformer functions is the complement of the event that both transformers fail simultaneously.

Key Concept

Independent Compound Events and the Complement Rule

Question 149Question →

A sports analyst recorded the number of points scored by a basketball player across five consecutive games as $12$ , $14$ , $15$ , $16$ , and $18$ . What is the mean deviation of these scores?

1.6

0.0

2.0

4.0

Show answer & explanation

Answer: $1.6$

Answer

The mean deviation of the scores is

1.6

.

The arithmetic mean of the five scores is

15

. The sum of the absolute differences between each score and the mean is

|12-15| + |14-15| + |15-15| + |16-15| + |18-15| = 3 + 1 + 0 + 1 + 3 = 8

. Dividing this total by the number of scores (

5

) gives

1.6

, which correctly represents the mean deviation.

Step-by-Step Solution

1

Calculate the mean (

\bar{x}

) of the data set.

\bar{x} = \frac{12 + 14 + 15 + 16 + 18}{5} = \frac{75}{5} = 15

The mean deviation measures dispersion relative to the arithmetic mean.

2

Compute the absolute deviations

|x - \bar{x}|

for each data point.

|12 - 15| = 3, \quad |14 - 15| = 1, \quad |15 - 15| = 0, \quad |16 - 15| = 1, \quad |18 - 15| = 3

Absolute values prevent positive and negative deviations from cancelling each other out.

3

Calculate the mean of the absolute deviations.

\text{Mean Deviation} = \frac{\sum |x - \bar{x}|}{n} = \frac{3 + 1 + 0 + 1 + 3}{5} = \frac{8}{5} = 1.6

Dividing the sum of absolute deviations by the number of observations yields the mean deviation.

Key Concept

Mean Deviation of Ungrouped Data

Estimated Time:1m 30s

Question 150Question →

In a quality control experiment, a technician tested $200$ light bulbs and found that $36$ were defective. According to the manufacturer's specification, the theoretical probability of a bulb being defective is $\frac{3}{20}$ . What is the difference between the theoretical probability and the experimental probability of selecting a non-defective light bulb?

\frac{3}{100}

\frac{67}{100}

\frac{33}{100}

\frac{17}{100}

Show answer & explanation

Answer: $\frac{3}{100}$

Answer

The correct answer is

\frac{3}{100}

(or

0.03

).

The experimental probability of obtaining a non-defective bulb is

1 - \frac{36}{200} = 0.82

. The theoretical probability of a non-defective bulb is

1 - 0.15 = 0.85

. Subtracting the two gives

0.85 - 0.82 = 0.03 = \frac{3}{100}

.

Step-by-Step Solution

1

Calculate the experimental probability of selecting a defective bulb

\text{P}_{\text{exp}}(\text{defective}) = \frac{36}{200} = 0.18

Relative frequency is determined by dividing the observed frequency of defective bulbs by the total number of trials.

2

Find the experimental probability of selecting a non-defective bulb

\text{P}_{\text{exp}}(\text{non-defective}) = 1 - 0.18 = 0.82

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

1

.

3

Find the theoretical probability of selecting a non-defective bulb

\text{P}_{\text{theo}}(\text{non-defective}) = 1 - \frac{3}{20} = 1 - 0.15 = 0.85

Subtracting the given theoretical probability of a defective bulb from

1

yields the theoretical probability of a non-defective bulb.

4

Calculate the difference between the two probabilities of selecting a non-defective bulb

0.85 - 0.82 = 0.03 = \frac{3}{100}

Subtract the experimental probability from the theoretical probability.

Key Concept

Experimental probability measures relative frequency from observed outcomes, while theoretical probability is based on expected mathematical outcomes. Complementary probabilities fulfill

P(E') = 1 - P(E)

.

Estimated Time:1m 30s

Question 151Question →

A international conference delegation of $6$ members is to be selected from $5$ diplomats and $4$ translators. In how many distinct ways can the delegation be formed if it must include at least $4$ diplomats?

Show answer & explanation

Answer: 34

Answer

The total number of distinct ways to form the delegation is 34.

To form a delegation of 6 with at least 4 diplomats out of 5 diplomats and 4 translators, we evaluate two mutually exclusive cases: selecting 4 diplomats and 2 translators (5C4 * 4C2 = 30 ways) and selecting 5 diplomats and 1 translator (5C5 * 4C1 = 4 ways). Summing these gives 34 distinct ways.

Step-by-Step Solution

1

Determine the valid combinations of diplomats and translators.

Two valid cases exist: (4 diplomats, 2 translators) and (5 diplomats, 1 translator).

The total delegation size is 6 and it must contain at least 4 diplomats out of the 5 available.

2

Compute the selection ways for Case 1 (4 diplomats and 2 translators).

5 * 6 = 30 ways

Selecting 4 diplomats out of 5 is 5C4 = 5 ways, and selecting 2 translators out of 4 is 4C2 = 6 ways.

3

Compute the selection ways for Case 2 (5 diplomats and 1 translator).

1 * 4 = 4 ways

Selecting 5 diplomats out of 5 is 5C5 = 1 way, and selecting 1 translator out of 4 is 4C1 = 4 ways.

4

Sum the combinations from both mutually exclusive cases.

30 + 4 = 34 ways

By the addition principle of counting, mutually exclusive scenarios are added together.

Key Concept

Combinations with constraints and mutually exclusive cases

Question 152Question →

In a manufacturing plant, two independent automated assembly units, $U_1$ and $U_2$ , undergo safety inspection. The probability that unit $U_1$ fails the inspection is $\frac{1}{5}$ , while the probability that unit $U_2$ fails the inspection is $\frac{1}{4}$ . What is the probability that at least one of the two units fails the inspection? Express your answer as a decimal.

Show answer & explanation

Answer: 0.4

Answer

0.4

The probability that at least one unit fails is determined using the addition law for non-mutually exclusive independent events:

P(U_1 \cup U_2) = P(U_1) + P(U_2) - P(U_1 \cap U_2) = 0.20 + 0.25 - (0.20 \times 0.25) = 0.45 - 0.05 = 0.40

. Alternatively, using the complement rule gives

1 - P(\text{neither fails}) = 1 - (1 - 0.20)(1 - 0.25) = 1 - 0.60 = 0.40

.

Step-by-Step Solution

1

Identify the individual failure probabilities

P(U_1) = 0.20

and

P(U_2) = 0.25

These values represent the single event failure probabilities given in the problem statement.

2

Calculate the probability that neither unit fails

P(U_1') \times P(U_2') = (1 - 0.20) \times (1 - 0.25) = 0.80 \times 0.75 = 0.60

Because the units operate independently, their complement events (passing inspection) are also independent.

3

Determine the probability of at least one unit failing

1 - 0.60 = 0.40

The event that at least one unit fails is the complementary event of neither unit failing.

Key Concept

Addition and Multiplication Laws of Probability for Independent Compound Events

Question 153Question →

For a science exhibition, a student coordinator needs to choose $4$ project displays from a pool of $6$ chemistry projects and $4$ physics projects. If the selection must include at least $2$ chemistry projects, in how many distinct ways can the projects be selected?

170

185

210

1,200

Show answer & explanation

Answer: $185$

Answer

The total number of distinct ways to select the projects is

185

.

The correct total number of ways is

185

. The condition requiring at least

2

chemistry projects splits the total selection into three mutually exclusive scenarios: selecting

2

chemistry and

2

physics projects (

90

ways),

3

chemistry and

1

physics project (

80

ways), or

4

chemistry projects (

15

ways). Adding these together yields

90 + 80 + 15 = 185

.

Step-by-Step Solution

1

Identify the distinct valid cases for selecting

4

projects with at least

2

chemistry projects.

Case 1:

2

chemistry and

2

physics; Case 2:

3

chemistry and

1

physics; Case 3:

4

chemistry and

0

physics.

The phrase 'at least 2 chemistry projects' means the number of chemistry projects can be

2

,

3

, or

4

.

2

Calculate the combinations for Case 1 (

2

chemistry,

2

physics).

$\binom{6}{2} \times \binom{4}{2} = 15 \times 6 = 90$ ways.

We select

2

projects out of

6

chemistry projects and

2

out of

4

physics projects using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

3

Calculate the combinations for Case 2 (

3

chemistry,

1

physics).

$\binom{6}{3} \times \binom{4}{1} = 20 \times 4 = 80$ ways.

We select

3

projects out of

6

chemistry projects and

1

out of

4

physics projects.

4

Calculate the combinations for Case 3 (

4

chemistry,

0

physics).

$\binom{6}{4} \times \binom{4}{0} = 15 \times 1 = 15$ ways.

We select

4

projects out of

6

chemistry projects and

0

out of

4

physics projects.

5

Sum the number of ways from all mutually exclusive cases.

$90 + 80 + 15 = 185$ ways.

By the addition principle of counting, the total number of ways is the sum of the ways from each independent case.

Key Concept

Combinations with Constraints (At Least / At Most)

Question 154Question →

A candidate takes a two-stage driving examination consisting of a theoretical test followed by a practical test. The probability that the candidate passes the theoretical test is $0.75$ . If the candidate passes the theoretical test, the probability of passing the practical test is $0.80$ . However, if the candidate fails the theoretical test, the probability of passing the practical test is $0.20$ . What is the probability that the candidate passes exactly one of the two tests?

Show answer & explanation

Answer: 0.2

Answer

The probability that the candidate passes exactly one of the two tests is 0.2.

To pass exactly one test, the candidate must either pass theory and fail practical (

0.75 \times 0.20 = 0.15

) or fail theory and pass practical (

0.25 \times 0.20 = 0.05

). Since these two outcomes cannot happen at the same time, their probabilities are added:

0.15 + 0.05 = 0.20

.

Step-by-Step Solution

1

Calculate the probability of passing the theoretical test and failing the practical test.

P(\text{Theory} \cap \text{Practical}') = 0.75 \times (1 - 0.80) = 0.15

The probability of failing practical given passing theory is

1 - 0.80 = 0.20

.

2

Calculate the probability of failing the theoretical test and passing the practical test.

P(\text{Theory}' \cap \text{Practical}) = (1 - 0.75) \times 0.20 = 0.05

The probability of failing theory is

1 - 0.75 = 0.25

.

3

Sum the probabilities of the two mutually exclusive outcomes.

0.15 + 0.05 = 0.20

Passing exactly one test is the union of two disjoint compound events.

Key Concept

Compound probability laws and conditional independence structure in sequential events

Estimated Time:1m 30s

Question 155Question →

Two independent events $A$ and $B$ have probabilities $P(A) = \frac{1}{4}$ and $P(B) = \frac{2}{3}$ . What is the probability that at least one of the events occurs, represented by $P(A \cup B)$ ?

\frac{3}{4}

\frac{11}{12}

\frac{1}{6}

\frac{5}{6}

Show answer & explanation

Answer: $\frac{3}{4}$

Answer

The probability that at least one of the events occurs is

\frac{3}{4}

.

The correct answer

\frac{3}{4}

is obtained by applying the multiplication law for independent events

P(A \cap B) = P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{6}

, and substituting into the general addition law

P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{2}{3} - \frac{1}{6} = \frac{9}{12} = \frac{3}{4}

.

Step-by-Step Solution

1

Calculate the intersection probability

P(A \cap B)

for independent events

P(A \cap B) = P(A) \times P(B) = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}

For independent events, the probability of both occurring together is the product of their individual probabilities.

2

Apply the general addition law of probability

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

The addition law accounts for double-counting the intersection when finding the union of two events.

3

Substitute the values and simplify the fraction

P(A \cup B) = \frac{1}{4} + \frac{2}{3} - \frac{1}{6} = \frac{3}{12} + \frac{8}{12} - \frac{2}{12} = \frac{9}{12} = \frac{3}{4}

Finding a common denominator of 12 allows exact fractional addition and simplification.

Key Concept

Probability Laws for Independent and Compound Events

Question 156Question →

In a meteorological study conducted over $120$ days in a coastal town, rain was observed on $45$ days. According to a theoretical weather model, the probability of rain on any given day during this period is $\frac{1}{3}$ . Calculate the absolute difference between the observed number of rainy days and the expected number of rainy days predicted by the theoretical model.

Show answer & explanation

Answer: 5

Answer

The absolute difference between the observed and expected number of rainy days is 5 days.

The expected number of rainy days based on theoretical probability is

\frac{1}{3} \times 120 = 40

days. The experimental observation recorded

45

rainy days. The absolute difference between the observed and expected number of rainy days is

|45 - 40| = 5

days.

Step-by-Step Solution

1

Calculate the theoretical expected number of rainy days

Expected rainy days = 40

Multiply the total number of trial days (120) by the theoretical probability of rain (1/3).

2

Identify the experimental (observed) number of rainy days

Observed rainy days = 45

The problem states that rain occurred on 45 days during the 120-day period.

3

Compute the positive absolute difference

Absolute difference = 5

Subtract the theoretical expected count (40) from the experimental observed count (45).

Key Concept

Difference Between Experimental Frequency and Theoretical Expectation

Question 157Question →

A farmer plants two types of crops, maize and cassava, in separate fields. The probability that the maize crop yields a successful harvest is $\frac{3}{8}$ , and the probability that the cassava crop yields a successful harvest is $\frac{1}{2}$ . Assuming that the harvest outcomes of the two crops are independent, what is the probability that at least one of the crops yields a successful harvest?

\frac{3}{16}

\frac{11}{16}

\frac{7}{8}

\frac{5}{16}

Show answer & explanation

Answer: $\frac{11}{16}$

Answer

The probability that at least one of the crops yields a successful harvest is

\frac{11}{16}

.

The probability of at least one successful harvest is given by the union of the two events,

P(M \cup C) = P(M) + P(C) - P(M \cap C)

. Since the harvests are independent, the joint probability is

P(M \cap C) = P(M) \times P(C) = \frac{3}{8} \times \frac{1}{2} = \frac{3}{16}

. Substituting these values gives

\frac{3}{8} + \frac{1}{2} - \frac{3}{16} = \frac{6}{16} + \frac{8}{16} - \frac{3}{16} = \frac{11}{16}

.

Step-by-Step Solution

1

Identify given probabilities and event relationship

Let

M

be maize success and

C

be cassava success:

P(M) = \frac{3}{8}

and

P(C) = \frac{1}{2}

. The events are independent.

Establishing the basic event probabilities and independence condition is essential for applying compound probability laws.

2

Calculate the joint probability of both events occurring

P(M \cap C) = P(M) \times P(C) = \frac{3}{8} \times \frac{1}{2} = \frac{3}{16}

.

For independent events, the multiplication law states that the probability of both occurring is the product of their individual probabilities.

3

Apply the addition law of probability for the union of two events

P(M \cup C) = P(M) + P(C) - P(M \cap C) = \frac{3}{8} + \frac{1}{2} - \frac{3}{16} = \frac{6}{16} + \frac{8}{16} - \frac{3}{16} = \frac{11}{16}

.

The probability that at least one crop succeeds corresponds to the union

P(M \cup C)

, which requires subtracting the intersection to prevent double counting.

Key Concept

Addition Law of Probability and Multiplication Law for Independent Events

Question 158Question →

Chidi and Zainab independently attempt to solve a mathematics problem. The probability that Chidi solves it is $\frac{3}{5}$ and the probability that Zainab solves it is $\frac{2}{3}$ . What is the probability that exactly one of them solves the problem?

\frac{2}{5}

\frac{4}{15}

\frac{7}{15}

\frac{13}{15}

Show answer & explanation

Answer: $\frac{7}{15}$

Answer

The probability that exactly one of them solves the problem is

\frac{7}{15}

.

The correct answer is

\frac{7}{15}

. 'Exactly one' means either Chidi solves the problem while Zainab fails (

P(C \cap Z') = \frac{3}{5} \times \frac{1}{3} = \frac{3}{15}

) OR Zainab solves it while Chidi fails (

P(C' \cap Z) = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15}

). Summing these mutually exclusive probabilities gives

\frac{3}{15} + \frac{4}{15} = \frac{7}{15}

.

Step-by-Step Solution

1

Find the complement probabilities of failure for each student.

Probability Chidi fails,

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

. Probability Zainab fails,

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

.

Required to determine individual non-occurrence probabilities.

2

Calculate the joint probability for each mutually exclusive scenario.

Scenario 1 (Chidi solves, Zainab fails):

\frac{3}{5} \times \frac{1}{3} = \frac{3}{15}

. Scenario 2 (Chidi fails, Zainab solves):

\frac{2}{5} \times \frac{2}{3} = \frac{4}{15}

.

Since the events are independent, individual probabilities multiply.

3

Sum the probabilities of the two mutually exclusive outcomes.

Total probability =

\frac{3}{15} + \frac{4}{15} = \frac{7}{15}

.

By the Addition Law of probability for mutually exclusive compound outcomes.

Key Concept

Probability of Compound Independent Events

Statistics and Probability

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