Class 8 Mathematics Chapter 4 Data Handling

Key Concept: Bar Graph

c) 60

[Solution Description]

The height of the bar is 12 cm, and 1 cm represents 5 students. To find the number of students, multiply the height of the bar by the number of students each centimeter represents. Therefore, $12 \times 5 = 60$ students.

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: Bar Graph Characteristics

b) A graph using bars of uniform width with heights proportional to values

[Solution Description]

A bar graph is a graphical representation of data where bars of uniform width are used. The height of each bar corresponds to the value it represents, making it easy to compare different categories or quantities.

Key Concept: Data Definition, Pictograph

b) 1150

[Solution Description]

To find the total number of cars sold, we first calculate the number of cars for each month by multiplying the number of symbols by 100. For July: $2.5 \times 100 = 250$ cars. For August: $3 \times 100 = 300$ cars. For September: $4 \times 100 = 400$ cars. For October: $2 \times 100 = 200$ cars. Adding these together gives the total number of cars sold: $250 + 300 + 400 + 200 = 1150$ cars.

Key Concept: Data Collection

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

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

Key Concept: Data 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 Interpretation

b) 4

[Solution Description]

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

Key Concept: 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: Data Representation

c) Pie chart

[Solution Description]

A pie chart is used to represent the percentage distribution of different categories as it divides a circle into segments proportional to the frequency of each category.

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: Circle Graph

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

[Solution Description]

The assertion states that a circle graph is used to represent the relationship between a whole and its parts. This is true because a circle graph (or pie chart) divides the whole circle into sectors, where each sector represents a part of the whole. The reason states that in a circle graph, the size of each sector is proportional to the activity or information it represents. This is also true because the angle of each sector in the circle graph is determined by the proportion of the part it represents relative to the whole. For example, if a child spends 8 hours sleeping out of 24 hours in a day, the sector for sleeping will be $\frac{1}{3}$ of the circle. Since both the assertion and reason are true, and the reason correctly explains the assertion, the correct answer is option a).

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 in a bar graph, the height of each bar is proportional to the value it represents. This is true as the primary purpose of a bar graph is to visually represent data where the length or height of the bars corresponds to the values of the data points. The reason states that bar graphs are used to compare quantities across different categories. This is also true as bar graphs are specifically designed to allow easy comparison between different categories by displaying their respective values side by side. However, the reason does not directly explain why the height of the bar is proportional to the value it represents; it only describes the purpose of using bar graphs. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Pictograph

c) 250

[Solution Description]

One symbol = 100 cars. July has 2 full symbols and half of a symbol. Calculation: $2 \times 100 + 0.5 \times 100 = 200 + 50 = 250$.

Key Concept: Pictograph, Data Interpretation

b) 300

[Solution Description]

First, calculate the number of cars produced in July: $2.5 \times 100 = 250$ cars. Next, calculate the number of cars produced in August: $3 \times 100 = 300$ cars. The total number of cars produced in these two months is $250 + 300 = 550$ cars. Since the total production for all three months is 850 cars, the number of cars produced in September is $850 – 550 = 300$ cars.

Key Concept: Bar Graph Interpretation

b) 2003

[Solution Description]

To find the year with the maximum increase in the number of students, we need to calculate the difference in the number of students for each consecutive year. Suppose the data is as follows:

2001: 500 students,

2002: 600 students,

2003: 750 students,

2004: 800 students,

2005: 900 students,

2006: 1000 students.

Now, calculate the differences:

2002 – 2001 = 600 – 500 = 100,

2003 – 2002 = 750 – 600 = 150,

2004 – 2003 = 800 – 750 = 50,

2005 – 2004 = 900 – 800 = 100,

2006 – 2005 = 1000 – 900 = 100.

The maximum increase is 150, which occurred in 2003.

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 Interpretation

b) Performance in Math has improved

[Solution Description]

The double bar graph shows the average marks for each subject in two exams. For Math, the average marks increased from 75 in Exam 1 to 85 in Exam 2. This indicates an improvement in the performance of students in Math.

Key Concept: Double Bar Graph Comparison

c) Product A outperformed Product B in all quarters except one

[Solution Description]

The double bar graph compares the sales of Product A and Product B over four quarters. Since Product A’s sales were higher in Q1, Q2, and Q4 but lower in Q3, it implies that Product A generally outperformed Product B, except in Q3 where Product B performed better.

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: Double Bar Graph, Interpretation

c) 200

[Solution Description]

The total sales for Product X over the four quarters is $200 + 250 + 300 + 350 = 1100$. The total sales for Product Y over the four quarters is $150 + 200 + 250 + 300 = 900$. The total difference in sales between Product X and Product Y over the four quarters is $1100 – 900 = 200$.

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: Pie Chart Representation

b) 144$^\circ$

[Solution Description]

To find the central angle for the sector representing ‘Mathematics’, we use the formula:

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

Given that the percentage for ‘Mathematics’ is 40%, we substitute this value into the formula:

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

Therefore, the central angle for ‘Mathematics’ is $144^\circ$.

Key Concept: Complex Data Interpretation, Pie Chart Representation

b) 60°

[Solution Description]

Step 1: Calculate the fraction of people who play sports. Fraction = $\frac{60}{360} = \frac{1}{6}$.

Step 2: Calculate the central angle for ‘Playing Sports’. Central angle = $\frac{1}{6} \times 360° = 60°$.

Thus, the central angle for the ‘Playing Sports’ category is 60°.

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

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: 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: Total Angle Verification

b) 70°

[Solution Description]

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

$360° – 90° – 60° = 210°$

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

$\frac{210°}{3} = 70°$

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

Key Concept: Pie Chart Construction

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

[Solution Description]

To construct a pie chart, we first calculate the central angle for each sector. The total angle at the center of a circle is 360°. For a sector representing 50% of the data, the central angle is calculated as follows:

$\text{Central Angle} = \frac{50}{100} \times 360° = 180°$

The assertion states that the central angle for a sector representing 50% of the data is 180°, which is correct. The reason explains that the total angle at the center of a circle is 360° and the central angle is calculated as a fraction of 360°, which is also correct. Moreover, 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: Interpretation of Pie Chart

b) $60^\circ$

[Solution Description]

To find the angle for “Play” , first calculate the total hours spent in a day: $8 + 6 + 4 + 4 + 2 = 24$ hours. The angle for each activity is calculated using the formula $\text{angle} = (\text{hours spent on activity} / \text{total hours}) \times 360^\circ$. For “Play”, the calculation is $(4 / 24) \times 360^\circ = 60^\circ$.

Key Concept: Pie Chart Representation, Data Interpretation

c) 6 hours

[Solution Description]

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

Given:

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

The fraction of time spent on Leisure is:

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

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

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

Therefore, the time spent on Leisure is 6 hours.

Key Concept: Pie Chart Representation

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

[Solution Description]

To verify the assertion, we first calculate the fraction of time spent on “Play”. The given data shows that a child spends 4 hours on “Play” out of a total of 24 hours in a day. Therefore, the fraction is:

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

Next, we determine the central angle for the “Play” sector. Since the total angle in a circle is 360°, the central angle for “Play” is:

$\frac{1}{6} \times 360° = 60°$

Thus, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Pie Chart – Equal Expenditure and Savings

b) 15%

[Solution Description] According to the syllabus, the expenditure on education is equal to the savings of the family, which is 15% each.

Key Concept: Pie Chart, Savings Calculation

c) \$50,000

[Solution Description]

Given that 15% of the total income is spent on education and it equals \$7,500, we can find the total income:

$15\% \times \text{Total Income} = 7,500 \\ \\ 0.15 \times \text{Total Income} = 7,500 \\ \\ \text{Total Income} = \frac{7,500}{0.15} = \$50,000$

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: Fraction Calculation

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

[Solution Description]

To find the fraction of sales for biscuits:

1. Divide the sales of biscuits by the total sales: $\frac{120}{720} = \frac{1}{6}$.

The fraction representing biscuit sales is $\frac{1}{6}$.

Key Concept: Central Angle Calculation

b) 108°

[Solution Description]

To find the central angle for the strawberry sector, we first convert the percentage to a fraction. 30% is equivalent to $\frac{30}{100} = \frac{3}{10}$. The total angle at the center of the circle is 360°. The central angle for the strawberry sector is calculated as $\frac{3}{10} \times 360° = 108°$.

Key Concept: Angle Comparison

c) 2:1

[Solution Description]

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

Key Concept: Fraction 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: 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: 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 a fair coin has two equally likely outcomes: head and tail. The reason given is that in a fair coin toss, both head and tail are equally likely outcomes. This is also true and directly explains why the probability of getting a head is $\frac{1}{2}$. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Equally likely outcomes, Events

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

[Solution Description]

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

The probability of drawing the first red ball is:

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

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

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

The combined probability of both events occurring is:

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

Key Concept: Chance and Probability, Real-Life Applications

b) 0.2

[Solution Description]

The probability that the train is on time is 80%, which can be written as $P(\text{On Time}) = 0.8$. Since the total probability must sum to 1, the probability that the train is late is calculated as:

$P(\text{Late}) = 1 – P(\text{On Time}) = 1 – 0.8 = 0.2$

Therefore, the probability that the train is late on any given day is 0.2 or 20%.

Key Concept: Chance and Probability related to real life

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that the probability of rain on the day you forget your raincoat is higher than on any other day. This is not necessarily true because forgetting a raincoat does not influence the actual weather conditions. The reason suggests that forgetting your raincoat increases the chance of rain by changing weather conditions, which is incorrect as personal actions do not affect meteorological events. Therefore, the assertion is false, and the reason is also false.

Key Concept: Equally Likely Outcomes

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

[Solution Description]

A fair die has six faces with numbers 1, 2, 3, 4, 5, and 6. The even numbers on the die are 2, 4, and 6, giving 3 favorable outcomes. Thus, the probability of getting an even number is $\frac{3}{6} = \frac{1}{2}$.

Key Concept: Probability, Equally Likely Outcomes

b) $\frac{5}{33}$

[Solution Description]

First, find the total number of ways to draw two balls from the bag. The total number of balls is 5 red + 7 blue = 12 balls. The number of ways to choose 2 balls out of 12 is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$. So, $C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66$. Next, find the number of favorable outcomes where both balls drawn are red. There are 5 red balls, so the number of ways to choose 2 red balls out of 5 is $C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$. Therefore, the probability of drawing two red balls is $\frac{10}{66} = \frac{5}{33}$.

Key Concept: Probability of an Event

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

[Solution Description]

There are 13 hearts in a deck of 52 playing cards. The total number of outcomes is 52. The number of favorable outcomes for drawing a heart is 13. Therefore, the probability is calculated as $\frac{13}{52} = \frac{1}{4}$.

Key Concept: Probability of Not Getting an Event

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

[Solution Description]

There are 4 possible outcomes, one for each color. The number of favorable outcomes for not landing on blue is 3 (yellow, green, and red). Therefore, the probability is calculated as $\frac{3}{4}$.

Key Concept: Equally Likely Outcomes

a) $\binom{100}{50} \left(\frac{1}{2}\right)^{100}$

[Solution Description]

For a fair coin, the probability of getting a head in a single toss is $\frac{1}{2}$. Since each toss is independent, the probability of getting exactly 50 heads in 100 tosses can be calculated using the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where $n = 100$, $k = 50$, and $p = \frac{1}{2}$. Plugging in the values:

$P(X = 50) = \binom{100}{50} \left(\frac{1}{2}\right)^{50} \left(\frac{1}{2}\right)^{50} = \binom{100}{50} \left(\frac{1}{2}\right)^{100}$

Therefore, the probability of getting exactly 50 heads is $\binom{100}{50} \left(\frac{1}{2}\right)^{100}$.

Key Concept: Equally Likely Outcomes, Probability of Multiple Events

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

[Solution Description]

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

Key Concept: Equally Likely Outcomes

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

[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: Probability of Getting a Tail

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 an event is given by the formula:

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

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:

$\text{Probability of Tail} = \frac{1}{2}$

Key Concept: Probability of Events

b) 0.5

[Solution Description]

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

Key Concept: Probability, Combinations

b) $\frac{5}{16}$

[Solution Description]

The total number of possible outcomes when tossing a coin 5 times is $2^5 = 32$. The number of ways to get exactly 3 heads is given by the combination formula $\binom{5}{3} = 10$. Therefore, the probability is $\frac{10}{32} = \frac{5}{16}$.

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: Equally likely outcomes, Probability of events

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

[Solution Description]

The total number of possible outcomes when a die is rolled three times is $6 \times 6 \times 6 = 216$. We need to find the number of favorable outcomes where the sum of the three rolls is exactly 10. This requires identifying all combinations of three numbers between 1 and 6 that add up to 10. After listing all valid combinations, we find there are 27 such combinations. Therefore, the probability is $\frac{27}{216} = \frac{1}{8}$.

Key Concept: Equally Likely Outcomes

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

[Solution Description]

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

Key Concept: Probability, 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 $3 + 1 + 1 = 5$. The number of unfavorable outcomes (blue sector) is 1. The number of favorable outcomes (not getting a blue sector) is $5 – 1 = 4$. The probability of not getting a blue sector is calculated as $\frac{\text{Number of non-blue sectors}}{\text{Total number of sectors}} = \frac{4}{5}$.

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

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

[Solution Description]

The bag contains a total of $(5 + 3) = 8$ balls. Out of these, the number of red balls is 5. The probability of drawing a red ball is the ratio of the number of red balls to the total number of balls in the bag. Therefore, the probability of drawing a red ball is $\frac{5}{8}$.

Key Concept: Probability as a Fraction, 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 drawing a red ball from a bag containing 4 red balls and 2 yellow balls is $\frac{2}{3}$. To verify this, we calculate the probability of drawing a red ball.

Step 1: Identify the total number of equally likely outcomes. Here, there are 4 red balls and 2 yellow balls, so the total number of outcomes is $4 + 2 = 6$.

Step 2: Identify the number of favorable outcomes for drawing a red ball. There are 4 red balls, so the number of favorable outcomes is 4.

Step 3: Calculate the probability using the formula:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{6} = \frac{2}{3}.$

Thus, the assertion is true.

The reason states that the probability of an event is given by the ratio of the number of favorable outcomes to the total number of equally likely outcomes. This is the correct definition of probability when outcomes are equally likely, which is true in this case. Therefore, the reason is true, and it correctly explains the assertion.

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

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

[Solution Description]

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

Key Concept: Events in Probability, Probability Calculation

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

[Solution Description] A fair die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The event of getting an even number includes the outcomes {2, 4, 6}. Therefore, the number of favorable outcomes for this event is 3. Since all outcomes are equally likely, the probability of getting an even number is calculated as $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

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: 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 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

b) $\frac{14}{45}$

[Solution Description]

To find the probability that both balls are of the same color, we need to consider three cases: both balls are red, both balls are blue, or both balls are green.

First, calculate the total number of ways to draw 2 balls out of 10:

$\text{Total ways} = \binom{10}{2} = \frac{10 \times 9}{2} = 45$

Next, calculate the number of favorable outcomes for each case:

1. Both balls are red:

$\binom{3}{2} = \frac{3 \times 2}{2} = 3$

2. Both balls are blue:

$\binom{2}{2} = \frac{2 \times 1}{2} = 1$

3. Both balls are green:

$\binom{5}{2} = \frac{5 \times 4}{2} = 10$

Now, add up all the favorable outcomes:

$\text{Favorable outcomes} = 3 + 1 + 10 = 14$

Finally, calculate the probability:

$\text{Probability} = \frac{14}{45}$

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 Compound Events

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

[Solution Description]

A standard die has 6 faces with numbers 1 through 6. The prime numbers on a die are 2, 3, and 5. Therefore, there are 3 favorable outcomes out of 6 possible outcomes. The probability of getting a prime number is $P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$.

Key Concept: Compound Events, Probability

c) $\frac{2}{9}$

[Solution Description]

The total number of balls initially is $5 + 3 + 2 = 10$. The probability of drawing the first red ball is $\frac{5}{10} = \frac{1}{2}$. After drawing one red ball, there are 4 red balls left and the total number of balls becomes 9. The probability of drawing a second red ball is $\frac{4}{9}$. Therefore, the probability of both events occurring in succession is $\frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9}$.

Key Concept: Probability of Compound Events

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

[Solution Description]

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

Key Concept: Probability, Real-Life Applications

a) 0.729

[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 the days is $\left(\frac{9}{10}\right)^3$. Calculating this gives:

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

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: Unequal Chances

c) A train being on time or late

[Solution Description]

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

Key Concept: 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: 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 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: Probability, Real-Life Application

c) $\frac{81}{100}$

[Solution Description]

The probability that it does not rain on a single day is $1 – \frac{1}{10} = \frac{9}{10}$. Since the occurrence of rain on different days is independent, the probability that it does not rain on two consecutive days is:

$\frac{9}{10} \times \frac{9}{10} = \frac{81}{100}$

Therefore, the probability that it does not rain on two consecutive days is $\frac{81}{100}$.

Key Concept: Probability of Independent Events

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

[Solution Description]

For independent events, the probability of both events occurring together is the product of their individual probabilities. Here, the probability of a voter being female is $\frac{1}{2}$ and the probability of supporting Candidate C is $\frac{1}{4}$. Thus, the probability of both events occurring together is $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

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, Random Experiment

b) 0.171

[Solution Description]

To find the probability of exactly 3 defective items in a batch of 100, we use the binomial probability formula:

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

Where:

– $n = 100$ (total items)

– $k = 3$ (defective items)

– $p = 0.02$ (probability of defect)

First, calculate $C(100, 3)$:

$C(100, 3) = \frac{100!}{3!(100-3)!} = \frac{100 \times 99 \times 98}{3 \times 2 \times 1} = 161700$

Next, calculate $p^k$:

$0.02^3 = 0.000008$

Then, calculate $(1-p)^{n-k}$:

$(1-0.02)^{97} = 0.98^{97} \approx 0.1326$

Finally, multiply all values:

$P(3) = 161700 \times 0.000008 \times 0.1326 \approx 0.171$

Key Concept: Probability of Defective Items

d) 0.188

[Solution Description]

To solve this problem, we use the binomial probability formula:

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

where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success on a single trial.

Here, $n = 20$, $k = 2$, and $p = 0.05$. Plugging in these values:

$P(X = 2) = C(20, 2) \times (0.05)^2 \times (0.95)^{18}$

Calculating each part:

$C(20, 2) = \frac{20!}{2! \times 18!} = 190$

$(0.05)^2 = 0.0025$

$(0.95)^{18} \approx 0.397$

Therefore:

$P(X = 2) = 190 \times 0.0025 \times 0.397 \approx 0.188$

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 in Real Life

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 particular day during the rainy season is $\frac{1}{10}$. This means, out of 10 days, it rains on one day on average. The reason given is that the probability of no rain on a particular day during the rainy season is $\frac{9}{10}$. This implies that out of 10 days, it does not rain on nine days. The reason correctly explains the assertion since if the probability of rain is $\frac{1}{10}$, then the probability of no rain is $1 – \frac{1}{10} = \frac{9}{10}$. Both statements are true, and the reason is the correct explanation of the assertion.

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: Probability of events

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

[Solution Description]

A fair die has six faces numbered from 1 to 6. The numbers less than 3 are 1 and 2. So, there are 2 favorable outcomes out of 6 possible outcomes. Therefore, the probability is given by the formula:

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

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: 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: Linking chances to probability

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

[Solution Description]

When a die is rolled once, there are six possible equally likely outcomes: 1, 2, 3, 4, 5, and 6. Out of these, the even numbers are 2, 4, and 6, which means there are three favorable outcomes. The probability of getting an even number is calculated as the ratio of favorable outcomes to total outcomes, which is $\frac{3}{6}$. Simplifying this fraction gives $\frac{1}{2}$. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Probability, Events

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

[Solution Description]

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

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

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

Key Concept: Probability, 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: Probability

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

[Solution Description]

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

Key Concept: Random Experiment 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: Random Experiment, Probability

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

[Solution Description]

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