Key Concept: Data Interpretation

b) 5

[Solution Description]

To find the average, add all the numbers together and then divide by the count of numbers. The sum of the numbers is $5 + 3 + 7 + 4 + 6 = 25$. There are 5 numbers, so the average is $\frac{25}{5} = 5$.

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: Definition of Data

a) Information collected for analysis

[Solution Description]

Data refers to the collection of information that can be used for analysis or interpretation. It can include numbers, facts, measurements, or observations gathered in various contexts.

Key Concept: Data Definition, Data Interpretation

c) 149.67 cm

[Solution Description]

First, calculate the total height of all students before the student leaves: $150 \text{ cm} \times 31 \text{ students} = 4650 \text{ cm}$. After the student leaves, the total height becomes $4650 \text{ cm} - 160 \text{ cm} = 4490 \text{ cm}$. Now, divide this by the number of remaining students to find the new average: $\frac{4490 \text{ cm}}{30 \text{ students}} \approx 149.67 \text{ cm}$.

Key Concept: Bar Graph

c) Respective values

[Solution Description]

In a bar graph, the heights of the bars are proportional to the respective values they represent.

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: Pictograph Interpretation

c) 100

[Solution Description]

To find the difference in the number of books produced in August and July, we first calculate the total books for each month using the given symbol value. In July, the symbol appears 3 times, so the total books are $3 \times 50 = 150$. In August, the symbol appears 5 times, so the total books are $5 \times 50 = 250$. The difference is $250 - 150 = 100$ books.

Key Concept: Bar Graph Analysis

b) 350

[Solution Description]

To find the average number of cars sold per month, we sum up the number of cars sold in all four months and divide by the number of months. The total sales are $200 + 300 + 400 + 500 = 1400$. The average is $\frac{1400}{4} = 350$ cars per month.

Key Concept: Data Representation

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

[Solution Description]

The assertion states that a bar graph is used to represent data where the height of each bar corresponds to the value it represents. This is true as in a bar graph, the height of the bars is proportional to the values they represent. The reason states that bar graphs are effective for comparing quantities across different categories. This is also true because bar graphs visually display the differences between categories, making comparisons straightforward. However, the reason does not explain why the height of the bar corresponds to the value; it only describes the utility of bar graphs. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Pie Chart

c) 8 hours

[Solution Description]

The angle for "Sleep" is 120 degrees out of the total 360 degrees in a circle. To find the time spent sleeping, calculate the fraction of the circle that 120 degrees represents and multiply it by the total time: $\frac{120}{360} \times 24 = 8$ hours.

Key Concept: Bar Graph

b) 150

[Solution Description]

To find the total number of students, add the number of students in each class. Class 6 has 40 students, Class 7 has 50 students, and Class 8 has 60 students. The total is $40 + 50 + 60 = 150$ students.

Key Concept: Bar Graph, Double Bar Graph

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

[Solution Description]

In a bar graph, the height of the bars corresponds to the values they represent, and their positions are independent of the data. Similarly, in a double bar graph, the heights of the bars for each category represent the quantities, and changing their positions does not alter the data. The assertion is true because the position of the bars does not change the value they represent. The reason is also true because the height of the bars, not their position, determines the quantity. Moreover, the reason correctly explains why the assertion is true.

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

Key Concept: Pictograph, Data Interpretation

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

[Solution Description]

The assertion states that a pictograph uses pictures to represent data, which is true as pictographs are designed to visually represent data using images or symbols. The reason given is that pictographs simplify complex information into a visual format, which is also true. However, the reason does not directly explain why pictographs make data visually appealing and easy to understand; rather, it explains another advantage of pictographs. Therefore, both the assertion and reason are true, but the reason is NOT the correct explanation of the assertion.

Key Concept: Pictograph

c) 400

[Solution Description]

One symbol = 100 cars. September has 4 full symbols. Calculation: $4 \times 100 = 400$.

Key Concept: Double Bar Graph, Comparison

b) 622.08 units

[Solution Description]

The initial total sales of product Y are 300 units. An increase of 20% in each quarter means that the sales in each quarter will be multiplied by 1.20. Since there are four quarters, the total sales after four quarters will be $300 \times (1.20)^4$. Calculating $(1.20)^4$ gives 2.0736. Therefore, the total sales of product Y after four quarters will be $300 \times 2.0736 = 622.08$ units.

Key Concept: Bar Graph, Data Interpretation

b) 60 units

[Solution Description]

The original heights of the bars for product A in January, February, and March are 50, 60, and 70 units respectively. Swapping the bars for January and March means that the new heights for January and March will be 70 and 50 units respectively. The height of the bar for February remains unchanged at 60 units.

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

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

[Solution Description]

The assertion states that in a double bar graph, the bars representing different categories are of equal width. This is true because, in a double bar graph, all bars must have the same width to maintain consistency in the visual representation of data. The reason provided is that equal width of bars ensures consistent visual representation and accurate comparison of data. This is also true because equal-width bars allow for an unbiased comparison between different categories. Moreover, the reason correctly explains the assertion as equal-width bars are essential for maintaining accuracy and consistency in data comparison.

Key Concept: Pie Chart, Bar Graph

d) Assertion is false, but Reason is true.

[Solution Description]

A pie chart is used to represent data where the whole is divided into parts, showing the proportion of each part relative to the whole. However, when comparing the production of food grains over multiple years, a bar graph or line graph is more appropriate because it allows for easy comparison of values across different categories (years). The pie chart does not effectively show trends or changes over time. Therefore, the assertion is false. The reason, while true in explaining the use of a pie chart, does not justify the assertion because the context of comparing yearly production requires a different type of graph.

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: Bar Graph, Comparison

c) 120

[Solution Description]

The average of the heights for January and February is $\frac{50 + 70}{2} = 60$. Therefore, the height of the bar for March is twice this average, which is $2 \times 60 = 120$.

Key Concept: Bar Graph Characteristics

c) Bars are of equal width with equal gaps

[Solution Description]

In a bar graph, bars are of equal width and have equal gaps between them. The height of each bar represents the quantity for the respective category.

Key Concept: Circle Graph Calculation

c) $90^\circ$

[Solution Description]

The total angle at the center of a circle is $360^\circ$. To find the central angle for the sector representing education:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360^\circ = \left(\frac{25}{100}\right) \times 360^\circ = 90^\circ$

Therefore, the central angle of the sector representing education is $90^\circ$.

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: Fraction of Circle

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

[Solution Description]

To find the fraction of the total data represented by a sector with a central angle of $72^\circ$, we use the formula:

$\text{Fraction} = \frac{\text{Central Angle}}{360^\circ}$

Substituting the given central angle:

$\text{Fraction} = \frac{72^\circ}{360^\circ} = \frac{1}{5}$

Therefore, the sector represents $\frac{1}{5}$ of the total data.

Key Concept: Central Angle Calculation

b) 108$^\circ$

[Solution Description]

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

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

Given that the percentage for 'Vanilla' is 30%, we substitute this value 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 'Vanilla' is $108^\circ$.

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, Central Angle Calculation

b) Rs. 30000

[Solution Description]

To find the total monthly expenditure, we first determine the percentage represented by the savings sector. The central angle for savings is 54°, and the total angle in a circle is 360°. The percentage of savings is calculated as:

$\frac{54°}{360°} \times 100 = 15\%$

Given that 15% corresponds to Rs. 4500, the total expenditure (100%) can be calculated as:

$\frac{4500}{15} \times 100 = 30000$

Therefore, the total monthly expenditure is Rs. 30000.

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: 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 determined by the fraction of the total data it represents. Given that 25% of the total data is represented, the central angle can be calculated as follows:

$\text{Central Angle} = \frac{25}{100} \times 360° = 90°$

Therefore, the assertion that the central angle is 90° for 25% of the data is true. Additionally, the reason provided states that the central angle is calculated by multiplying the percentage by 360°, which is also true and correctly explains the assertion.

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: Pie Chart, Central Angle Calculation

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

[Solution Description]

To calculate the central angle for a category in a pie chart, we use the formula:

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

For a category representing 30% of the total data:

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

The assertion states that the central angle for 30% is $108°$, which is correct. The reason explains the general method to calculate the central angle, which is also correct. However, the reason does not specifically explain why the central angle for 30% is $108°$; it only provides the formula. Therefore, both the assertion and reason are true, but the reason is NOT the correct explanation of the assertion.

Key Concept: Pie Chart, Interpretation of Data

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

[Solution Description] To determine whether the assertion and reason are true, let's analyze the data provided. The total time in a day is 24 hours. The time spent on "Play" is 4 hours, and the time spent on "Homework" is also 4 hours. The central angle for each activity in the pie chart is calculated as follows:

Central angle for an activity = $\frac{\text{Time spent on the activity}}{\text{Total time}} \times 360^\circ$

For "Play":

$\text{Central angle} = \frac{4}{24} \times 360^\circ = 60^\circ$

For "Homework":

$\text{Central angle} = \frac{4}{24} \times 360^\circ = 60^\circ$

Since both "Play" and "Homework" have the same central angle of $60^\circ$ in the pie chart, the assertion that the sectors for "Play" and "Homework" will always be equal is true. The reason given is also correct because both activities take the same amount of time (4 hours). Therefore, the reason correctly explains the assertion.

Key Concept: Pie Chart Interpretation, Central Angle Calculation

b) 54°

[Solution Description]

To find the central angle for the rent sector:

1. The total angle in a circle is 360°.

2. The proportion of rent in total expenditure is 15%.

3. The central angle for rent = $\frac{15}{100} \times 360° = 54°$.

Key Concept: Central Angle Calculation, Pie Chart Interpretation

b) $60^\circ$

[Solution Description]

To find the central angle for the Bicycle sector, first determine the fraction of people who prefer bicycles out of the total surveyed. Let $N$ be the number of people who prefer bicycles and $T$ be the total number of people surveyed.

Given:

$N = 100, \quad T = 600$

The fraction of people who prefer bicycles is:

$\frac{N}{T} = \frac{100}{600} = \frac{1}{6}$

The central angle $\theta$ for the Bicycle sector is calculated as:

$\theta = \frac{1}{6} \times 360^\circ = 60^\circ$

Therefore, the central angle for the Bicycle sector is $60^\circ$.

Key Concept: Pie Chart Representation

b) $90^\circ$

[Solution Description]

To find the central angle for "School", first calculate the fraction of the day spent in school. Total hours in a day are 24. The fraction for school is $\frac{6}{24} = \frac{1}{4}$. The total angle in a circle is $360^\circ$. So, the central angle for "School" is $\frac{1}{4} \times 360^\circ = 90^\circ$.

Key Concept: Pie Chart - Expenditure Calculation

b) Rs.2000

[Solution Description] Given that 15% represents Rs.3000, we can find 1% by dividing Rs.3000 by 15. Thus, 1% = Rs.3000 / 15 = Rs.200. Therefore, 10% would be Rs.200 * 10 = Rs.2000.

Key Concept: Pie Chart, Expenditure Comparison

d) Transport and Savings

[Solution Description]

Calculate the expenditure for each category:

$\text{Food} = 35\% \times 60,000 = 0.35 \times 60,000$

$= \$21,000 \\ \\ \text{Rent} = 20\% \times 60,000$

$= 0.20 \times 60,000 = \$12,000 \\ \\ \text{Education}$

$= 10\% \times 60,000 = 0.10 \times 60,000 = \$6,000 \\ \\ \text{Transport}$

$= 5\% \times 60,000 = 0.05 \times 60,000$

$= \$3,000 \\ \\ \text{Savings} = 25\% \times 60,000 = 0.25 \times 60,000 = \$15,000$

Compare the values to find which two categories have the same monetary value.

Key Concept: Sales Calculation, Pie Chart Interpretation

b) 400Rs.

[Solution Description]

The central angle for "Ordinary Bread" is $160^\circ$ and for "Fruit Bread" is $40^\circ$. The combined central angle is $160^\circ + 40^\circ = 200^\circ$. The combined sales can be calculated using the formula $\text{Sales} = \left(\frac{\text{Central Angle}}{360^\circ}\right) \times \text{Total Sales}$. Thus, Combined Sales = $\left(\frac{200}{360}\right) \times 720 = 400$Rs..

Key Concept: Pie Chart Central Angle Calculation

c) 40°

[Solution Description]

To find the central angle for "Fruit Bread," we first calculate the fraction of total sales it represents. The total sales are \$720. The sales for "Fruit Bread" are \$80. The fraction is $\frac{80}{720} = \frac{1}{9}$. The central angle is then calculated by multiplying this fraction by 360°: $\frac{1}{9} \times 360° = 40°$.

Key Concept: Fraction to Percent, Central Angle Calculation

b) 45$^\circ$

[Solution Description]

The original percentage for "Other flavours" is 25%. Splitting it into two equal parts gives 12.5\% each. To find the central angle for each new flavor, we calculate $\frac{12.5}{100}$ of $360^\circ$.

$\frac{12.5}{100} \times 360^\circ = 45^\circ$

Therefore, the central angle for each new flavor is $45^\circ$.

Key Concept: Pie Chart Central Angles

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:

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

Given that the percentage of students preferring Chocolate flavour is 50\%, we substitute this into the formula:

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

Therefore, the central angle for the Chocolate flavour is indeed 180°. The assertion is true, and the reason correctly explains why the assertion is true.

Key Concept: Percentage Calculation

b) 25%

[Solution Description]

The total angle at the center of the circle is 360°. To find the percentage of students who prefer vanilla flavor, we use the formula: $\text{Percentage} = \left(\frac{\text{Central Angle}}{\text{Total Angle}}\right) \times 100$. Substituting the given values, we get $\text{Percentage} = \left(\frac{90°}{360°}\right) \times 100 = 25\%$.

Key Concept: Probability of Equally Likely Outcomes

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

[Solution Description]

A fair die has six faces numbered from 1 to 6. The numbers greater than 4 are 5 and 6. So, there are 2 favorable outcomes. The total number of possible outcomes is 6. Therefore, the probability is calculated as:

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

Key Concept: Probability of Multiple Outcomes

c) $\frac{5}{6}$

[Solution Description]

The total number of balls in the bag is $3 + 2 + 1 = 6$. The number of balls that are not blue is $3 + 2 = 5$. Therefore, the probability of drawing a ball that is not blue is:

$\text{Probability} = \frac{\text{Number of non-blue balls}}{\text{Total number of balls}} = \frac{5}{6}$

Key Concept: Equally likely outcomes, Probability

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

[Solution Description]

To find the probability that the sum of two rolls of a fair six-sided die is exactly 7, we first determine the total number of possible outcomes. Since each die has 6 faces, there are $6 \times 6 = 36$ total possible outcomes when rolling the die twice.

Next, we identify the number of favorable outcomes where the sum is 7. These pairs are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 such pairs.

The probability $P$ of the sum being 7 is then calculated as:

$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$

Key Concept: Real-Life Probability

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

[Solution Description]

The probability that it rains on a day is $\frac{1}{10}$. The probability that it does not rain is calculated by subtracting the probability of raining from 1. Hence, the probability of not raining is $1 - \frac{1}{10} = \frac{9}{10}$.

Key Concept: Concept of Chance in Daily Life

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion describes a common scenario where chance plays a role in daily life. The reason states that chance events are unpredictable and have equal probability. However, the assertion does not imply that all chance events have equal probability. Therefore, the assertion is true, but the reason is false because chance events do not necessarily have equal probability.

Key Concept: Probability in Real Life

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

[Solution Description]

The probability of an event and its complement always add up to 1. Given the probability of rain is $\frac{1}{5}$, the probability of no rain is calculated as follows:

$P(\text{No Rain}) = 1 - P(\text{Rain}) = 1 - \frac{1}{5} = \frac{4}{5}$

Therefore, the probability that it does not rain is $\frac{4}{5}$.

Key Concept: Probability of Equally Likely Outcomes

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

[Solution Description]

The total number of outcomes is the sum of red and green balls, which is $5 + 3 = 8$. The number of favorable outcomes for drawing a green ball is 3. Therefore, the probability is calculated as $\frac{3}{8}$.

Key Concept: Probability of an Event

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

[Solution Description]

Total number of marbles = 3 (blue) + 5 (green) = 8. Number of outcomes that make the event (drawing a blue marble) = 3. Probability = $\frac{3}{8}$.

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: Outcomes as Events

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

[Solution Description]

When a fair coin is tossed twice, there are four possible outcomes: HH, HT, TH, TT. To find the probability of getting at least one head, we can calculate the probability of the complementary event (getting no heads, i.e., TT) and subtract it from 1. The probability of getting TT is $\frac{1}{4}$. Therefore, the probability of getting at least one head is:

$P(\text{at least one head}) = 1 - P(\text{TT}) = 1 - \frac{1}{4} = \frac{3}{4}$

Key Concept: Outcomes as Events

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

[Solution Description]

A fair die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The even numbers are 2, 4, and 6. The probability of getting an even number is calculated as:

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

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: 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: Probability of Events

c) 0.75

[Solution Description]

The possible outcomes when tossing a coin twice are HH, HT, TH, and TT. Out of these four equally likely outcomes, three include at least one head (HH, HT, TH). Therefore, the probability is $\frac{3}{4}$.

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: Probability of a Specific Outcome

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

[Solution Description]

A die has six faces with numbers from 1 to 6. Each outcome is equally likely. The probability of getting any specific number, such as 3, is calculated using the formula:

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

Here, the number of favorable outcomes is 1 (getting a 3), and the total number of possible outcomes is 6 (numbers 1 through 6). Thus, the probability is:

$\frac{1}{6}$

Therefore, the correct answer is $\frac{1}{6}$.

Key Concept: Events and Probability

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

[Solution Description]

A fair die has six faces numbered 1 through 6. The numbers greater than 4 are 5 and 6. There are 2 favorable outcomes (5, 6) out of 6 possible outcomes. The probability of getting a number greater than 4 is calculated as $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Equally Likely Outcomes

d) $\frac{1}{36}$

[Solution Description]

The probability of getting a 6 in a single roll of a fair die is $\frac{1}{6}$. Since the two rolls are independent events, the probability of getting a 6 on both rolls is calculated by multiplying the probabilities of each event: $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.

Key Concept: Probability, Outcomes

b) $\frac{2}{3}$

[Solution Description]

The total number of sectors is 12. The number of unfavorable outcomes (orange) is 4. Therefore, the probability of not landing on an orange sector is:

$\text{Probability} = 1 - \frac{\text{Number of unfavorable outcomes}}{\text{Total number of outcomes}} = 1 - \frac{4}{12} = \frac{8}{12} = \frac{2}{3}$

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 Not Getting a Blue Sector

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

[Solution Description]

The total number of sectors on the wheel is 3 (green) + 1 (blue) + 1 (red) = 5. The probability of not getting a blue sector is calculated by subtracting the probability of getting a blue sector from 1. So, $P(\text{Not Blue}) = 1 - P(\text{Blue}) = 1 - \frac{1}{5} = \frac{4}{5}$.

Key Concept: Probability as a Fraction

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

[Solution Description]

The assertion states that the probability of drawing a red ball is $\frac{2}{3}$. This is calculated using the formula for probability: $P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Here, the number of favorable outcomes (red balls) is 4, and the total number of outcomes (total balls) is 6. Therefore, $P(\text{Red ball}) = \frac{4}{6} = \frac{2}{3}$. The reason correctly identifies the total number of outcomes as 6 and the number of favorable outcomes as 4, which explains why the probability is $\frac{2}{3}$. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Probability of an Event

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

[Solution Description]

The total number of pencils in the box is $4 + 6 = 10$. The number of favorable outcomes for picking a black pencil is 6. The probability of picking a black pencil is:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{10} = \frac{3}{5}$

Therefore, the probability of picking a black pencil is $\frac{3}{5}$.

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: Events in Probability

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

[Solution Description]

When a coin is tossed twice, there are 4 possible outcomes: HH, HT, TH, TT. The event "at least one head" includes HH, HT, and TH. Therefore, there are 3 favorable outcomes out of 4. The probability is $\frac{3}{4}$.

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: Outcomes as Events

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 rolling a die is an event. This is true because, in the context of probability, an event is any outcome or a collection of outcomes from an experiment. In this case, the outcomes 2, 4, and 6 are all even numbers, so getting an even number qualifies as an event. The reason states that an event is defined as one or more outcomes of an experiment, which is also true and directly explains why getting an even number is considered an event. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Probability and real-life application

c) $\frac{16}{25}$

[Solution Description]

The probability that it will not rain on a single day is $1 - \frac{1}{5} = \frac{4}{5}$.

Since the weather on consecutive days is independent, the probability that it does not rain on two consecutive days is:

$\frac{4}{5} \times \frac{4}{5} = \frac{16}{25}$

Key Concept: Probability of equally likely events

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

[Solution Description]

When a fair die is rolled twice, there are $6 \times 6 = 36$ possible outcomes.

Now, we need to count the number of outcomes where the sum of the two rolls is exactly 7. These outcomes 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: Probability of Real-Life Events

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

[Solution Description]

The probability that it rains on any given day is $P(\text{Rain}) = \frac{1}{10}$. Therefore, the probability that it does not rain on a particular day is calculated as follows: $P(\text{Not Rain}) = 1 - P(\text{Rain}) = 1 - \frac{1}{10} = \frac{9}{10}$

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 Compound Events

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

[Solution Description] The total number of balls in the bag is 6 (4 red + 2 yellow). The probability of getting a red ball is calculated by dividing the number of red balls by the total number of balls. Therefore, the probability of getting a red ball is $\frac{4}{6} = \frac{2}{3}$. This calculation aligns with the formula for probability, which is the number of favorable outcomes divided by the total number of possible outcomes. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Real-Life Applications of Probability, Equal Likelihood

c) Assertion is true, but Reason is false.

[Solution Description]

To solve this assertion and reason question, let's analyze both statements:

1. **Assertion (A):** The probability that it rains on a particular day during the rainy season is $\frac{1}{10}$. This means that out of 10 days, it is expected to rain once. This is a valid probability statement based on the given information.

2. **Reason (R):** It is stated that raining or not raining on a day are equally likely events. However, if raining has a probability of $\frac{1}{10}$, then not raining must have a probability of $1 - \frac{1}{10} = \frac{9}{10}$. This contradicts the idea that both events are equally likely. Therefore, the reason is false.

Since the Assertion is true but the Reason is false, the correct option is (c).

Key Concept: Real-Life Application of Probability

b) 0.07776

[Solution Description]

The probability that a single voter favors Candidate A is 60%, or 0.60. For 5 independent voters, the probability that all 5 favor Candidate A is $0.60^5$.

$0.60^5 = 0.60 \times 0.60 \times 0.60 \times 0.60 \times 0.60 = 0.07776$

Key Concept: Real-Life Application of Probability

b) 0.80

[Solution Description]

The probability of an event and its complement always add up to 1. Here, the probability of rain is 20%, or 0.20. Therefore, the probability that it does not rain is $1 - 0.20 = 0.80$.

Key Concept: Chance and Probability

c) $\frac{9}{50}$

[Solution Description]

The probability that it rains on exactly one day out of two consecutive days can be calculated using the formula for independent events. The probability of rain on one day is $\frac{1}{10}$, and the probability of no rain on the other day is $\frac{9}{10}$. Since either day could be the rainy day, we multiply by 2: $2 \times \left(\frac{1}{10} \times \frac{9}{10}\right) = \frac{18}{100} = \frac{9}{50}$.

Key Concept: Probability of Rain

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

[Solution Description]

The probability that it does not rain is calculated by subtracting the probability of rain from 1. Given that the probability of rain is $\frac{1}{10}$, the probability of no rain is $1 - \frac{1}{10} = \frac{9}{10}$.

Key Concept: Probability of Rain

b) 0.58

[Solution Description]

To find the probability that it will rain on at least one of the days, we use the formula for the probability of at least one event occurring: $P(\text{At least one}) = 1 - P(\text{None})$. The probability that it does not rain on Monday is $1 - 0.30 = 0.70$. The probability that it does not rain on Tuesday is $1 - 0.40 = 0.60$. Since the events are independent, the probability that it does not rain on both days is $0.70 \times 0.60 = 0.42$. Therefore, the probability that it will rain on at least one of the days is $1 - 0.42 = 0.58$ or 58%.

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: Probability in Real Life

c) 0.9

[Solution Description]

The probability that it does not rain on any day is the complement of the probability that it rains. Given that the probability of rain is $\frac{1}{10}$, the probability of no rain is:

$\text{Probability of no rain} = 1 - \frac{1}{10} = \frac{9}{10}$

Since you carry a raincoat only when it rains, the probability that it does not rain on a day when you do not carry a raincoat is the same as the probability of no rain, which is $\frac{9}{10}$ or 0.9.

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: Probability of No Defects

d) 0.366

[Solution Description]

The probability that none of the items are defective can be calculated using the binomial probability formula where $k = 0$:

$P(X = 0) = C(n, 0) \times p^0 \times (1-p)^{n}$

Here, $n = 100$ and $p = 0.01$. Since $C(n, 0) = 1$ and $p^0 = 1$, the equation simplifies to:

$P(X = 0) = (1 - 0.01)^{100} = (0.99)^{100}$

Calculating $(0.99)^{100}$:

$(0.99)^{100} \approx 0.366$

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: Probability Basics

c) 6

[Solution Description]

A standard die has six faces numbered from 1 to 6. Therefore, when a die is thrown, there are six possible outcomes.

Key Concept: Probability, Real-life Application

c) 0.441

[Solution Description]

To solve this problem, we use the binomial probability formula, which is given by:

$P(k) = C(n, k) \times p^k \times (1-p)^{n-k}$

where:

- $n$ is the number of trials (days),

- $k$ is the number of successful trials (days it rains),

- $p$ is the probability of success (rain).

Here, $n = 3$, $k = 1$, and $p = 0.3$.

Step 1: Calculate the combination $C(3, 1)$:

$C(3, 1) = 3$

Step 2: Calculate $p^k$:

$0.3^1 = 0.3$

Step 3: Calculate $(1-p)^{n-k}$:

$(1-0.3)^{3-1} = 0.7^2 = 0.49$

Step 4: Multiply all the terms together:

$P(1) = 3 \times 0.3 \times 0.49 = 0.441$

So, the probability that it rains on exactly one of the three days is 0.441.

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: Weather Prediction

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

[Solution Description]

The probability that it rains is given as $P(\text{Rain}) = 20\% = \frac{20}{100} = \frac{1}{5}$. The probability that it does not rain can be calculated as:

$P(\text{No Rain}) = 1 - P(\text{Rain}) = 1 - \frac{1}{5} = \frac{4}{5}$

Therefore, the probability that it does not rain tomorrow is $\frac{4}{5}$.

Key Concept: Outcomes as events

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

[Solution Description]

The assertion states that getting a number greater than 4 when throwing a die is an event. This is true because in the experiment of throwing a die, the outcomes 5 and 6 are greater than 4, and these form a collection of outcomes, which by definition is an event. The reason states that an event is defined as any outcome or collection of outcomes of an experiment. This is also true and correctly explains why getting a number greater than 4 is considered an event. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

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 of events

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that in the experiment of rolling a fair six-sided die, the event of getting a number greater than 4 is an outcome. However, this is incorrect because getting a number greater than 4 is an event consisting of the outcomes {5, 6} and not a single outcome. The reason correctly defines an event as a collection of one or more outcomes of an experiment. Therefore, the assertion is false, but the reason is true.

Key Concept: Probability, Complementary Events

b) $\frac{25}{28}$

[Solution Description]

The probability of drawing at least one red ball can be found using complementary probability. Instead of calculating the probability directly, we calculate the probability of the complementary event (drawing no red balls, i.e., both balls are green) and subtract it from 1. The probability of drawing a green ball on the first draw is $\frac{3}{8}$. If the first ball drawn is green, the probability of drawing another green ball is $\frac{2}{7}$. Therefore, the probability of drawing two green balls is $\frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$. Subtracting this from 1 gives the probability of drawing at least one red ball: $1 - \frac{6}{56} = \frac{50}{56} = \frac{25}{28}$.

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: Chance and Probability

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

[Solution Description]

The probability that it will rain on a given day is $\frac{1}{5}$. Since raining and not raining are complementary events, the probability that it will not rain can be found by subtracting the probability of rain from 1:

$\text{Probability of not raining} = 1 - \frac{1}{5} = \frac{4}{5}$

Therefore, the probability that it will not rain on a particular day is $\frac{4}{5}$.

Key Concept: Probability of an Event

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

[Solution Description]

Total number of balls in the bag = 3 (red) + 5 (blue) = 8.

Number of red balls = 3.

Probability of drawing a red ball = $\frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{8}$.

Therefore, the probability is $\frac{3}{8}$.

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: Equally Likely Outcomes

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

[Solution Description]

When a coin is tossed twice, the possible outcomes are: HH, HT, TH, TT.

Total number of outcomes = 4.

Outcomes with at least one head are HH, HT, TH.

Number of favorable outcomes = 3.

Probability of getting at least one head is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$.

Therefore, the probability is $\frac{3}{4}$.