Key Concept: Data Interpretation, Pictograph

b) 45

[Solution Description]

Each symbol in the pictograph represents 5 books. The first student is represented by 2 symbols, which equals $2 * 5 = 10$ books. The second student is represented by 3 symbols, which equals $3 * 5 = 15$ books. The third student is represented by 4 symbols, which equals $4 * 5 = 20$ books. Adding these together, the total number of books read by the three students is $10 + 15 + 20 = 45$ books.

Key Concept: Data Handling

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

[Solution Description] A pictograph is indeed a graphical representation of data using symbols, as mentioned in the syllabus. The reason provided states that pictographs are used to make data easy to understand and visually appealing, which is also true. Additionally, the reason correctly explains why pictographs are used, making it the correct explanation for the assertion.

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: Data Representation

c) 4

[Solution Description] To find the number of symbols required to represent 200 books when one symbol represents 50 books, divide the total number of books by the number of books per symbol. $200 \div 50 = 4$. So, 4 symbols are needed.

Key Concept: Data Collection, Pictograph

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

[Solution Description]

The assertion states that a pictograph is an effective way to represent data because it uses symbols to depict information visually. This is true as a pictograph simplifies complex data into visual symbols, making it easier to understand. The reason states that in a pictograph, one symbol represents a fixed quantity, making it easier to interpret the data at a glance. This is also true because the use of symbols with fixed quantities allows for quick comparisons and interpretations. Furthermore, the reason correctly explains why a pictograph is effective, as the fixed quantity representation by symbols enhances clarity and ease of interpretation.

Key Concept: Data Collection

c) Data

[Solution Description]

The information collected in the context of a situation that we want to study is called data. For example, a teacher may collect the heights of all students in her class to find the average height.

Key Concept: Bar Graph

b) 150

[Solution Description]

Each unit in the bar graph represents 50 cars. The height of the bar for August is 3 units. Therefore, the number of cars represented by the bar is calculated as $3 \times 50 = 150$ cars.

Key Concept: Organization of Data

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

[Solution Description]

The assertion states that organizing data systematically is essential for drawing meaningful inferences, which is true because without proper organization, it is difficult to analyze and interpret data effectively. The reason given is that systematic organization helps in identifying patterns and trends in the data, which is also true and directly explains why systematic organization is essential. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Bar Graph, Double Bar Graph

a) School A

[Solution Description]

Calculate the combined number for each school. School A: 40 walking + 45 cycling = 85 students. School B: 55 walking + 25 cycling = 80 students. School C: 15 walking + 35 cycling = 50 students. The highest number is 85 students for School A.

Key Concept: Pictograph, Bar Graph

b) 950 cars

[Solution Description]

First, calculate the number of cars for each month. For July: 2.5 symbols $\times$ 100 cars/symbol = 250 cars. For August: 3 symbols $\times$ 100 cars/symbol = 300 cars. For September: 4 symbols $\times$ 100 cars/symbol = 400 cars. Now, add the numbers together: 250 + 300 + 400 = 950 cars.

Key Concept: Pie Chart

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

[Solution Description]

A pie chart represents the whole data as a circle. Since 25% means $\frac{1}{4}$ of the total, the sector representing 25% of the data covers $\frac{1}{4}$ of the circle.

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, Comparison

b) Second Quarter

[Solution Description]

Calculate the number of units sold in each quarter: First quarter: $4 \times 500 = 2000$ units. Second quarter: $6 \times 500 = 3000$ units. Third quarter: $5 \times 500 = 2500$ units. Fourth quarter: $3 \times 500 = 1500$ units. The quarter with the highest sales is the second quarter with 3000 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: 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: Bar Graph

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

[Solution Description]

The assertion states that the height of each bar in a bar graph is proportional to the value it represents. This is true because the height of the bar directly corresponds to the quantity or value of the category it represents. The reason states that the uniform width of bars ensures consistency in data representation. This is also true because having bars of equal width allows for an accurate and fair comparison between different categories. However, the reason does not explain why the height of the bar is proportional to the value; it merely explains why the width is uniform. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of 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: Double Bar Graph Analysis

c) The gender distribution in HR is balanced

[Solution Description]

The double bar graph shows the number of male and female employees in various departments. In the HR department, the number of male employees is equal to the number of female employees. This indicates a balanced gender distribution in the HR department.

Key Concept: Circle Graph, Pictograph

b) 7

[Solution Description]

First, calculate the amount spent on Rent from the total monthly expense of \$2000. Since Rent occupies 35% of the expenses, the amount spent on Rent is $35\% \times \$2000 = \$700$. In the pictograph, each symbol represents \$100, so the number of symbols for Rent is $\$700 / \$100 = 7$.

Key Concept: Bar Graph

c) Assertion is true, but Reason is false.

[Solution Description]

A bar graph is a graphical representation of data using bars of uniform width, where the height of each bar is proportional to the value it represents. The assertion correctly states this fact. However, the reason claims that the width of the bars can vary, which is incorrect. In a standard bar graph, all bars must have the same width to ensure fair comparison and accurate interpretation of the data. Therefore, the assertion is true, but the reason is false.

Key Concept: Double Bar Graph

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

[Solution Description]

A double bar graph is specifically designed to display two sets of data side by side, which facilitates comparison between them. The height of each bar corresponds to the quantity or value of the category it represents. This visual representation makes it straightforward to compare the data sets directly. Therefore, both the assertion and the reason are true, and the reason correctly explains why double bar graphs are used for comparison.

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: Pie Chart, Real-World Application

b) \$600

[Solution Description]

The central angle for the sector representing Cakes is $60^\circ$. Since the total angle in a circle is $360^\circ$, the fraction of the circle representing Cakes is $\frac{60}{360} = \frac{1}{6}$. Therefore, the sales amount for Cakes is $\frac{1}{6} \times \$3600 = \$600$.

Key Concept: Circle Graph Proportion

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

[Solution Description]

To find the fraction of the circle graph representing the time spent in school, we use the formula:

$\text{Fraction} = \frac{\text{Number of school hours}}{\text{Total hours in a day}} = \frac{6}{24} = \frac{1}{4}$

Therefore, the fraction of the circle graph representing the time spent in school is $\frac{1}{4}$.

Key Concept: Drawing Pie Charts

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

[Solution Description]

To verify the assertion, we need to check if the central angle for a sector representing 25% of the total data is indeed $90^\circ$. According to the reason, the central angle is calculated as $\frac{25}{100} \times 360^\circ = 90^\circ$. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

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

c) 90°

[Solution Description] The central angle of a sector in a pie chart is calculated by multiplying the percentage of the data by 360°. Here, the sector represents 25% of the data. So, the central angle is $25\% \times 360° = 90°$.

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: Fraction Representation in Pie Chart

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

[Solution Description]

To find the fraction of the circle representing item A:

1. Total sales = \$1000.

2. Sales of item A = \$200.

3. Fraction = $\frac{200}{1000} = \frac{1}{5}$.

Key Concept: Circle Graph, Representation of Data in Pie Charts

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

[Solution Description]

To verify the assertion, we first calculate the total angle in a circle, which is $360^\circ$. The proportion of the sector representing sleep is given as $120^\circ$. The time spent sleeping is 8 hours. We can check if the angle corresponds to the time by using the formula:

$\text{Angle} = \left(\frac{\text{Time spent on activity}}{\text{Total time}}\right) \times 360^\circ$

Substituting the values:

$120^\circ = \left(\frac{8 \text{ hours}}{24 \text{ hours}}\right) \times 360^\circ = 120^\circ$

This confirms that the assertion is true. The reason states that the angle of each sector is directly proportional to the time spent on that activity, which is also true and correctly explains the assertion.

Key Concept: Pie Chart Construction

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

[Solution Description]

To find the central angle for the sector representing $\textbf{Sleep}$, follow these steps:

1. Calculate the total hours in a day, which is 24 hours.

2. Determine the fraction of time spent on sleep: $\frac{8}{24} = \frac{1}{3}$.

3. Multiply this fraction by 360° to find the central angle: $\frac{1}{3} \times 360° = 120°$.

The assertion states that the central angle is 120°, which is correct as per the calculation. The reason explains the method to calculate the central angle using the fraction of time spent on sleep, which is also correct and directly 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, 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: Interpretation of Pie Chart

c) 33.33%

[Solution Description]

First, calculate the total number of people: $30 + 40 + 25 + 25 = 120$. The percentage for South Indian food is calculated using the formula $(\text{number of people choosing South Indian} / \text{total number of people}) \times 100$. So, $(40 / 120) \times 100 = 33.\overline{3}\%$.

Key Concept: Pie Chart Central Angle Calculation

c) $90^\circ$

[Solution Description]

To find the central angle for the vanilla sector, first convert the percentage to a fraction. 25% is equivalent to $\frac{25}{100} = \frac{1}{4}$. The central angle is then calculated as $\frac{1}{4} \times 360^\circ = 90^\circ$.

Key Concept: Pie Chart Representation, 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 activity "Play", we first determine the fraction of time spent in "Play" out of the total day. The total day consists of 24 hours. The time spent in "Play" is given as 4 hours. Therefore, the fraction of time spent in "Play" is calculated as follows:

$\text{Fraction of Play} = \frac{\text{Time spent in Play}}{\text{Total time in a day}} = \frac{4}{24} = \frac{1}{6}$

Since the total angle at the center of a circle is $360^\circ$, the central angle for "Play" is calculated as:

$\text{Central Angle for Play} = \frac{1}{6} \times 360^\circ = 60^\circ$

The assertion states that the central angle for "Play" is $60^\circ$, which matches our calculation. The reason correctly explains that the fraction of time spent in "Play" is $\frac{1}{6}$. Both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Pie Chart, Percentage Calculation

c) \$2,500

[Solution Description]

First, calculate the expenditure on education:

$\text{Education} = 15\% \times 50,000 = 0.15 \times 50,000 = \$7,500$

Next, calculate the expenditure on transport:

$\text{Transport} = 10\% \times 50,000 = 0.10 \times 50,000 = \$5,000$

Finally, find the difference between education and transport:

$\text{Difference} = 7,500 - 5,000 = \$2,500$

Key Concept: Circle Graph or Pie Chart

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

[Solution Description]

To verify the assertion, we use the given formula to calculate the central angle for food. The percentage expenditure on food is 40%. Applying the formula:

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

The calculation confirms that the central angle for food is indeed $144^\circ$, making the assertion true. The reason provided is also correct as it accurately describes the formula used to calculate the central angle in a pie chart.

Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Central Angle Calculation

d) 160°

[Solution Description]

To find the central angle for ordinary bread:

1. Calculate the fraction of sales: $\frac{320}{720} = \frac{4}{9}$.

2. Multiply by 360° to get the central angle: $\frac{4}{9} \times 360° = 160°$.

The central angle for ordinary bread is 160°.

Key Concept: Pie Chart Analysis, Central Angle Comparison

b) 4:3

[Solution Description]

The central angle for "Cakes and Pastries" is $80^\circ$ and for "Biscuits" is $60^\circ$. The sales can be calculated using the formula $\text{Sales} = \left(\frac{\text{Central Angle}}{360^\circ}\right) \times \text{Total Sales}$. Thus, Sales of "Cakes and Pastries" = $\left(\frac{80}{360}\right) \times 720 = 160$Rs.. Similarly, Sales of "Biscuits" = $\left(\frac{60}{360}\right) \times 720 = 120$Rs.. The ratio is $\frac{160}{120} = \frac{4}{3}$.

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: Central Angle Calculation

c) 90°

[Solution Description]

The percentage of students preferring Vanilla is 25%. To find the central angle, we calculate 25% of 360°. This is done by multiplying 360° by $\frac{25}{100}$ or $\frac{1}{4}$. So, $360° \times \frac{1}{4} = 90°$.

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: 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: 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: 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: Chance and Probability, Real-Life Applications

c) 0.7

[Solution Description]

The probability of rain is given as 30%, which can be written as $P(\text{Rain}) = 0.3$. Since the total probability must sum to 1, the probability of no rain is calculated as:

$P(\text{No Rain}) = 1 - P(\text{Rain}) = 1 - 0.3 = 0.7$

Therefore, the probability that it does not rain tomorrow is 0.7 or 70%.

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 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: 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: 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: Real-Life Probability

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

[Solution Description]

Total number of students = 30. Number of girls = 12. Probability = $\frac{12}{30} = \frac{2}{5}$.

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 a head when a fair coin is tossed is $\frac{1}{2}$. This is true because in a fair coin toss, there are only two possible outcomes - head and tail - and both are equally likely. The reason given is that in a fair coin toss, there are two equally likely outcomes - head and tail. This is also true. Moreover, the reason correctly explains why the probability of getting a head is $\frac{1}{2}$. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

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: Probability of Tossing a Coin

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

[Solution Description]

When a fair coin is tossed, there are two possible outcomes: Head and Tail. Both outcomes are equally likely because the coin has an equal chance of landing on either side. Therefore, the probability of getting a Head is calculated as the number of favorable outcomes divided by the total number of possible outcomes, which is $\frac{1}{2}$. The Reason correctly explains why the probability of getting a Head is $\frac{1}{2}$.

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: Random Experiment

b) No, it cannot be controlled

[Solution Description]

Tossing a coin is a random experiment where the outcome cannot be controlled. It is impossible to predict or control whether the result will be a head or a tail. Each outcome is equally likely, and the result depends purely on chance.

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

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 of Not Getting a Prime Number

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

[Solution Description]

The prime numbers on a die are 2, 3, and 5. There are 3 prime numbers out of 6 possible outcomes. The number of non-prime numbers is therefore $6 - 3 = 3$. The probability of not getting a prime number is calculated as:

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

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

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 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, Spinning a Wheel

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

[Solution Description]

To determine if both the Assertion (A) and Reason (R) are true and whether R correctly explains A, let's analyze the problem step by step.

Step 1: Identify the total number of sectors on the wheel.

The wheel has 3 green sectors, 1 blue sector, and 1 red sector.

Total sectors = 3 + 1 + 1 = 5.

Step 2: Calculate the total number of possible outcomes.

Since each sector represents a unique outcome, the total number of possible outcomes is equal to the total number of sectors, which is 5.

Step 3: Determine the probability of landing on a green sector.

Probability of an event = (Number of favorable outcomes) / (Total number of possible outcomes).

Number of favorable outcomes for a green sector = 3.

Thus, probability of green sector = $\frac{3}{5}$.

Step 4: Evaluate the Assertion (A) and Reason (R).

Assertion (A): The probability of landing on a green sector is $\frac{3}{5}$. This is true, as calculated above.

Reason (R): The total number of possible outcomes is equal to the number of sectors. This is also true, as we determined that the total number of possible outcomes is 5, which is equal to the number of sectors.

Conclusion: Both Assertion and Reason are true, and Reason correctly explains the Assertion since the probability calculation depends on the total number of possible outcomes (which is determined by the number of sectors).

Key Concept: Probability as a Fraction

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

[Solution Description]

When a fair coin is tossed once, there are two equally likely outcomes: Heads (H) or Tails (T). The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes. In this case, the number of favorable outcomes for getting a Tail is 1, and the total number of possible outcomes is 2. Therefore, the probability of getting a Tail is $\frac{1}{2}$.

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: Probability of an Event

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

[Solution Description]

When a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, and 6. The even numbers in these outcomes are 2, 4, and 6. So, there are 3 favorable outcomes out of 6. The probability of getting an even number is:

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

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

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: Real-life Probability

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

[Solution Description]

The probability that it rains is $\frac{1}{5}$. Therefore, the probability that it does not rain is:

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

Key Concept: Events in Probability

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

[Solution Description]

Total number of balls in the bag is 3 (red) + 2 (blue) = 5. Drawing a ball that is not red means drawing a blue ball. There are 2 blue balls. Therefore, the probability is $\frac{2}{5}$.

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 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 Simple Events

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

[Solution Description]

The total number of balls in the bag is $3 + 5 = 8$. The number of red balls is $3$. Therefore, the probability of drawing a red ball is calculated as follows: $P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{8}$

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: Compound Events, Probability

b) $\frac{16}{33}$

[Solution Description]

The total number of marbles is $8 + 4 = 12$. The probability of drawing a white marble followed by a black marble is $\frac{8}{12} \times \frac{4}{11} = \frac{32}{132}$. Similarly, the probability of drawing a black marble followed by a white marble is $\frac{4}{12} \times \frac{8}{11} = \frac{32}{132}$. Adding these probabilities gives the total probability of drawing marbles of different colors: $\frac{32}{132} + \frac{32}{132} = \frac{64}{132} = \frac{16}{33}$.

Key Concept: Compound Events

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

[Solution Description]

The problem involves calculating the probability of drawing a red ball from a bag containing 4 red balls and 2 yellow balls. The total number of balls in the bag is 6. The number of favorable outcomes for drawing a red ball is 4. The probability of an event is given by the formula:

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

Substituting the values, we get:

$\text{Probability of getting a red ball} = \frac{4}{6} = \frac{2}{3}$

This confirms that the Assertion is true. The Reason is also true because the total number of outcomes is indeed 6, and 4 of these outcomes are favorable for drawing a red ball. Moreover, the Reason correctly explains why the probability of getting a red ball is $\frac{2}{3}$.

Key Concept: Probability, Real-Life Applications

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

[Solution Description]

The total number of chapters is 5, and the student has prepared 4 chapters. The probability of a question being asked from any one chapter is $\frac{1}{5}$. Since only one chapter is left unprepared, the probability that the major question comes from the unprepared chapter is $\frac{1}{5}$.

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) $\frac{729}{1000}$

[Solution Description]

The probability that it does not rain on a single day is $1 - \frac{1}{10} = \frac{9}{10}$. For three consecutive days, the probability that it does not rain on any of these days is calculated by multiplying the individual probabilities: $\left(\frac{9}{10}\right)^3$.

$\left(\frac{9}{10}\right)^3 = \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} = \frac{729}{1000} = 0.729$

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: Meteorological Prediction

b) Historical data and trends

[Solution Description]

The Meteorological Department uses historical data and observed trends from previous years to predict future weather conditions. This method relies on the analysis of patterns and recurring phenomena.

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: 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: Exit Polls, Probability

c) 600,000

[Solution Description]

The exit poll indicates that 60% of the respondents favored Candidate A. To estimate the total number of votes for Candidate A, we apply this percentage to the total number of voters.

$0.60 \times 1,000,000 = 600,000$

Therefore, the estimated number of votes for Candidate A is 600,000.

Key Concept: Chance and Probability

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

[Solution Description]

Assertion states that exit polls are highly accurate in predicting election results, while Reason explains that exit polls use probability to estimate the voting behavior of the entire population based on a small sample. Although probability is used in exit polls to make predictions, the accuracy of these predictions can be influenced by various factors such as sampling bias, non-response bias, and changes in voter behavior after the poll is conducted. Therefore, while both Assertion and Reason are true, Reason does not fully explain the high accuracy claimed in the Assertion.

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: 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: Expected Number of Defects

c) 10

[Solution Description]

The expected number of defective widgets can be calculated using the formula for expectation in a binomial distribution:

$E(X) = n \times p$

where $n$ is the number of trials and $p$ is the probability of a defect. Here, $n = 500$ and $p = 0.02$. Plugging in these values:

$E(X) = 500 \times 0.02 = 10$

Key Concept: Probability, Real-life Application

b) 0.0596

[Solution Description]

This problem can be solved using the binomial probability formula:

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

Here, $n = 20$, $k = 3$, $p = 0.05$.

Step 1: Calculate $C(20, 3)$:

$C(20, 3) = \frac{20!}{3!(20-3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$

Step 2: Calculate $p^k$:

$0.05^3 = 0.000125$

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

$(1-0.05)^{17} = 0.95^{17} \approx 0.4181$

Step 4: Multiply all the terms together:

$P(3) = 1140 \times 0.000125 \times 0.4181 \approx 0.0596$

So, the probability that exactly 3 out of 20 bulbs are defective is approximately 0.0596.

Key Concept: Application of Probability in Real Life

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

[Solution Description]

The probability is estimated by the ratio of the number of favorable outcomes to the total number of observations. Here, 60 out of 100 voters voted for Candidate A, so the estimated probability is $\frac{60}{100} = \frac{3}{5}$.

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: 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 in a die roll is an event. This is true because the possible outcomes for this event are 5 and 6, which are specific outcomes of the experiment. The reason states that an event is defined as one or more outcomes of an experiment, which is also true according to the syllabus. Moreover, the reason correctly explains the assertion because the assertion is based on the definition provided in the reason.

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: Probability of Events

c) $\frac{4}{13}$

[Solution Description]

Total number of cards in a deck = 52

Number of hearts in a deck = 13

Number of queens in a deck = 4

However, there is 1 card which is both a heart and a queen (the queen of hearts), so we need to subtract this overlap to avoid double-counting.

Number of favorable outcomes = Number of hearts + Number of queens - Overlap = 13 + 4 - 1 = 16

Therefore, the probability $P$ of drawing a heart or a queen is calculated as:

$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{16}{52} = \frac{4}{13}$

Key Concept: Probability of Complementary Event

b) $0.4$

[Solution Description]

The probability of the complementary event (event not occurring) is calculated as $1 - P(\text{event})$. Given $P(\text{event}) = 0.6$, the probability of the event not occurring is $1 - 0.6 = 0.4$.

Key Concept: Probability, Die Roll

c) $\frac{1}{8}$

[Solution Description]

To find the probability that the sum of three dice rolls is exactly 10, we need to count the number of favorable outcomes where the sum is 10 and divide by the total number of possible outcomes. The total number of outcomes when rolling a die three times is $6^3 = 216$. The number of ways to get a sum of 10 can be calculated by finding all combinations $(a, b, c)$ where $a + b + c = 10$ and $1 \leq a, b, c \leq 6$. There are 27 such combinations. Therefore, the probability is $\frac{27}{216} = \frac{1}{8}$.

Key Concept: Probability, Coin Toss

b) $2p(1-p)$

[Solution Description]

The probability of getting exactly one head in two tosses can be calculated by considering the two scenarios where either the first toss is a head and the second is a tail, or the first toss is a tail and the second is a head. The probability of the first scenario is $p \times (1-p)$, and the probability of the second scenario is $(1-p) \times p$. Adding these together gives the total probability: $p(1-p) + (1-p)p = 2p(1-p)$.

Key Concept: Random Experiment and Probability

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

[Solution Description]

In a fair coin toss, there are two possible outcomes: Head and Tail. Since the coin is fair, both outcomes are equally likely. The probability of an event is calculated as the number of favorable outcomes divided by the total number of possible outcomes. Here, the favorable outcome is getting a Head, and the total outcomes are 2 (Head and Tail). Therefore, the probability of getting a Head is $\frac{1}{2}$. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

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