Key Concept: Data Handling

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that a teacher collected the heights of all students in her class to find the average height. This is true because collecting such data is a common method to compute averages. The reason states that representing data graphically helps in better understanding and interpretation of the information. This is also true as graphical representations like bar graphs or pictographs make it easier to analyze and interpret data visually. However, the reason does not directly explain why the teacher collected the heights of the students; it explains a general benefit of data representation. Therefore, both assertions are true, but the reason is not the correct explanation of the assertion.

Key Concept: Bar Graph

c) The value of the data

[Solution Description]

In a bar graph, the height of each bar is proportional to the value it represents. This means that taller bars indicate higher values, while shorter bars indicate lower values.

Key Concept: Data Definition, Bar Graph

c) 8 cm

[Solution Description]

Since the height of the bar is proportional to the number of books, we can set up a proportion to find the height for 16 books. Given that 10 books correspond to 5 cm, the height for 16 books can be found using the formula: $\frac{5 \text{ cm}}{10 \text{ books}} = \frac{x \text{ cm}}{16 \text{ books}}$. Solving for $x$, we get $x = \frac{5 \times 16}{10} = 8$ cm.

Key Concept: Pictograph Interpretation

c) 300 cars

[Solution Description]

In a pictograph, each symbol represents a specific quantity. Here, one symbol stands for 100 cars. For August, there are 3 symbols. Therefore, the total number of cars produced in August is calculated as follows:

$3 \times 100 = 300 \text{ cars}$

Key Concept: Pictograph Interpretation

b) 200

[Solution Description]

To find the number of bicycles produced in June, multiply the number of symbols by the value each symbol represents. Here, each symbol stands for 50 bicycles, and there are 4 symbols:

$4 \times 50 = 200$

Thus, 200 bicycles were produced in June.

Key Concept: Data Collection

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that data collection is important for making informed decisions, which is true because data provides the necessary information to analyze situations accurately. The reason also states that data provides accurate information that helps in analyzing and interpreting situations effectively, which is true and correctly explains why data collection is important. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Data Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that a bar graph uses bars of uniform width, which is correct. The reason explains that the height of each bar is proportional to the value it represents, which is also true. Furthermore, the reason correctly explains why bar graphs are effective for data representation. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Data Interpretation

b) 4

[Solution Description]

To find the median, we first arrange the numbers in ascending order: 2, 3, 4, 5, 6. Since there are an odd number of data points, the median is the middle value, which is 4.

Key Concept: Pie Chart

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

A pie chart is indeed suitable for representing the time spent by a child during a day because it effectively displays how different activities contribute to the total time. The reason provided explains that a pie chart divides the whole into sectors, with each sector's size proportional to the activity it represents. This makes it a clear and accurate choice for visualizing such data. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Pictograph, Bar Graph

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The Assertion claims that a pictograph is more effective than a bar graph for representing data with large numbers. This is true because in a pictograph, one symbol can represent a large quantity (e.g., 1 symbol = 100 cars), which simplifies visualization. The Reason states that in a pictograph, one symbol can represent a large quantity, making it easier to interpret data involving large numbers. This is also true. Moreover, the Reason correctly explains why the Assertion is true, as the ability of a single symbol to represent a large quantity is what makes pictographs more effective in such cases.

Key Concept: Pie Chart

c) 50%

[Solution Description]

In a pie chart, the entire circle represents 360 degrees, which corresponds to 100% of the data. The sector for chocolate is 180 degrees. To find the percentage, divide the sector angle by the total degrees and multiply by 100: $\frac{180}{360} \times 100 = 50\%$. Thus, 50% of students prefer chocolate.

Key Concept: Pictograph, Bar Graph

c) 8 units

[Solution Description]

To solve this, first, calculate the number of cars sold in September using the pictograph. Since one symbol represents 100 cars, 4 symbols represent $4 \times 100 = 400$ cars. In the bar graph, each unit on the y-axis represents 50 cars. Therefore, the height of the bar for September would be $\frac{400}{50} = 8$ units.

Key Concept: Pictograph

c) 300

[Solution Description]

For August, the number of symbols is 3, and each symbol represents 100 cars. Thus, the total number of cars produced in August is $3 \times 100 = 300$ cars.

Key Concept: Pictograph

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that in a pictograph, one symbol represents 100 cars. This is true as per the syllabus example provided where one symbol stands for 100 cars. The reason states that pictographs use symbols to represent data in a simple and visually appealing manner. This is also true as pictographs are indeed used to simplify data representation using symbols. Moreover, the reason correctly explains why one symbol represents 100 cars in the given context. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Double Bar Graph Comparison

a) Mathematics

[Solution Description]

To determine the subject where the performance of both schools is at par, we need to compare the heights of the bars for each subject. If the heights are equal for a particular subject, then the performance is at par. Suppose the data shows that in Mathematics, both schools have the same number of students scoring above 80% marks.

Key Concept: Bar Graph, Information Conveyance

d) The information conveyed by the bar graph remains unchanged.

[Solution Description]

Swapping the positions of the bars representing Class 1 and Class 3 does not change the heights of the bars or the data they represent. The information conveyed by the bar graph remains the same; only the visual arrangement of the bars is altered. Therefore, none of the numerical data or comparisons between the classes change.

Key Concept: Double Bar Graph, Information Conveyed

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

A double bar graph is used to compare two sets of data across the same categories. The height of each bar represents the quantity for that category, and the relative positions of the bars help in comparing the two datasets. If the position of any bar is changed, the visual comparison between the two datasets would be disrupted, thereby altering the information conveyed by the graph. Thus, the Assertion is true. The Reason states that the height of the bars represents the quantity for each category, which is correct, but it also mentions that the relative positions are crucial for comparison, which is also true. Therefore, the Reason correctly explains the Assertion.

Key Concept: Double Bar Graph Information

c) To compare two sets of data

[Solution Description] A double bar graph is used to compare two sets of data simultaneously. It helps in understanding the relationship or difference between the two datasets.

Key Concept: Bar Graph

c) 60

[Solution Description]

Each centimeter represents 5 apples. For a bar height of 12 cm, the total number of apples sold is calculated as follows:

$12 \times 5 = 60$

Therefore, 60 apples were sold.

Key Concept: Pictograph

b) 300

[Solution Description]

Each symbol represents 100 cars. For 3 symbols, the total number of cars is calculated as follows:

$3 \times 100 = 300$

Therefore, 300 cars were produced in July.

Key Concept: Double Bar Graph

b) To compare two sets of data

[Solution Description]

A double bar graph is used to compare two sets of data simultaneously. It helps in understanding the relationship or difference between the two datasets easily.

Key Concept: Comparison of data using bar graphs

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The Assertion states that a bar graph is more effective than a pie chart for comparing sales of different products over a year. This is true because bar graphs are particularly useful for comparing discrete categories, such as different products, by using rectangular bars whose lengths are proportional to the values they represent. The Reason explains that bar graphs provide a clearer comparison of discrete categories due to their visual structure, which supports the Assertion. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Circle Graph or Pie Chart

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] A pie chart is a circular statistical graphic that is divided into sectors to illustrate numerical proportions. The assertion states that a pie chart is used to represent data as parts of a whole, which is true. The reason explains that in a pie chart, the central angle of each sector is proportional to the quantity it represents, which is also true and correctly explains why the pie chart is used to represent data as parts of a whole. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Pie Chart - Central Angle Calculation

c) 90°

[Solution Description]

To find the central angle for Vanilla, first convert the percentage to a fraction. The percentage of students who prefer Vanilla is 25%, which is $\frac{25}{100} = \frac{1}{4}$. The total angle at the center of a circle is 360°. So, the central angle for Vanilla is calculated as:

$\frac{1}{4} \times 360° = 90°$

Therefore, the central angle of the sector representing Vanilla is $90°$.

Key Concept: Percentage to Central Angle Conversion

b) 72°

[Solution Description]

To find the central angle for the rent sector, we use the formula:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360°$

Given that 20% of the total expenditure is on rent, we substitute the value into the formula:

$\text{Central Angle} = \left(\frac{20}{100}\right) \times 360° = 72°$

Therefore, the central angle for the rent sector is 72°.

Key Concept: Central Angle Calculation

b) 108°

[Solution Description]

To find the central angle for the strawberry sector, we use the formula:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360°$

Given that 30% of students prefer strawberry flavour, we substitute the value into the formula:

$\text{Central Angle} = \left(\frac{30}{100}\right) \times 360° = 108°$

Therefore, the central angle for the strawberry sector is 108°.

Key Concept: Pie Chart Central Angle Calculation

b) 144 degrees

[Solution Description]

To find the central angle for the sector representing apples, we use the formula:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360^\circ$

For apples:

$\text{Central Angle} = \left(\frac{40}{100}\right) \times 360^\circ = 0.4 \times 360^\circ = 144^\circ$

Thus, the central angle for apples is $144^\circ$.

Key Concept: Pie Chart Percentage Calculation

b) 20%

[Solution Description]

To find the percentage represented by the sector, we use the formula:

$\text{Percentage} = \left(\frac{\text{Sector Angle}}{360^\circ}\right) \times 100$

For the given sector:

$\text{Percentage} = \left(\frac{72^\circ}{360^\circ}\right) \times 100 = 0.2 \times 100 = 20\%$

Thus, the sector represents 20% of the whole.

Key Concept: Pie Chart Representation

b) $108^\circ$

[Solution Description]

To find the central angle for the sector representing "School", we use the formula:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360^\circ$

Given that the percentage for "School" is $30\%$, we substitute into the formula:

$\text{Central Angle} = \left(\frac{30}{100}\right) \times 360^\circ = 0.3 \times 360^\circ = 108^\circ$

Therefore, the central angle for "School" is $108^\circ$

Key Concept: Pie Chart Representation

b) $\frac{1}{4}$

[Solution Description]

To find the fraction of the circle representing "Comedy", we divide the percentage by \$100.

$\text{Fraction} = \frac{25}{100} = \frac{1}{4}$

Therefore, the fraction representing "Comedy" is $\frac{1}{4}$

Key Concept: Total Angle Verification

b) 70°

[Solution Description]

The total angle in a circle is 360°. Subtract the given angles:

$360° - 90° - 60° = 210°$

Since the remaining sectors have equal central angles, divide 210° by the number of remaining sectors (let's assume there are 3 remaining sectors):

$\frac{210°}{3} = 70°$

Therefore, the central angle for one of the remaining sectors is 70°.

Key Concept: Interpreting Pie Charts

b) \$500

[Solution Description] The central angle for food is $90°$, which corresponds to $\frac{90°}{360°} = \frac{1}{4}$ of the total expenditure. To find the expenditure on food, we calculate:

$\frac{1}{4} \times \$2000 = \$500$

Therefore, the family spends \$500 on food.

Key Concept: Interpretation of Pie Chart

c) $76.5^\circ$

[Solution Description]

To find the angle for the year 2006, use the formula $\text{angle} = (\text{production in 2006} / \text{total production}) \times 360^\circ$. So, $(85 / 400) \times 360^\circ = 76.5^\circ$.

Key Concept: Pie Chart Interpretation

b) $\frac{1}{3}$

[Solution Description]

The total angle in a circle is $360^\circ$. The angle for "Sleep" is given as $120^\circ$. To find the fraction, divide the angle for "Sleep" by the total angle: $\frac{120^\circ}{360^\circ} = \frac{1}{3}$. So, the child spends $\frac{1}{3}$ of the day sleeping.

Key Concept: Pie Chart Representation, Data Interpretation

c) 6 hours

[Solution Description]

To find the time spent on Leisure, first determine the fraction of the circle that the Leisure sector occupies. Let $\theta_L$ be the central angle for the Leisure sector and $T$ be the total degrees in a circle.

Given:

$\theta_L = 90^\circ, \quad T = 360^\circ$

The fraction of time spent on Leisure is:

$\frac{\theta_L}{T} = \frac{90^\circ}{360^\circ} = \frac{1}{4}$

Since the total time in a day is 24 hours, the time spent on Leisure is:

$\text{Leisure Time} = \frac{1}{4} \times 24 = 6 \text{ hours}$

Therefore, the time spent on Leisure is 6 hours.

Key Concept: Pie Chart Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The central angle of a sector in a pie chart is calculated as a fraction of 360°. If 50% of students prefer chocolate flavour, this can be represented as $\frac{50}{100} = \frac{1}{2}$. Therefore, the central angle for the chocolate sector is $\frac{1}{2} \times 360° = 180°$. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Circle Graph, Expenditure of a Family

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine the amount spent on education, we use the formula: Amount = (Percentage of expenditure / 100) $\times$ Total expenditure. Given that the percentage of expenditure on education is 20% and the total expenditure is \$50,000, we substitute these values into the formula: Amount = (20 / 100) $\times$ \$50,000 = \$10,000. Therefore, the assertion is correct. The reason correctly explains that the percentage of expenditure on an item in a pie chart can be used to calculate the exact amount spent on that item when the total expenditure is known. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Proportional Expenditure

b) \$750

[Solution Description]

First, determine the total income using the expenditure on groceries:

$\text{Total income} = \frac{1250}{25} \times 100 = 5000$

Now, calculate the expenditure on utilities:

$\text{Expenditure on utilities} = \frac{15}{100} \times 5000 = 750$

Key Concept: Pie Chart Central Angle Calculation

c) 80°

[Solution Description]

To find the central angle for "Cakes and Pastries," we first calculate the fraction of total sales it represents. The total sales are \$720. The sales for "Cakes and Pastries" are \$160. The fraction is $\frac{160}{720} = \frac{2}{9}$. The central angle is then calculated by multiplying this fraction by 360°: $\frac{2}{9} \times 360° = 80°$.

Key Concept: Pie Chart Central Angle Calculation

c) \$120

[Solution Description]

To find the sale amount for "Biscuits," we use the formula for the central angle: $\text{Central Angle} = \left(\frac{\text{Sales}}{\text{Total Sales}}\right) \times 360°$. We know the central angle is 60° and the total sales are \$720. Rearranging the formula to solve for sales: $\text{Sales} = \frac{60°}{360°} \times 720 = \frac{1}{6} \times 720 = \$120$.

Key Concept: Angle Comparison

c) 2:1

[Solution Description]

Given the central angle for chocolate flavor is 180° and for vanilla flavor is 90°, the ratio of the central angles of chocolate to vanilla is $\frac{180°}{90°} = 2:1$.

Key Concept: Fraction of Circle

b) $\frac{1}{2}$

[Solution Description]

The percentage of students preferring Chocolate is 50%. This can be written as a fraction of the whole circle. Since the total angle of a circle is 360°, the fraction is $\frac{50}{100}$ or $\frac{1}{2}$.

Key Concept: Pie Chart Central Angle Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To find the central angle for the 'Chocolate' flavour, we use the formula:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360^\circ$

For 'Chocolate', the percentage is $50\%$. Plugging in the values:

$\text{Central Angle} = \left(\frac{50}{100}\right) \times 360^\circ = 0.5 \times 360^\circ = 180^\circ$

Thus, the assertion that the central angle for 'Chocolate' is $180^\circ$ is correct. The reason given correctly explains how this value is derived from the percentage and the total angle of a circle.

Key Concept: Probability of an Event

c) $\frac{1}{6}$

[Solution Description]

When a die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. All outcomes are equally likely. The probability of getting the number 5 is calculated using the formula:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Here, the number of favorable outcomes for getting the number 5 is 1, and the total number of outcomes is 6. Therefore,

$\text{Probability of getting the number 5} = \frac{1}{6}$

Thus, the probability of getting the number 5 when a die is thrown once is $\frac{1}{6}$.

Key Concept: Equally likely outcomes, Events

b) $\frac{2}{15}$

[Solution Description]

The bag contains a total of 10 balls (4 red and 6 blue). We are to draw two balls without replacement and find the probability that both are red.

The probability of drawing the first red ball is:

$P(\text{First red}) = \frac{4}{10} = \frac{2}{5}$

After drawing one red ball, there are now 3 red balls and 6 blue balls left, totaling 9 balls. The probability of drawing a second red ball is:

$P(\text{Second red}) = \frac{3}{9} = \frac{1}{3}$

The combined probability of both events occurring is:

$P(\text{Both red}) = P(\text{First red}) \times P(\text{Second red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$

Key Concept: Equally Likely Outcomes

c) $\frac{1}{2}$

[Solution Description]

When a fair coin is tossed, there are two possible outcomes: Head or Tail. Both outcomes are equally likely. The probability of getting a head is calculated using the formula:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Here, the number of favorable outcomes for getting a head is 1, and the total number of outcomes is 2. Therefore,

$\text{Probability of getting a head} = \frac{1}{2}$

Thus, the probability of getting a head when a fair coin is tossed once is $\frac{1}{2}$.

Key Concept: Chance in Daily Life

b) $\frac{1}{5}$

[Solution Description]

There are 5 chapters in total, and the student prepared 4 chapters. The number of chapters left unprepared is 1. The probability of selecting a question from the unprepared chapter is the ratio of the number of unprepared chapters to the total number of chapters:

$P(\text{Unprepared Chapter}) = \frac{1}{5}$

Thus, the probability is $\frac{1}{5}$.

Key Concept: Probability of an Event

b) $\frac{3}{8}$

[Solution Description]

The total number of balls in the bag is 5 (red) + 3 (green) = 8. The number of favorable outcomes for drawing a green ball is 3. Therefore, the probability of getting a green ball is calculated as $\frac{3}{8}$.

Key Concept: Probability and Chance

b) $\frac{3}{10}$

[Solution Description]

The probability of the train being on time is $\frac{7}{10}$. The probability of the train being late is the complement of it being on time:

$P(\text{Late}) = 1 - P(\text{On Time}) = 1 - \frac{7}{10} = \frac{3}{10}$

Hence, the probability that the train will be late is $\frac{3}{10}$.

Key Concept: Equally Likely Outcomes

b) $\frac{1}{3}$

[Solution Description]

Total possible outcomes = 6 (numbers 1 to 6). Favorable outcomes = 2 (numbers 5 and 6). Probability = $\frac{2}{6} = \frac{1}{3}$.

Key Concept: Probability and Random Experiment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of getting a head when tossing a fair coin is $\frac{1}{2}$. This is true because a fair coin has two equally likely outcomes: Head or Tail. The reason explains that a fair coin has two equally likely outcomes, which is correct. Since the reason correctly explains why the assertion is true, both are true, and the reason is the correct explanation of the assertion.

Key Concept: Probability, Equally Likely Outcomes

b) $\frac{5}{36}$

[Solution Description]

First, find the total number of possible outcomes when a die is rolled twice. Each roll has 6 possible outcomes, so two rolls have $6 \times 6 = 36$ possible outcomes. Next, find the number of favorable outcomes where the sum of the two numbers is 8. The pairs that give a sum of 8 are: (2,6), (3,5), (4,4), (5,3), and (6,2). There are 5 such pairs. Therefore, the probability of getting a sum of 8 when rolling a die twice is $\frac{5}{36}$.

Key Concept: Equally Likely Outcomes

b) $\frac{1}{2}$

[Solution Description]

A fair coin has two equally likely outcomes: Head and Tail. The probability of getting a head is calculated as:

$\text{Probability of Head} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{2}$

Key Concept: Equally Likely Outcomes, Probability in Real-Life Context

b) $\frac{2}{5}$

[Solution Description]

There are two scenarios where the contestant wins exactly one \$100 prize:

1. Winning on the first try and losing on the second try.

2. Losing on the first try and winning on the second try.

The probability of scenario 1: $\frac{1}{5} \times \frac{4}{4} = \frac{4}{20}$.

The probability of scenario 2: $\frac{4}{5} \times \frac{1}{4} = \frac{4}{20}$.

Adding these probabilities together: $\frac{4}{20} + \frac{4}{20} = \frac{8}{20} = \frac{2}{5}$.

Key Concept: Equally Likely Outcomes, Probability of Multiple Events

b) $\frac{2}{15}$

[Solution Description]

First, determine the total number of balls: $4 + 6 = 10$. The probability of drawing a red ball on the first draw is $\frac{4}{10}$. After one red ball is drawn, there are 3 red balls left out of 9 total balls. So, the probability of drawing a second red ball is $\frac{3}{9}$. Multiply these probabilities to find the combined probability: $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$.

Key Concept: Equally Likely Outcomes

b) It remains constant at $\frac{1}{2}$

[Solution Description]

As the number of tosses increases, the ratio of heads to tails approaches 1:1, which implies that the probability of getting a head in a single toss is $\frac{1}{2}$.

Key Concept: Probability of Events

b) 0.5

[Solution Description]

Since there are two equally likely outcomes when a coin is tossed, the probability of getting a tail is $\frac{1}{2}$.

Key Concept: Conditional Probability

b) $\frac{1}{3}$

[Solution Description]

The possible outcomes when tossing a coin twice are HH, HT, TH, TT. The condition is that at least one head appears, so we exclude TT. The favorable outcome for both tosses being heads is HH. Thus, the probability is $\frac{1}{3}$.

Key Concept: Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of getting an even number on a fair die is $\frac{1}{2}$. On a fair die, the possible even numbers are 2, 4, and 6. Since there are 6 equally likely outcomes, the probability of getting an even number is $\frac{3}{6}$, which simplifies to $\frac{1}{2}$. The reason correctly explains that an even number can be 2, 4, or 6, and there are six equally likely outcomes. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Probability of an Event

b) $\frac{1}{2}$

[Solution Description]

An even number on a die can be 2, 4, or 6. So, there are 3 favorable outcomes out of the total 6 possible outcomes. The probability is calculated as:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$

Therefore, the correct answer is $\frac{1}{2}$.

Key Concept: Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

When rolling a fair die, each of the six outcomes (1, 2, 3, 4, 5, 6) is equally likely, with a probability of $\frac{1}{6}$. This probability does not change based on who is rolling the die or the results of previous rolls. Both the assertion and reason are true, and the reason correctly explains why the assertion is true.

Key Concept: List of Possible Outcomes

b) Green, Blue, Red

[Solution Description]

The possible outcomes when spinning the wheel are the sectors at which the pointer stops. In this case, the possible outcomes are Green, Blue, and Red.

Key Concept: Probability, Outcomes

c) $\frac{3}{4}$

[Solution Description]

The total number of sectors is 8. The number of favorable outcomes (green or red) is $3 + 3 = 6$. Therefore, the probability is calculated as:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{8} = \frac{3}{4}$

Key Concept: Probability of Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the probability of getting a green sector, we use the formula for probability:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Here, the number of favorable outcomes is the number of green sectors, which is 3. The total number of possible outcomes is the total number of sectors, which is 5. Therefore, the probability of getting a green sector is:

$\frac{3}{5}$

The assertion correctly calculates the probability, and the reason provides the correct explanation by stating the total number of sectors and the number of green sectors.

Key Concept: Probability as a Fraction, Real-life Application

b) $\frac{1}{2}$

[Solution Description]

Prime numbers between 1 and 8 are 2, 3, 5, and 7. There are 4 prime numbers and 8 total sectors. So, the probability of landing on a prime number is $\frac{4}{8} = \frac{1}{2}$.

Key Concept: Probability as a Fraction, Multi-step Solution

b) $\frac{1}{3}$

[Solution Description]

First, calculate the total number of balls in the bag: $6 + 4 = 10$. The probability of drawing the first green ball is $\frac{6}{10} = \frac{3}{5}$. After drawing one green ball, there are 5 green balls left out of 9 total balls. Now, the probability of drawing the second green ball is $\frac{5}{9}$. Multiply both probabilities to get the combined probability: $\frac{3}{5} \times \frac{5}{9} = \frac{15}{45} = \frac{1}{3}$.

Key Concept: Probability as a Fraction

b) $\frac{1}{6}$

[Solution Description]

A die has six faces, each with a distinct number from 1 to 6. When the die is rolled once, there are six equally likely outcomes. To find the probability of getting the number 3, we note that there is only one favorable outcome (the face with the number 3) out of six possible outcomes. Therefore, the probability of getting the number 3 is $\frac{1}{6}$.

Key Concept: Events in Probability, Outcomes as events

b) $\frac{1}{6}$

[Solution Description]

The total number of possible outcomes when rolling a die twice is $6 \times 6 = 36$. The favorable outcomes where the sum is 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such outcomes. Therefore, the probability is $\frac{6}{36} = \frac{1}{6}$.

Key Concept: Chance and probability related to real life, Probability

b) $\frac{3}{10}$

[Solution Description]

The total number of people surveyed is 100. Out of these, 70 prefer tea, so the number of people who prefer coffee is $100 - 70 = 30$. Therefore, the probability is $\frac{30}{100} = \frac{3}{10}$.

Key Concept: Events in Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that getting an even number when a die is thrown is an event. This is true because an event can be a single outcome (like getting a 2) or a collection of outcomes (like getting 2, 4, or 6). The reason states that an event is defined as one or more outcomes of an experiment, which is also true and correctly explains why getting an even number is an event.

Key Concept: Probability of Simple Events

b) $\frac{7}{10}$

[Solution Description]

The total number of balls in the box is $5 + 3 + 2 = 10$. The number of balls that are not blue is $5 + 2 = 7$. Therefore, the probability of drawing a ball that is not blue is $\frac{7}{10}$.

Key Concept: Probability of Equally Likely Outcomes

c) $\frac{1}{2}$

[Solution Description]

A fair die has six equally likely outcomes: $1, 2, 3, 4, 5,$ and $6$. The even numbers among these are $2, 4,$ and $6$, which are three outcomes. Therefore, the probability of getting an even number is calculated as follows: $P(\text{Even}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$

Key Concept: Probability, Simple Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The bag contains 5 red balls and 3 blue balls, making the total number of balls 8. The probability of drawing a red ball is calculated by dividing the number of red balls by the total number of balls, which is $\frac{5}{8}$. The reason states the general formula for calculating probability, which is correct. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Probability of Compound Events

c) $\frac{1}{2}$

[Solution Description]

There are 4 equally likely outcomes when the spinner is spun. The favorable outcomes for landing on blue or green are 2 (blue and green). The probability of landing on blue or green is $P(\text{Blue or Green}) = \frac{2}{4} = \frac{1}{2}$.

Key Concept: Probability of Compound Events

a) $\frac{1}{6}$

[Solution Description]

There are 36 possible outcomes when two dice are rolled. The favorable outcomes where the sum is 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 favorable outcomes. Thus, the probability is $\frac{6}{36} = \frac{1}{6}$.

Key Concept: Probability of Compound Events

c) $\frac{3}{5}$

[Solution Description]

Total number of balls in the bag = 3 (red) + 2 (blue) = 5. The probability of drawing a red ball is given by the formula $P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$. Substituting the values, we get $P(\text{Red}) = \frac{3}{5}$.

Key Concept: Probability of Rain

b) $\frac{9}{10}$

[Solution Description]

The probability that it rains on a given day is $\frac{1}{10}$. Since the total probability must sum to 1, the probability that it does not rain is calculated as:

$1 - \frac{1}{10} = \frac{9}{10}$

Key Concept: Unequal Chances

c) A train being on time or late

[Solution Description]

In real-life situations, some events do not have equal chances of occurring. For example, a train being on time or late does not have equal probabilities, as delays can occur due to various factors.

Key Concept: Real-Life Applications of Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of rain on a day during the rainy season is $\frac{1}{10}$. This implies that in a span of 10 days, it is expected to rain on one of those days. The reason explains that this probability indicates that out of 10 days, it will rain only once. Both the assertion and the reason are true, and the reason correctly explains the assertion by interpreting the probability in terms of expected occurrences over a period of 10 days.

Key Concept: Chance and Probability, Weather Forecasting

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of rainfall on a given day is $\frac{1}{10}$ and claims that the chance of it raining on a day when you forget to carry a raincoat is also $\frac{1}{10}$. This is correct because the probability of rainfall is independent of whether you carry a raincoat or not. The reason correctly explains the assertion by stating that the probability of an event (rainfall) remains constant regardless of past occurrences, as each day's weather is independent. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Expected Value

d) 420 kg

[Solution Description]

The expected harvest is calculated by taking the weighted average of the possible outcomes based on their probabilities. The expected harvest if it rains is $500 \times 0.60 = 300$ kg. The expected harvest if it does not rain is $300 \times 0.40 = 120$ kg. Therefore, the total expected harvest is $300 + 120 = 420$ kg.

Key Concept: Chance and Probability - Weather Forecasting

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of rain on a given day during the rainy season is $\frac{1}{10}$. This is based on the assumption that raining or not raining on a day are equally likely events. The reason provided is that this probability is derived from historical weather data indicating that it rains one in every ten days. Historical weather data is indeed used by meteorological departments to predict weather patterns, and assuming equal likelihood, the probability calculation is correct. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Probability of Event

b) 0.4

[Solution Description]

The probability of an event is given by the number of favorable outcomes divided by the total number of outcomes. Here, the number of voters who voted for Candidate A is 200 out of a total of 500 voters surveyed.

$\text{Probability} = \frac{\text{Number of voters for Candidate A}}{\text{Total number of voters surveyed}} = \frac{200}{500} = \frac{2}{5} = 0.4$

Therefore, the probability that a randomly selected voter voted for Candidate A is 0.4.

Key Concept: Exit Polls and Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Exit polls involve asking a small, representative group of voters whom they voted for after they have cast their votes. This sample is used to estimate the probability of each candidate winning the election. The assertion correctly states that exit polls are used for predicting election results. The reason explains that the probability of winning is estimated based on the proportion of votes received in the exit poll, which is also true. Furthermore, the reason correctly explains the assertion because the estimation of probability is directly tied to the exit poll results.

Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Complementary Probability

c) 0.6

[Solution Description]

The complementary probability of an event not happening is calculated by subtracting the probability of the event from 1. Here, the probability that a voter voted for Candidate A is $\frac{200}{500} = 0.4$. The probability that a voter did not vote for Candidate A is:

$\text{Probability} = 1 - \frac{200}{500} = 1 - 0.4 = 0.6$

Alternatively, since 300 out of 500 voters did not vote for Candidate A, the probability is directly:

$\frac{300}{500} = 0.6$

Therefore, the probability that a randomly selected voter did not vote for Candidate A is 0.6.

Key Concept: Random Experiment

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that tossing a coin is a random experiment because the result cannot be controlled. This is true as the outcome of a coin toss (head or tail) is unpredictable and cannot be influenced. The reason given is that in a random experiment, all possible outcomes have an equal chance of occurring. This is also true because in a fair coin toss, both head and tail have an equal probability of $\frac{1}{2}$. However, the reason does not directly explain why the result cannot be controlled; it only explains the nature of the outcomes in a random experiment. Therefore, while both the assertion and reason are true, the reason is not the correct explanation of the assertion.

Key Concept: Random Experiment Outcomes

b) Head and Tail

[Solution Description]

When a fair coin is tossed, there are two equally likely outcomes: Head or Tail. These are the only possible results of the toss.

Key Concept: Quality Control in Manufacturing, Random Experiment

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that exactly 2 items will be defective out of 100 produced, given a defect probability of 0.02. However, this is not necessarily true because probability does not guarantee exact outcomes; it only provides the likelihood. The reason correctly defines probability as the ratio of favorable outcomes to total possible outcomes, which is true and aligns with the concept of probability in random experiments. However, the reason does not explain why exactly 2 items will be defective, as probability deals with likelihood, not certainty. Thus, the assertion is false, while the reason is true.

Key Concept: Probability and Weather Prediction

b) $\frac{8}{25}$

[Solution Description]

The probability of rain on a single day is $20\%$ or $\frac{1}{5}$, and the probability of no rain is $80\%$ or $\frac{4}{5}$. The probability of rain on exactly one day in a two-day period can occur in two scenarios: rain on the first day and no rain on the second day, or no rain on the first day and rain on the second day. Thus, the probability is $\left(\frac{1}{5} \times \frac{4}{5}\right) + \left(\frac{4}{5} \times \frac{1}{5}\right) = \frac{8}{25}$.

Key Concept: Probability, Real-Life Applications

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that exit polls can accurately predict the probability of a candidate winning an election because they provide a random sample of voters. This is partially true, as exit polls do use random sampling to gather data. However, the accuracy of predictions depends on the randomness and representativeness of the sample. The reason explains that exit polls are based on the principle of sampling, which is correct. Sampling allows us to infer characteristics of a large population from a smaller subset. However, while sampling is a valid principle, the assertion overstates the accuracy of exit polls by implying they are always accurate, which is not necessarily true due to potential biases or sampling errors. Therefore, both the assertion and reason are true, but the reason does not fully explain the assertion.

Key Concept: Chance and Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of rain on a given day during the rainy season is $\frac{1}{10}$. This means that out of 10 days, it is expected to rain on one day. The reason explains the general method of calculating probability, which is dividing the number of favorable outcomes by the total number of possible outcomes. In this case, the favorable outcome is raining on a day, and the total number of possible outcomes is the number of days considered. Since the assertion aligns with the general method of probability calculation explained in the reason, both are true, and the reason correctly explains the assertion.

Key Concept: Outcomes as events

a) Getting a prime number

[Solution Description]

An event is defined as one or more outcomes of an experiment. Rolling a die has six possible outcomes: 1, 2, 3, 4, 5, and 6. Each of these outcomes individually can be considered an event. Additionally, combinations like getting an even number (2, 4, 6) are also events.

Key Concept: Outcomes as events

c) Events can consist of multiple outcomes

[Solution Description]

In probability, an event is defined as a collection of one or more outcomes from an experiment. For example, when tossing a coin, getting a head is an event, and getting a tail is another event. Events can also be combinations of outcomes, such as getting an even number when rolling a die. The correct statement must align with this definition.

Key Concept: Outcomes as events, Probability

b) $\frac{5}{12}$

[Solution Description]

First, list all possible outcomes when a die is rolled twice. There are 6 outcomes for the first roll and 6 outcomes for the second roll, making a total of $6 \times 6 = 36$ possible outcomes.

Next, identify all pairs $(a, b)$ where $a$ is the outcome of the first roll and $b$ is the outcome of the second roll such that $a + b$ is a prime number. The possible sums that are prime numbers are 2, 3, 5, 7, 11.

Now, count the number of ways to achieve each of these prime sums:

- For sum = 2: Only one pair (1,1). Number of ways = 1.

- For sum = 3: Pairs (1,2), (2,1). Number of ways = 2.

- For sum = 5: Pairs (1,4), (2,3), (3,2), (4,1). Number of ways = 4.

- For sum = 7: Pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Number of ways = 6.

- For sum = 11: Pairs (5,6), (6,5). Number of ways = 2.

Total number of favorable outcomes = $1 + 2 + 4 + 6 + 2 = 15$.

Therefore, the probability is $\frac{15}{36} = \frac{5}{12}$.

Key Concept: Probability of an event

b) $\frac{5}{8}$

[Solution Description]

The total number of balls in the bag is $3 + 5 = 8$. The number of favorable outcomes (drawing a blue ball) is 5. Using the probability formula:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Substituting the values:

$\text{Probability} = \frac{5}{8}$

Therefore, the probability of drawing a blue ball is $\frac{5}{8}$.

Key Concept: Probability, Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

When a die is thrown, the total number of possible outcomes is 6 (1, 2, 3, 4, 5, 6). An even number can be obtained in 3 ways (2, 4, 6). Therefore, the probability of getting an even number is calculated as:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$

Hence, the Assertion is true. The Reason correctly states that there are 3 favorable outcomes out of 6, which directly explains why the probability is $\frac{1}{2}$. Thus, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Probability of an event

b) $\frac{1}{2}$

[Solution Description]

A fair die has 6 possible outcomes: 1, 2, 3, 4, 5, or 6. Out of these, the odd numbers are 1, 3, and 5, which are 3 in total. The probability of getting an odd number is calculated using the formula:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Substituting the values:

$\text{Probability} = \frac{3}{6} = \frac{1}{2}$

Therefore, the probability of getting an odd number is $\frac{1}{2}$.

Key Concept: Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the probability of getting a head in a single toss of a fair coin is $\frac{1}{2}$. This is true because in a fair coin toss, there are only two possible outcomes: head or tail. Since both outcomes are equally likely, the probability of getting a head is calculated as $\frac{1}{2}$. The reason provided also correctly explains the assertion by stating that there are two equally likely outcomes: head and tail. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Random Experiment, Probability

b) $\frac{1}{6}$

[Solution Description]

First, determine the total number of possible outcomes when a die is rolled twice. Since each roll has 6 possible outcomes, the total number of outcomes is $6 \times 6 = 36$. Next, identify the number of favorable outcomes where the sum of the two rolls is exactly 7. The pairs are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 such pairs. Therefore, the probability is the number of favorable outcomes divided by the total number of outcomes, which is $\frac{6}{36} = \frac{1}{6}$.

Key Concept: Probability, Complementary Events

d) $\frac{4}{5}$

[Solution Description]

First, find the number of tickets that are multiples of 5 or 10. Multiples of 5 between 1 and 100 are 5, 10, 15, ..., 100. There are $\frac{100}{5} = 20$ multiples of 5. Multiples of 10 are also multiples of 5, so we do not count them separately. Thus, there are 20 unfavorable outcomes. The total number of tickets is 100. The number of favorable outcomes is $100 - 20 = 80$. The probability is $\frac{80}{100} = \frac{4}{5}$.