Data Handling

A cricket scorecard, a school report card, a weather bulletin, an election result — every one of these is a page of raw numbers waiting to become a picture a person can actually read. Data Handling is the chapter that teaches you how to make that picture, and then how to reason sensibly about situations where the outcome cannot be predicted in advance. The chapter has two halves that feel different but share one habit of mind: organise first, then read off the answer. The first half is about representing data you already have. You revise the pictograph and the bar graph from earlier classes, meet the double bar graph for comparing two data sets side by side, and then learn the star of this chapter — the circle graph, or pie chart — which shows how a whole quantity splits into parts by turning percentages or fractions into slices of a circle, each slice carved out by a central angle that is a fraction of the full 360 degrees around the centre. The second half turns to chance. Tossing a coin, throwing a die, spinning a wheel, drawing a ball from a bag without looking — these are random experiments, and although you cannot control any single result, Class 8 mathematics gives you exact language for the situation: outcomes, events, and when every outcome is just as likely as every other, a precise number called probability, computed as the count of favourable outcomes divided by the total count of outcomes. By the end of this chapter you will be able to build a pie chart from a percentage table using nothing more than the fraction-of-360-degrees idea, and you will be able to compute the probability of an event for coins, dice, spinners and balls-in-a-bag with total confidence, because both skills come down to the same underlying question: out of everything that could happen, what fraction does the part I care about occupy?

Key takeaways

Data is information collected for a purpose, and it must be organised before it can be read or interpreted.
A pictograph uses repeated symbols and a bar graph uses bars of equal width and equal spacing to represent frequencies; a double bar graph places two bars side by side per category to compare two data sets at once.
A circle graph (pie chart) shows how a whole splits into parts; the whole circle stands for the total, and each part gets a sector whose central angle is that part's fraction of $360°$ .
The central-angle rule: central angle of a sector $= \dfrac{\text{value of that part}}{\text{total value}} \times 360°$ , and the angles of all sectors always add up to exactly $360°$ .
A random experiment is one whose exact outcome cannot be predicted in advance, such as tossing a coin or throwing a die; the possible results are its outcomes.
Outcomes are equally likely when each one has exactly the same chance of occurring, which is what makes coins, dice, well-shuffled cards, and evenly divided spinners fair.
One or more outcomes together form an event; for a die, "getting an even number" is the event made of the outcomes 2, 4 and 6.
The probability of an event, when outcomes are equally likely, is $P(E) = \dfrac{\text{number of outcomes favourable to } E}{\text{total number of equally likely outcomes}}$ , a number always between 0 and 1.
Key idea: turning a whole into readable parts and turning uncertainty into a number both reduce to the same fraction — favourable part over total — expressed as a central angle out of 360 degrees for pie charts, and as a count of outcomes out of all outcomes for probability.

The core insight

Whether you are slicing a circle into a pie chart or counting outcomes for a probability, the entire chapter is one recurring computation — the fraction that one part occupies of the whole — read off as a central angle out of $360°$ for data, or as a count out of the total number of equally likely outcomes for chance.

What this chapter covers

This chapter builds in three connected stages, exactly as it does in the NCERT textbook. Section 4.1, Looking for Information, revises how raw facts become data and how that data is displayed using a pictograph, a bar graph, and a double bar graph for comparing two related sets of figures side by side — the tools you already met in earlier classes, now used as a springboard. Section 4.2, Circle Graph or Pie Chart, is the new geometric idea of the chapter: representing a whole and its parts using the sectors of a circle, computing each sector's central angle as a fraction of $360°$ , and actually drawing a pie chart using a protractor. Section 4.3, Chance and Probability, moves from certain data to uncertain events: what a random experiment is, what makes outcomes equally likely, what an event is, and how to attach an exact number — the probability — to the chance of an event happening.

The core ideas

Organising data: pictographs, bar graphs and double bar graphs

Data is information collected in the context of some situation we want to study — runs scored by a batsman, marks in a test, the number of story books read by each of your friends. Raw data by itself is hard to read; it becomes useful only once it is organised systematically, often as a table of values against categories, and then displayed graphically so a reader can grasp the pattern at a glance. A pictograph represents data using repeated picture symbols, where one symbol stands for a fixed quantity — one car icon standing for 100 cars, so 250 cars would be shown as two and a half icons. A bar graph displays data using bars of uniform width with equal gaps between them, where the height of each bar is directly proportional to the value it represents. A double bar graph places two bars per category side by side, the natural choice whenever you must compare two related data sets over the same categories — say, student numbers in a school across two years. The key reading skill is the same for all three: identify what one unit of the graph represents, then compare quantities using that common scale.

Circle graphs (pie charts) and the central angle

A circle graph, or pie chart, shows the relationship between a whole and its parts. The entire circle represents the total quantity, and it is divided into sectors, one for each part, with the size of each sector made proportional to the fraction of the whole that part represents. Since a full circle subtends an angle of $360°$ at its centre, the fraction that a part occupies of the whole is converted directly into an angle at the centre — the central angle of that sector.

\text{central angle of a sector} = \left(\dfrac{\text{value of the part}}{\text{total value of all parts}}\right) \times 360°. \qquad (1)

where: the value of the part and the total value must be in the same unit (rupees with rupees, students with students, percentage with percentage), so the fraction in the brackets is a pure number between 0 and 1 and the central angle comes out in degrees. Because every part's fraction is taken out of the same total, the central angles of all sectors in one pie chart must always add up to exactly $360°$ — the single most useful self-check for any pie chart you draw or read. As a first illustration, take the favourite ice-cream flavours of a group of students: Chocolate is preferred by $50\%$ , Vanilla by $25\%$ , and other flavours together by the remaining $25\%$ . Writing each percentage as a fraction of the whole and multiplying by $360°$ gives Chocolate $= \tfrac{1}{2}\times 360° = 180°$ , Vanilla $= \tfrac{1}{4}\times 360° = 90°$ , and Other flavours $= \tfrac{1}{4}\times 360° = 90°$ . These three angles add to $180° + 90° + 90° = 360°$ , confirming the split accounts for the whole class.

Figure

A pie chart of ice-cream flavour preferences with chocolate occupying half the circle at 180 degrees and vanilla and other flavours each occupying a quarter at 90 degrees

Random experiments, equally likely outcomes and events

A random experiment is a trial whose exact result cannot be predicted in advance, even though every possible result is known beforehand — tossing a coin, throwing a die, or drawing a ball from a bag without looking. Each possible result is called an outcome. Tossing a coin has two outcomes, Head and Tail; throwing an ordinary die has six outcomes, 1 through 6; spinning a wheel with coloured sectors has one outcome per sector. Outcomes are called equally likely when each one has exactly the same chance of occurring as every other. A fair coin and a fair die give equally likely outcomes by construction, confirmed experimentally: toss a coin a very large number of times and the number of heads and tails drift closer together as the tosses grow — the chapter's own tally-mark table shows this directly, moving from 27 heads out of 50 tosses to 52 heads out of 100, converging toward a fifty-fifty split. One outcome, or a collection of outcomes, makes an event. In throwing a die, getting the outcome 4 is an event, and so is "getting an even number," which is the collection of the three outcomes 2, 4 and 6. Recognising which outcomes belong to an event is the whole skill needed for the next idea, probability.

Figure

A spinning wheel divided into five equal sectors of which three are green one is blue and one is red used to illustrate equally likely outcomes

Probability as a fraction of equally likely outcomes

When the outcomes of a random experiment are equally likely, the probability of an event is defined as the number of outcomes that make up the event divided by the total number of possible outcomes.

P(E) = \dfrac{\text{number of outcomes favourable to the event } E}{\text{total number of equally likely outcomes}}. \qquad (2)

where: $P(E)$ denotes the probability of event $E$ , always a number satisfying $0 \le P(E) \le 1$ ; $P(E) = 0$ means the event can never happen and $P(E) = 1$ means it is certain to happen. For a single coin toss, getting a Head has probability $\tfrac{1}{2}$ because 1 outcome out of 2 equally likely outcomes favours it. For a single die throw, getting a "2" has probability $\tfrac{1}{6}$ , while getting an even number has probability $\tfrac{3}{6} = \tfrac{1}{2}$ , since 3 of the 6 equally likely outcomes (2, 4, 6) favour that event. This one formula is the entire content of the probability half of this chapter — every question reduces to correctly counting the favourable outcomes and the total outcomes.

Derivation AFinding the central angle of a pie-chart sector

The goal of this method is to turn a table of raw values — rupee amounts, vote counts, percentages, anything that is one part of a larger total — into the angle marked off at the centre of a circle, so a protractor can draw the pie chart while guaranteeing the finished chart is internally consistent, meaning its sector angles add up to exactly one full turn. We use the bakery-shop sales data from the textbook, where the total sale for one day across five items is ₹720, to carry the method through completely and check the arithmetic at the end.

Step 1. Find the total of all the values, since every sector's angle is measured as a fraction of this one number. Here the sales were Ordinary Bread ₹320, Biscuits ₹120, Cakes and pastries ₹160, Fruit Bread ₹80 and Others ₹40, giving a total of $320 + 120 + 160 + 80 + 40 = 720$ rupees.

Step 2. Write each part's value as a fraction of the total, simplifying where convenient. Ordinary Bread is $\tfrac{320}{720} = \tfrac{4}{9}$ of the sales, Biscuits is $\tfrac{120}{720} = \tfrac{1}{6}$ , Cakes and pastries is $\tfrac{160}{720} = \tfrac{2}{9}$ , Fruit Bread is $\tfrac{80}{720} = \tfrac{1}{9}$ , and Others is $\tfrac{40}{720} = \tfrac{1}{18}$ .

Step 3. Multiply each fraction by $360°$ to convert it into a central angle, using equation (1). Ordinary Bread gets $\tfrac{4}{9}\times 360° = 160°$ ; Biscuits gets $\tfrac{1}{6}\times 360° = 60°$ ; Cakes and pastries gets $\tfrac{2}{9}\times 360° = 80°$ ; Fruit Bread gets $\tfrac{1}{9}\times 360° = 40°$ ; and Others gets $\tfrac{1}{18}\times 360° = 20°$ .

Step 4. Verify the angles by adding them: $160° + 60° + 80° + 40° + 20° = 360°$ , exactly one full turn, which confirms no arithmetic slip was made. Only after this check should the sectors be marked off with a protractor, starting from any radius and laying off each angle in turn around the circle so that the last sector closes back onto the starting radius. Method summary: the four steps — total, fraction, angle, check-that-angles-sum-to-360 — apply unchanged to any pie-chart question, whether the raw data is in rupees, percentages, or a plain head-count.

Go deeper · Derivation B · 1 minThe probability formula for equally likely outcomes

Derivation BThe probability formula for equally likely outcomes

The goal of this derivation is to show, rather than merely state, why the probability of an event equals the count of favourable outcomes divided by the total count of outcomes whenever those outcomes are equally likely — and to confirm, as a built-in check, that this rule makes the probabilities of all outcomes of an experiment add up to exactly 1, matching the certainty that some outcome must occur.

Step 1. Suppose a random experiment has $n$ possible outcomes, and suppose they are equally likely, meaning each one carries exactly the same chance $p$ of occurring. Since something must happen every time the experiment is performed, the chances of all $n$ outcomes together must account for total certainty, so $n \times p = 1$ , which gives $p = \tfrac{1}{n}$ for each single outcome.

Step 2. Now consider an event $E$ that consists of $k$ of these $n$ outcomes (for instance, "getting an even number" on a die consists of the 3 outcomes 2, 4 and 6, out of $n = 6$ total outcomes). Because the $k$ outcomes making up $E$ are mutually exclusive — only one outcome happens on any single trial — the chance of $E$ is the sum of the chances of its $k$ individual outcomes: $P(E) = \underbrace{p + p + \cdots + p}_{k \text{ times}} = k \times p = k \times \dfrac{1}{n} = \dfrac{k}{n}$ , which is exactly equation (2).

Step 3. Test the formula on a coin: $n = 2$ equally likely outcomes (Head, Tail), and the event "getting a Head" has $k = 1$ favourable outcome, so $P(\text{Head}) = \tfrac{1}{2}$ , matching the known fair-coin result.

Step 4. Test it on a die and check the complementary-event property. For "getting an even number," $n = 6$ and $k = 3$ , so $P(\text{even}) = \tfrac{3}{6} = \tfrac{1}{2}$ . The complementary event "getting an odd number" also has $k = 3$ outcomes (1, 3, 5), so $P(\text{odd}) = \tfrac{1}{2}$ as well, and $P(\text{even}) + P(\text{odd}) = \tfrac{1}{2} + \tfrac{1}{2} = 1$ — every outcome is either even or odd, so the two probabilities must add to certainty, confirming the formula is internally consistent.

Worked examples

Board level - Finding an unknown expenditure from a household pie chart

A family's monthly expenditure and savings, shown as a pie chart, break up as follows: Food 30%, House Rent 20%, Clothes 10%, Education of children 15%, Other expenses 10%, and Savings 15%. The monthly savings of the family is ₹3000.

Step 1. Identify the item on which expenditure was maximum by comparing all six percentages directly: 30%, 20%, 10%, 15%, 10%, 15%. The largest is 30%, for Food, so the expenditure was maximum on Food.

Step 2. Identify which expenditure item equals the family's savings. Savings is 15% and Education of children is also 15%, so the expenditure on Education of children equals the family's total savings.

Step 3. Use the given rupee value of one percentage slice to find another. Savings is 15% of the monthly amount and equals ₹3000, so 1% of the monthly amount is $\tfrac{3000}{15} = 200$ rupees.

Step 4. Multiply to get the expenditure on Clothes, which is 10% of the monthly amount: $10 \times 200 = 2000$ rupees. The monthly expenditure on clothes is ₹2000.

Board level - Probability of drawing a red ball from a bag

A bag has 4 red balls and 2 yellow balls, identical in every way except colour. A ball is drawn from the bag without looking. Find the probability of getting a red ball, and state whether it is more or less than the probability of getting a yellow ball.

Step 1. Count the total number of outcomes. There are $4 + 2 = 6$ balls in all, and since the balls are identical except for colour, drawing any one of them is equally likely, so there are 6 equally likely outcomes.

Step 2. Count the outcomes favourable to "getting a red ball." There are 4 red balls, so 4 outcomes are favourable.

Step 3. Apply the probability formula: $P(\text{red}) = \dfrac{4}{6} = \dfrac{2}{3}$ .

Step 4. Repeat for yellow: 2 outcomes are favourable out of 6, so $P(\text{yellow}) = \dfrac{2}{6} = \dfrac{1}{3}$ . Comparing $\tfrac{2}{3}$ with $\tfrac{1}{3}$ , the probability of getting a red ball, 2/3, is more than the probability of getting a yellow ball, 1/3.

Higher-order thinking (HOTS / olympiad-style) - Combining a pie chart with a probability question

At a sports day, 720 students were asked to name their favourite sport, and the results were displayed as a pie chart with these central angles: Cricket 150°, Football 110°, Basketball 60°, and the remaining students preferring Badminton. Find the number of students who prefer each sport, and find the probability that a student picked at random from this group prefers Cricket.

Step 1. Find the central angle for Badminton using the full-circle check from the derivation above: the three given angles add to $150° + 110° + 60° = 320°$ , so Badminton's angle is $360° - 320° = 40°$ .

Step 2. Convert each central angle back into a number of students using the same fraction-of-360 idea in reverse: number of students $= \dfrac{\text{central angle}}{360°} \times 720$ . For Cricket, $\dfrac{150}{360} \times 720 = 300$ students. For Football, $\dfrac{110}{360} \times 720 = 220$ students. For Basketball, $\dfrac{60}{360}\times 720 = 120$ students. For Badminton, $\dfrac{40}{360}\times 720 = 80$ students.

Step 3. Check the four counts add back to the total: $300 + 220 + 120 + 80 = 720$ , matching the number of students surveyed, so the conversion is consistent.

Step 4. Treat picking a student at random as a random experiment with 720 equally likely outcomes (any one of the 720 students), of which 300 are favourable to "prefers Cricket." So $P(\text{Cricket}) = \dfrac{300}{720} = \dfrac{5}{12}$ . Notice this equals the central-angle fraction directly, $\dfrac{150°}{360°} = \dfrac{5}{12}$ — the probability of a randomly chosen student preferring Cricket is 5/12, and it could have been read straight off the pie chart's angle without ever finding the head-count first. Timing note: this problem has three linked parts — recovering a missing angle, converting angles to counts, and converting a count to a probability — so budget about 4 to 5 minutes in an exam; if time is short, note that the probability answer can be obtained directly from the central angle as (angle/360°) without computing the head-count at all.

Common traps

Forgetting to check that a completed pie chart's central angles add up to $360°$ — a single arithmetic slip otherwise goes unnoticed until the drawing visibly fails to close.
Mixing units when finding a fraction for a central angle, such as dividing a rupee value by a total given in percentage, instead of first converting both to the same unit.
In bar graphs, misreading the axis scale — assuming each grid square equals one unit when the graph's key states it equals five or ten.
Confusing "outcomes" with "events": an outcome is one single result (getting a 4), while an event can be a collection of outcomes (getting an even number).
Applying $P(E) = \tfrac{\text{favourable}}{\text{total}}$ to outcomes that are not actually equally likely, such as a biased coin — the formula only holds when every outcome has the same chance.
Miscounting the total number of outcomes, especially with dice and cards — a die has 6 faces, a standard deck has 52 cards.
Writing a probability as a percentage or raw count instead of the required fraction between 0 and 1.

Common trapThe single biggest misconception: treating every random situation as if all outcomes were automatically equally likely

The probability formula $P(E) = \tfrac{\text{favourable outcomes}}{\text{total outcomes}}$ only works when the outcomes genuinely have the same chance of occurring. It is tempting to apply it blindly — for instance, to say a cricket team has a $\tfrac{1}{2}$ chance of winning simply because there are two possible results, win or lose. But win and lose are not equally likely outcomes; one team may be far stronger than the other. This chapter's formula is reserved for situations built to be fair — a balanced coin, an unbiased die, a well-shuffled deck, identical balls in a bag, or a spinner whose sectors are equal in size. Before writing down the formula, always check, or be told, that the outcomes are equally likely; if they are not, this simple counting rule does not apply.

You should be able to…

Read and interpret a pictograph, bar graph, and double bar graph, correctly identifying the scale each graph uses.
Explain what a circle graph or pie chart represents and why its sectors' angles must add up to 360 degrees.
Compute a sector's central angle from raw data or a percentage, using central angle = (part/total) × 360°.
Draw a pie chart accurately with a compass and protractor from a table of values or percentages.
Read a pie chart back into numbers, recovering percentages, angles, or head-counts from one another.
Identify whether an experiment is random, and list its outcomes correctly.
Decide whether outcomes are equally likely, and explain why coins, dice and well-shuffled cards qualify.
Identify the outcomes making up a given event, distinguishing a single outcome from a multi-outcome event.
Compute an event's probability as P(E) = favourable outcomes / total outcomes, as a fraction between 0 and 1.
Connect a pie chart's central angle directly to a probability, since angle/360° gives the same fraction.

Ask the tutor: Want to work through drawing your own pie chart from a data table, or practise a few more probability problems with dice, coins and bags of coloured balls?

What is the actual difference between a bar graph and a pie chart, since both show quantities visually?

A bar graph compares separate quantities against a common scale using the heights of bars, and it works well even when the categories do not add up to any meaningful total — like comparing rainfall in five different cities. A pie chart, on the other hand, always shows how one single whole splits into parts, and every sector's size only makes sense relative to that one total, which is why the central angles of a pie chart must always add up to 360 degrees. If your data is "parts of a whole" (like a family's monthly budget), use a pie chart; if your data is a set of independent comparable values, use a bar graph.

Why is the total angle of a circle always 360 degrees, and does that ever change for a pie chart?

A full circle always measures 360 degrees at its centre, no matter what data the pie chart represents — this is a fixed geometric fact, not something that depends on the numbers in the table. What changes from one pie chart to another is only how that fixed 360 degrees gets divided among the sectors, based on the fractions each part contributes to the whole. So while the individual central angles change with the data, their sum never does; it is always 360 degrees, and that is exactly the check you should run on every pie chart you draw.

If I throw a die and get a 6, does that mean I am now less likely to get a 6 on my next throw?

No. Each throw of a fair die is an independent random experiment: the die has no memory of what happened before, so the probability of getting a 6 is exactly 1/6 on every single throw, regardless of what was rolled previously. This is exactly the kind of question the chapter raises directly — "does it mean the third player would not have a chance of getting a six?" — and the answer is that every player's throw is a fresh, independent experiment with the same six equally likely outcomes.

Can probability ever be a negative number, or greater than 1?

No, never. Probability is always a fraction of the total number of outcomes, so it can range only from 0 (the event is impossible, like getting a 7 on an ordinary die) up to 1 (the event is certain, like getting a number from 1 to 6 on that same die). If your calculation for a probability ever comes out negative or bigger than 1, it is a sure sign that a favourable count or a total count was set up incorrectly, and you should recheck which outcomes actually belong to the event and how many outcomes the experiment has in total.