Introduction to Probability, titled The Mathematics of Maybe in the new NCERT book, treats probability as a measurement of how likely an event is, on a scale from 0 (impossible) to 1 (certain). It explains randomness, then two objective ways to estimate probability: experimental probability from trials and survey data, and theoretical probability from equally likely outcomes. Students learn to write sample spaces and events, use samples to estimate for a population, avoid the Gambler’s Fallacy and list multi-step outcomes with tree diagrams. These counting skills return in Class 10.
Key Concepts
1. Probability, Randomness and Random Experiments
NCERT opens the chapter by calling probability a type of measurement. We measure length, area and volume for physical quantities. Probability measures something different: the likelihood of an event, or how confident we can be that it will happen.
Think of three everyday questions from the chapter:
- Is it going to rain today?
- Will our school win the inter-school hockey match tomorrow?
- Will I be chosen in the monthly lucky draw to perform at the school assembly? (The names of all students are written on slips of paper and one slip is picked at random.)
In each case we know the possible outcomes: it rains or it does not; the team wins, draws or loses; one student is chosen. What we do not know is which outcome will actually occur. Such events are called random events, because there is an element of chance every time they take place.
Randomness describes a situation or action where you cannot predict exactly what will happen, even though you may know all the possible outcomes.
- Tossing a coin: it could be heads or tails, and you cannot be sure which one will come up in a single toss.
- Rolling a die: the possible outcomes are 1, 2, 3, 4, 5 or 6, and you cannot say which number will appear on a particular roll.
Observations like these are called experiments or trials. NCERT defines a random experiment as something you can repeat, where the result might be different each time and you cannot know the outcome in advance.
Compare this with “Which day comes after Sunday?” The answer is always Monday. Nothing is uncertain, so it is not a random experiment.
Probability is the area of mathematics that studies randomness and how likely a specific outcome is in a random situation. When you toss a coin, the probability of heads is 1/2 and the probability of tails is 1/2, because each is equally likely. In a random experiment you can find the likelihood of each outcome, but you cannot know the exact result of the next trial.
Rain as a random event. Rain depends on many atmospheric factors (temperature, humidity, wind patterns, pressure) and is so sensitive to them that it cannot be predicted with total certainty. We can only estimate its likelihood from patterns in data.
Randomness can be useful. A coin toss decides which cricket team bats first. Nobody can predict the result and each captain has an equal chance, so it is fair.
Subjective probability. One friend says, “The sun is shining brightly, so it’s unlikely to rain today.” Another says, “It’s very hot, which makes me think it could rain later.” A likelihood based on personal judgement like this is the subjective probability of the event. The chapter then shows how to measure probability objectively, from evidence.
2. The Probability Scale
Probability is measured on a scale from 0 to 1. For any event E, the chapter summary writes this as 0 ≤ P(E) ≤ 1. A probability can also be written as a percentage by multiplying by 100.
| Probability | Word on the scale | Meaning for “our school wins the hockey match” |
|---|---|---|
| 0 | Impossible | Winning cannot happen, for example winning without playing |
| Between 0 and 0.5 | Less likely | Not winning is more likely than winning |
| 0.5 | Even chance (equally likely) | A 50% chance: winning and losing are equally likely |
| Between 0.5 and 1 (for example 0.75) | More likely | A 75% chance: more likely than not to win |
| 1 | Certain | The school is sure to win |
Most events have probabilities strictly between 0 and 1.
The deck of six cards. Imagine six cards, some purple and some green, with the split unknown. If none is purple, picking a purple card is impossible (probability 0). If all six are purple, it is certain (probability 1). As purple cards increase, the probability moves along the scale: 0, 1/6, 2/6, 3/6 (even chance), 4/6, 5/6, 1.
The chapter places these five events on the scale:
| Event | Place on the scale | Reason |
|---|---|---|
| Getting a number greater than 6 on a die | Impossible | A die only has the numbers 1 to 6 |
| Rolling a 3 on a standard die | Less likely | Possible, since one face shows 3, but only one face of six |
| Flipping a coin and getting heads | Even chance | Heads and tails are equally likely |
| Drawing any number from 2 to 10 from a deck of 52 cards | More likely | 36 of the 52 cards carry numbers from 2 to 10 |
| Choosing a red sweet from a bag of all red sweets | Certain | Every sweet is red |
Check: 36/52 ≈ 0.69 (more likely) and 1/6 ≈ 0.17 (less likely).
Two checks that catch errors. A probability can never be negative, and it can never be greater than 1. If your answer is −0.2 or 7/5, the working has gone wrong.
3. Experimental Probability and Relative Frequency
NCERT gives two main ways of estimating probability objectively:
- Using evidence from experience: collect data, either by performing an experiment many times or by analysing statistical data from past observations. The probability is the relative frequency of the event in that data. This is experimental probability (the CBSE course structure also calls it empirical probability).
- Using theoretical methods: assume all possible outcomes are equally likely and reason about favourable outcomes out of possible outcomes. This is theoretical probability (Concept 5).
In an experiment, each result is called an outcome, and the set of all possible outcomes is called the sample space. It is written inside curly brackets with outcomes separated by commas:
- A coin is tossed: possible outcomes Heads (H) or Tails (T). Sample space {H, T}.
- A die is rolled: the top face shows 1, 2, 3, 4, 5 or 6 dots. Sample space {1, 2, 3, 4, 5, 6}.
Experimental Probability = Number of times the event occurred / Total number of trials
NCERT Example 2. You roll a die 50 times and it lands on 4 exactly 8 times. The experimental probability of rolling a 4 is 8/50 = 0.16, or 16%. We also say the relative frequency of rolling a 4 is 8/50 = 0.16.
Keep two words apart. The frequency is the count (8 times). The relative frequency is the count divided by the total (8/50).
Activities from the chapter. Toss a coin 20 times, or toss a paper cup 100 times and record whether it lands on its bottom, its top or its side. The three landing positions are not alike, so there is no reason to call them equally likely; the cup’s probabilities must come from experiment.
4. Analysing Statistical Data and Sampling
Collecting evidence through statistical data is used widely in business (marketing, forecasting sales, insurance) and in science and social science research. The probability here is again a relative frequency, found from survey or record data.
NCERT Example 5: favourite fruit. 50 students in a class name their favourite fruit: 20 mango, 15 apple, 10 banana, 5 grapes. One student is picked at random.
Probability that the favourite fruit is mango = Number of students who like mango / Total number of students = 20/50 = 0.4, a 40% chance.
| Fruit | Number of students | Probability |
|---|---|---|
| Mango | 20 | 20/50 = 0.4 |
| Apple | 15 | 15/50 = 0.3 |
| Banana | 10 | 10/50 = 0.2 |
| Grapes | 5 | 5/50 = 0.1 |
| Total | 50 | 1.0 |
From sample to population. Now suppose you must buy fruit for the whole school of 1500 students. Asking every student is usually impractical. The whole school is the population (1500 students) and the class of 50 is a sample. Using the sample estimate, about 0.4 × 1500 = 600 students would prefer mango, so you would buy approximately 600 mangoes, and the rest would be apples, bananas and grapes.
Estimate for the population = probability from the sample × population size
Sampling. To be more confident about the estimate, choose a larger sample (say 100 students) and make it more representative, for example by including students from different classes and grades. This process is called sampling. Sample size and fairness both affect accuracy.
5. Theoretical Probability
Theoretical probability studies the likelihood of an event when all possible outcomes are equally likely. It describes what we expect in an ideal, perfectly fair situation, and needs no experiment or data. It is written P(Event) or P(Outcome).
Theoretical Probability P(E) = Number of favourable outcomes / Number of possible outcomes
A favourable outcome is an outcome that makes the event happen.
NCERT Example 3. A standard 6-sided die is rolled. Favourable outcomes for “getting a 4”: 1 (only the number 4). Possible outcomes: 6. P(rolling a 4) = 1/6 = 0.1666… ≈ 0.167, or 16.7%.
NCERT Example 4. A letter is picked at random from the word PROBABILITY. The word has 11 letters (P, R, O, B, A, B, I, L, I, T, Y), and 2 of them are B. P(picking B) = 2/11 = 0.1818… ≈ 0.182, or 18.2%. Every letter position is counted, so a repeated letter counts as many times as it appears.
Fair and unbiased. When we speak of a coin, we assume it is fair: it is symmetrical, so there is no reason for it to land more often on one side. This property is called being unbiased. A random toss means the coin falls freely without any bias or interference. Theoretical probability depends on this assumption.
Checking for equally likely outcomes. The formula works only when every outcome in the list has the same chance.
| Experiment | Outcomes listed | Equally likely? |
|---|---|---|
| Tossing a fair coin once | H, T | Yes, by symmetry |
| Rolling a fair 6-sided die | 1, 2, 3, 4, 5, 6 | Yes, by symmetry |
| A driver attempts to start a car | Starts, does not start | No: a working car usually starts |
| A marble from 3 red and 7 blue | Red, Blue | No: 7 blue marbles against 3 red |
For the marble bag, each of the 10 marbles is equally likely, so count marbles: P(red) = 3/10 and P(blue) = 7/10. Counting the colours as two outcomes would wrongly give 1/2 each.
The event not happening. Every outcome either makes E happen or does not. So the favourable outcomes for “not E” = n(S) − favourable outcomes for E, and P(not E) = 1 − P(E). For a die, P(not 4) = 5/6 = 1 − 1/6. Use this as a shortcut and as a check.
Extremes: on a die, P(number less than 7) = 6/6 = 1 (certain) and P(8) = 0/6 = 0 (impossible).
6. Experimental vs Theoretical, the Law of Large Numbers and the Gambler’s Fallacy
| Point of comparison | Experimental probability | Theoretical probability |
|---|---|---|
| Based on | Actual data from trials or records | Reasoning about equally likely outcomes |
| Needs an experiment? | Yes | No |
| Formula | Times event occurred / Total trials | Favourable outcomes / Possible outcomes |
| Typical use | Paper cup landing, tyre life, survey data | Fair coin, fair die, well-mixed cards |
Even in a perfectly fair situation, experimental probability can differ from theoretical probability, especially when the number of trials is small. In Example 2 the die gave 0.16 for a 4 against the theoretical 1/6 ≈ 0.167: close, but not equal.
Law of Large Numbers. As the number of trials increases, the experimental probability tends to get closer to the theoretical probability. Ten tosses of a fair coin might give 7 heads (0.7); thousands of tosses usually give a proportion close to 0.5.
Gambler’s Fallacy. Many people think that if one outcome happens many times in a row, the opposite outcome becomes “due”. A fair coin that has shown heads six times in a row still has probability 1/2 (50%) of tails on the next flip. The coin has no memory: every flip starts afresh. This mistaken belief is called the Gambler’s Fallacy.
NCERT Example 6. In Snakes and Ladders you roll three 6s in a row and think, “There is no way I will get a 6 again.” This is the Gambler’s Fallacy. Each roll is an independent event, and the probability of a 6 on the next roll is still 1/6. The textbook writes this as ≈ 0.166 (16.6%); rounded correctly to three decimal places, 1/6 = 0.1666… ≈ 0.167 (16.7%), as in NCERT Example 3. Probability describes what happens in the long run; it cannot tell you the next result.
7. Sample Space and Events
The sample space, written S, is the list of all possible outcomes of a random experiment. Each outcome is an element of the sample space. NCERT gives three rules:
- S must include every possible outcome.
- No outcome should be listed more than once.
- The number of elements in S is the sample size, written n(S).
| Experiment | Sample space S | n(S) |
|---|---|---|
| Whether it rains tomorrow | {Rain, No Rain} | 2 |
| Result of a team’s match | {Win, Lose, Draw} | 3 |
| Tossing a fair coin once | {H, T} | 2 |
| Rolling a standard 6-sided die | {1, 2, 3, 4, 5, 6} | 6 |
| Tossing two coins together | {HH, HT, TH, TT} | 4 |
For two coins, HT (coin 1 heads, coin 2 tails) and TH are different outcomes, so both are listed.
Choose the level of detail to match the question. {Rain, No Rain} is enough to ask whether it will rain. To ask about how much rain, expand it to {No Rain, Drizzle, Light Rain, Heavy Rain}. The sample space must be detailed enough for the problem being studied.
Events. An event is any single possible result, or a combination of results, of a random action. It is a choice of particular outcomes from all that could occur, so an event is a subset of the sample space.
| Experiment | Sample space | Event | Event as a set |
|---|---|---|---|
| Tossing two coins | {HH, HT, TH, TT} | At least one coin shows head | E = {HH, HT, TH} |
| Rolling a 6-sided die | {1, 2, 3, 4, 5, 6} | The number rolled is greater than 4 | E = {5, 6} |
| Picking fruit from a basket | {Apple, Banana, Orange} | Picking a fruit that is yellow | E = {Banana} |
With equally likely outcomes, P(E) = (number of elements in E) / n(S). For “at least one head” with two coins, P = 3/4.
Recording only the number of heads. For three coins this gives S = {0, 1, 2, 3}, whose outcomes are not equally likely: 1 head happens in 3 ways (HTT, THT, TTH), 0 heads in 1 way (TTT). Find probabilities from the 8 equally likely outcomes instead.
8. Tree Diagrams and Tables for Multi-step Experiments
A multi-step experiment involves a series of trials, such as tossing a coin two times or rolling a die three times. A tree diagram is a visual way to list all possible outcomes of such an experiment. Each branch is a possible outcome of one step, and branches split again to show what can happen at the next step. Each complete path from the start to the end of the tree is one outcome of the whole experiment.
NCERT Example 7: toss a fair coin two times. From a starting point draw two branches for the first toss, H and T. From the end of each of these draw two more branches for the second toss, H and T. Reading along each path:
| First toss | Second toss | Outcome (path) | Probability |
|---|---|---|---|
| H | H | HH | 1/4 |
| H | T | HT | 1/4 |
| T | H | TH | 1/4 |
| T | T | TT | 1/4 |
The tree shows 4 possible outcomes, so S = {HH, HT, TH, TT}. Each branch of a fair coin carries probability 1/2. P(HH) = 1/4 = 0.25, or 25%. The event “one head and one tail” is {HT, TH}, so its probability is 2/4 = 1/2.
Counting paths quickly. If step 1 has a choices and step 2 has b choices for each of them, the tree ends in a × b paths. Two coins: 2 × 2 = 4. Three coins: 2 × 2 × 2 = 8. A coin and a die: 2 × 6 = 12. Two dice: 6 × 6 = 36.
Tables do the same job. For two dice, a 6 by 6 table with the first die down the side and the second die across the top gives all 36 outcomes. Writing the sum in each cell makes it easy to count events such as “sum is 7”: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), which is 6 outcomes, so P(sum 7) = 6/36 = 1/6.
With and without replacement. A box has 4 balls numbered 1 to 4 and two balls are drawn one after another.
- With replacement (the first ball goes back): each draw has 4 choices, so the tree has 4 × 4 = 16 paths, from (1, 1) to (4, 4). Repeats like (2, 2) are possible.
- Without replacement (the first ball is kept out): the second draw has only 3 choices, so the tree has 4 × 3 = 12 paths. Pairs like (2, 2) cannot happen.
Label items of the same colour (R1, R2, …) so that every path is equally likely, then count. A branch probability is the chance of that step given the path so far.
Formula and Theorem Sheet
| Result | Statement | When to use it |
|---|---|---|
| Probability scale | 0 ≤ P(E) ≤ 1; P(impossible) = 0, P(certain) = 1 | Check every answer |
| Experimental probability | Number of times the event occurred / Total number of trials | Data from repeated trials |
| Estimate for a population | Probability from the sample × Population size | Scaling a sample result up |
| Theoretical probability | P(E) = Number of favourable outcomes / Number of possible outcomes | Equally likely outcomes only |
| Event not happening | P(not E) = 1 − P(E) | “Not red”, “not an E” |
| Counting paths | Total outcomes = a × b × … (choices at each step) | Tree diagrams, tables |
| Two draws from n items | With replacement n × n; without replacement n × (n − 1) | Ordered draws |
| Law of Large Numbers | More trials: experimental probability gets closer to theoretical | Explaining why results differ |
Read the rest of the chapter →Hide the rest ↑
Important Definitions
| Term | Definition |
|---|---|
| Probability | A measurement of the likelihood of an event, on a scale from 0 (impossible) to 1 (certain) |
| Randomness | A situation where the possible outcomes are known but the exact result cannot be predicted |
| Random experiment | Something you can repeat, where the result may differ each time and cannot be known in advance |
| Sample space (S) | The set of all possible outcomes, each listed exactly once |
| Event | One outcome or a group of outcomes; a subset of the sample space |
| Experimental probability | Probability found from actual trials or data, as a relative frequency |
| Theoretical probability | Probability found by reasoning, assuming all outcomes are equally likely |
| Equally likely outcomes | Outcomes with no reason for any one to occur more often than another |
| Fair (unbiased) | Symmetrical, so no side or outcome is favoured |
| Gambler’s Fallacy | The false belief that past results of independent trials change the next result |
Solved Examples (NCERT-Based)
Example 1: Ranking events on the scale (Exercise Set 7.1)
Rank the events on a scale from 0 (Impossible) to 1 (Certain), label each, and give reasons: (i) The next Monday will come after Sunday. (ii) It will snow in Mumbai in July. (iii) An elephant will walk through your classroom today. (iv) You will greet at least one friend at school tomorrow.
Solution: (i) Certain (1): Monday always follows Sunday. (ii) Impossible, or as close to 0 as an event gets: Mumbai is warm in July, its monsoon month. (iii) Close to 0 (nearly impossible): it could happen only in a very unusual situation. (iv) More likely: on a normal school day you meet friends, though illness or a holiday could prevent it. In (ii) to (iv) the reason matters as much as the label.
Example 2: Sweets sample (Exercise Set 7.2, Q1)
A sample of 30 sweets has 10 red, 8 green, 7 yellow and 5 blue. (i) Find the probability that a randomly picked sweet from the sample is green. (ii) If there are 600 sweets in the large bag, estimate how many are yellow.
Solution: Check the total: 10 + 8 + 7 + 5 = 30. (i) P(green) = 8/30 = 4/15 ≈ 0.267. (ii) P(yellow) = 7/30. Estimated yellow sweets = (7/30) × 600 = 7 × 20 = 140.
Example 3: Favourite club survey (Exercise Set 7.2, Q2)
Of 40 students sampled, 14 prefer Science Club, 11 Arts Club, 9 Sports Club and 6 Debate Club. The school has 800 students. (i) Find the probability that a student from the sample prefers the Arts Club. (ii) Estimate how many students in the school prefer the Sports Club.
Solution: 14 + 11 + 9 + 6 = 40. (i) P(Arts) = 11/40 = 0.275. (ii) (9/40) × 800 = 9 × 20 = 180 students.
Example 4: Experimental against theoretical (Exercise Set 7.2, Q6)
A 6-sided die is rolled 12 times and shows ‘3’ three times. (i) Find the experimental probability of a ‘3’. (ii) Find the theoretical probability. (iii) Why might they differ, and what happens with 60, 600 or 6000 rolls?
Solution: (i) 3/12 = 1/4 = 0.25. (ii) One favourable face out of 6: 1/6 ≈ 0.167. (iii) Twelve rolls is a small number of trials, and chance variation in a small set is large. By the Law of Large Numbers, as the rolls go up to 60, 600 and 6000, the experimental probability should get closer to 1/6.
Example 5: Writing sample spaces (Exercise Set 7.3, Q2)
Write the sample space for: (i) rolling a die and tossing a coin together; (ii) choosing a random integer between −5 and +5; (iii) drawing one ball from a box of 5 green and 7 red balls.
Solution: (i) 6 × 2 = 12 outcomes: S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.
(ii) Taking both ends as included: S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}, with n(S) = 11. If “between” is read as excluding −5 and 5, S = {−4, …, 4} with 9 elements; state which reading you use.
(iii) Recording colour: S = {Green, Red}. These two are not equally likely. Labelling the balls, S = {G1, G2, G3, G4, G5, R1, R2, R3, R4, R5, R6, R7}, 12 equally likely outcomes, so P(green) = 5/12.
Example 6: Snacks and drinks at the fair (Exercise Set 7.3, Q3)
Snacks: Samosa, Pakora, Bhaji. Drinks: Chai or Lassi. (i) List the sample space of all snack and drink combinations. (ii) List the event ‘Selecting Samosa as a snack.’
Solution: (i) 3 × 2 = 6 combinations: S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}. (ii) E = {(Samosa, Chai), (Samosa, Lassi)}.
Example 7: Two fruit baskets (Exercise Set 7.4, Q1)
Basket A has one apple and two oranges. Basket B has one banana and one mango. One fruit is picked at random from each basket. Draw the tree, list the sample space, and find the probability of one apple and one banana.
Solution: Label the oranges O1 and O2 so that the three fruits in A are equally likely. Tree: three first branches (Apple, O1, O2); from each, two branches (Banana, Mango). Paths: 3 × 2 = 6.
S = {(Apple, Banana), (Apple, Mango), (O1, Banana), (O1, Mango), (O2, Banana), (O2, Mango)}.
Only (Apple, Banana) is favourable, so P = 1/6. Branch check: P(apple) × P(banana) = 1/3 × 1/2 = 1/6.
Example 8: Pens with replacement (Exercise Set 7.4, Q2)
A box has 3 red, 4 black and 2 green pens. You pick a pen without looking and put it back; then your friend does the same. Find the possible colour outcomes and the probability that both pick the same colour.
Solution: Colour outcomes (you, friend): 3 × 3 = 9 paths: RR, RB, RG, BR, BB, BG, GR, GB, GG. They are not equally likely: each step has branches red 3/9, black 4/9, green 2/9.
Count with individual pens: 9 pens, replaced, so 9 × 9 = 81 equally likely pairs. Same colour: red 3 × 3 = 9, black 4 × 4 = 16, green 2 × 2 = 4, total 29. P(same colour) = 29/81 ≈ 0.358.
Example 9: Five quick probabilities (End-of-Chapter, Q4)
(i) Two coins tossed; at least one head. S = {HH, HT, TH, TT}; E = {HH, HT, TH}. P = 3/4.
(ii) Cards 1 to 10; an even number. E = {2, 4, 6, 8, 10}. P = 5/10 = 1/2.
(iii) A die; a number greater than 4. E = {5, 6}. P = 2/6 = 1/3.
(iv) 3 red, 2 blue, 1 green; not red. Not red = 2 + 1 = 3 of 6 balls. P = 3/6 = 1/2.
(v) Three coins; exactly two heads. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; E = {HHT, HTH, THH}. P = 3/8.
Example 10: Tyre life data (End-of-Chapter, Q7)
A tyre company records distances before replacement in 1000 cases: less than 4000 km: 20; 4001 to 9000 km: 210; 9001 to 14000 km: 325; more than 14000 km: 445. Find the probability that a randomly chosen tyre lasts (i) less than 4000 km, (ii) between 4000 and 14000 km, (iii) more than 14000 km.
Solution: Check: 20 + 210 + 325 + 445 = 1000. (i) 20/1000 = 0.02. (ii) 210 + 325 = 535, so 535/1000 = 0.535. (iii) 445/1000 = 0.445. Check: 0.02 + 0.535 + 0.445 = 1. This is experimental probability from records.
Example 11: PEACE cards and the spinner (End-of-Chapter, Q8 and Q9)
PEACE: the 5 cards are P, E, A, C, E. (i) P, E or C: 1 + 2 + 1 = 4 cards, P = 4/5. (ii) Not an E: 5 − 2 = 3 cards, P = 3/5.
Spinner with equally likely outcomes 1 to 8: (i) 8: 1/8. (ii) Odd {1, 3, 5, 7}: 4/8 = 1/2. (iii) Greater than 2 {3, 4, 5, 6, 7, 8}: 6/8 = 3/4. (iv) Less than 9: all 8, P = 1 (certain). (v) Multiple of 3 {3, 6}: 2/8 = 1/4.
Example 12: Two balls without replacement (End-of-Chapter, Q10)
A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Find (i) P(red then blue) and (ii) P(two blue).
Solution: First step: red 4/9, blue 5/9. After a red is removed, 3 red and 5 blue remain: red 3/8, blue 5/8. After a blue is removed, 4 red and 4 blue remain: red 4/8, blue 4/8.
| Path | Probability |
|---|---|
| Red, Red | 4/9 × 3/8 = 12/72 |
| Red, Blue | 4/9 × 5/8 = 20/72 |
| Blue, Red | 5/9 × 4/8 = 20/72 |
| Blue, Blue | 5/9 × 4/8 = 20/72 |
(i) P(red then blue) = 20/72 = 5/18. (ii) P(two blue) = 20/72 = 5/18. Counting check: with labelled balls there are 9 × 8 = 72 ordered draws; red then blue can happen in 4 × 5 = 20 ways. The four paths add to 72/72 = 1.
Example 13: Harder counting (End-of-Chapter, Q12)
(i) Two dice; the sum is a prime number greater than 5. n(S) = 36. Possible sums are 2 to 12, and the primes greater than 5 are 7 and 11. Sum 7: 6 outcomes; sum 11: (5, 6), (6, 5). P = 8/36 = 2/9.
(ii) 4 red, 3 green, 2 blue; two drawn without replacement; different colours. Ordered draws: 9 × 8 = 72. Same colour: 4 × 3 + 3 × 2 + 2 × 1 = 20. Different: 72 − 20 = 52. P = 52/72 = 13/18.
(iii) Three coins; first coin heads and exactly two heads. E = {HHT, HTH}. P = 2/8 = 1/4.
(iv) Four-digit number from 1, 2, 3, 4 without repetition; even. Total: 4 × 3 × 2 × 1 = 24. Even means the last digit is 2 or 4; for each, the other three digits fill the first three places in 3 × 2 × 1 = 6 ways. Favourable 12. P = 1/2.
(v) 3 questions, 4 options each, guessing; exactly 2 correct. Total answer patterns: 4 × 4 × 4 = 64. Choose the one wrong question (3 ways) and its wrong option (3 ways): 9. P = 9/64.
Example 14: Which list is a sample space? (End-of-Chapter, Q13 and Q15)
Q13. A box has 4 balls numbered 1 to 4 and two are drawn. With replacement: 4 × 4 = 16 outcomes, (1, 1) to (4, 4). Without replacement: 4 × 3 = 12 outcomes, the pairs with two different numbers.
Q15. Three coins are tossed and the number of heads is recorded. The sample space is (iv) {0, 1, 2, 3}. (i) {1, 2, 3} misses 0 heads (TTT). (ii) {0, 1, 2} misses 3 heads (HHH). (iii) {0, 1, 2, 3, 4} contains 4, which cannot happen with three coins.
Competency-Based Questions (with answers)
1. Case-based: The canteen survey
A student asks 60 students of Class 9 about their favourite canteen item: 24 choose idli, 18 choose sandwich, 12 choose poha and 6 choose fruit chaat. The school has 900 students.
(a) Find the probability that a student from the sample chooses sandwich. 18/60 = 0.3.
(b) Estimate how many idlis the canteen should plan for the whole school. P(idli) = 24/60 = 0.4, so 0.4 × 900 = 360.
(c) Give one way to make the estimate more reliable. Take a larger sample, and include students from all classes so the sample represents the whole school.
2. Case-based: The board game
In a board game a fair 6-sided die is rolled. A player needs a 5 or a 6 to enter the board.
(a) Write the event “player enters” as a set and find its probability. E = {5, 6}; P = 2/6 = 1/3.
(b) The player failed on four rolls in a row. Is the next roll more likely to succeed? No. Each roll is independent, so P stays 1/3. Expecting a success to be “due” is the Gambler’s Fallacy.
(c) Find the probability that the player does not enter on one roll. 1 − 1/3 = 2/3.
3. Source-based: Reading about the Gambler’s Fallacy
Read: “The coin does not keep track of what happened before. Every time you flip, the chance of heads or tails stays exactly the same.”
(a) Which word describes flips whose results do not affect each other? Independent.
(b) After six heads in a row, what is P(tails) on the next flip of a fair coin? 1/2, or 50%.
(c) How does the Law of Large Numbers fit with this? It describes the long run: over many flips the proportion of heads settles near 1/2. A streak creates no debt that tails must repay soon.
4. Assertion-Reason
Options for Questions 4 to 6: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Assertion (A): When a bag has 3 red and 7 blue marbles, the probability of drawing a red marble is 1/2. Reason (R): The theoretical probability formula applies only when the outcomes counted are equally likely.
Answer: (d). A is false: there are 10 equally likely marbles, so P(red) = 3/10. R is true, and it is exactly why counting the two colours as outcomes fails.
5. Assertion-Reason
Options: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Assertion (A): When two coins are tossed, the sample space has 4 elements. Reason (R): HT and TH are different outcomes because the coins are distinct.
Answer: (a). S = {HH, HT, TH, TT}, and R explains why HT and TH both appear, giving 4.
6. Assertion-Reason
Options: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Assertion (A): A die rolled 50 times showed 4 eight times, so the experimental probability of a 4 is 0.16. Reason (R): The probability of any event lies between 0 and 1.
Answer: (b). 8/50 = 0.16, so A is true, and R is true. A follows from the formula (times occurred / total trials), and R does not explain it.
7. Data reasoning: The paper cup experiment
A group tosses a paper cup 100 times. It lands on its side 62 times, on its bottom 26 times and upside down on its top 12 times.
(a) Assign probabilities to the three outcomes. Side 0.62, bottom 0.26, top 0.12. They add to 1.
(b) Why can theoretical probability not be used here? The three landing positions have different shapes and areas, so there is no reason to treat them as equally likely.
(c) Another group gets side 58 times in 100 tosses. Is either group wrong? No. Experimental results vary between sets of trials; pooling both gives 120/200 = 0.6, a better estimate from more trials.
8. Simulation: Guessing a true-false quiz
A student guesses all answers of a 2-question true-false quiz. To model this, she tosses two coins, with H for a correct guess and T for a wrong one.
(a) Why is a coin a fair model for one guess? Each guess has 2 equally likely results, like a fair coin.
(b) Find P(both correct) and P(exactly one correct). S = {HH, HT, TH, TT}. P(HH) = 1/4; P(exactly one) = 2/4 = 1/2.
(c) She repeats the two tosses 40 times and gets HH 13 times. Compare with theory. Experimental 13/40 = 0.325, against theoretical 0.25. With only 40 repeats this gap is normal; more repeats should bring it closer to 0.25.
Important Questions for Exams
1-Mark Questions
- “The probability of an impossible event is _______.” 0.
- “The set of all possible outcomes of a random experiment is called the __________.” Sample space.
- “Tossing a fair coin has a probability of ______ for getting heads.” 1/2.
- “When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?” 6.
- “A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?” 1/3.
2-Mark Questions
- “In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________.” Relative frequency; 15/50 = 3/10 = 0.3.
- “What is the probability of getting an even number when rolling a fair 6-sided die?” Favourable {2, 4, 6}: 3/6 = 1/2.
- “I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.” Probability 0: “the sum is 13” (the largest sum is 12). Probability 1: “the sum is between 2 and 12, inclusive”, which covers all 36 outcomes.
3-Mark Questions
- “Which of the following experiments have equally likely outcomes? Explain.” (i) Car starts or does not start: not equally likely, since it depends on the car’s condition. (ii) Fair coin: equally likely. (iii) Fair die: equally likely. (iv) Red or blue from 3 red and 7 blue: not equally likely (3/10 against 7/10). (v) Baby boy or girl: treated as equally likely, since there is no reason to expect one more than the other (real birth records show nearly half boys and half girls).
- “A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits… in a table format.” Rows red and blue, columns jeans, khakis and shorts: Red-Jeans, Red-Khakis, Red-Shorts, Blue-Jeans, Blue-Khakis, Blue-Shorts. 2 × 3 = 6 outfits.
- “List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.” S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}; n(S) = 12. For example, P(head and an even card) = 3/12 = 1/4.
5-Mark Questions
- A coin is tossed three times. (a) Draw a tree diagram and list S. (b) Find P(exactly two heads), P(at least one head) and P(first toss heads and exactly two heads). (a) Each toss doubles the branches: 8 paths, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. (b) Exactly two heads {HHT, HTH, THH}: 3/8. At least one head: all except TTT, 7/8. First heads and exactly two heads {HHT, HTH}: 2/8 = 1/4.
- A survey of 80 households in a colony of 2000 households records the main cooking fuel: LPG 56, PNG 16, electric induction 8. (a) Find each probability. (b) Estimate the number of LPG households in the colony. (c) Name two conditions for the estimate to be reliable. (d) Is this experimental or theoretical probability? (a) LPG 56/80 = 0.7, PNG 16/80 = 0.2, induction 8/80 = 0.1; sum 1. (b) 0.7 × 2000 = 1400. (c) The sample should be large enough and representative, drawn from all parts of the colony. (d) Experimental (statistical), since it comes from collected data.
- Two dice are rolled. Using a 6 by 6 table, find (a) P(sum 8), (b) P(doublet), (c) P(sum less than 5), (d) P(sum is not 7). n(S) = 36. (a) (2, 6), (3, 5), (4, 4), (5, 3), (6, 2): 5/36. (b) (1, 1) to (6, 6): 6/36 = 1/6. (c) Sum 2: 1, sum 3: 2, sum 4: 3, total 6: 6/36 = 1/6. (d) Sum 7 has 6 outcomes, so 1 − 6/36 = 5/6.
Common Mistakes and Examiner Tips
- Answer outside 0 to 1. A probability of 1.2 or −0.3 signals an error. Fix: check that the favourable count is not larger than the total.
- Counting colours as equally likely outcomes. 3 red and 7 blue does not give P(red) = 1/2. Fix: count the individual items: 3/10.
- Listing HT and TH once. Two coins give 4 outcomes. Fix: name coin 1 and coin 2, or draw the tree.
- Missing an outcome in S. For the number of heads with three coins, 0 is forgotten. Fix: check the smallest and largest possible values.
- Mixing frequency and relative frequency. 15 is the frequency; 15/50 is the relative frequency.
- Dividing by the wrong total in survey questions. Use the sample size for the probability, then multiply by the population size for the estimate.
- Gambler’s Fallacy. After a streak, P for the next trial of a fair coin or die is unchanged. Say “each trial is independent”.
- Treating replacement and no replacement alike. Without replacement the second step has one item fewer, and the branch probabilities change.
- Ignoring repeated letters. In PROBABILITY, B counts twice: 2/11.
- Expecting experimental and theoretical results to match exactly. Small samples differ; name the Law of Large Numbers in your reason.
Quick Revision Points
- Probability measures the likelihood of an event, from 0 (impossible) to 1 (certain): 0 ≤ P(E) ≤ 1.
- 0.5 is an even chance; below 0.5 is less likely; above 0.5 is more likely.
- A random experiment can be repeated, its outcomes are known, and one result cannot be predicted.
- Subjective probability is a personal judgement; this chapter measures probability objectively.
- Experimental probability = times the event occurred / total trials, also called relative frequency.
- A die rolled 50 times showing 4 eight times: 8/50 = 0.16.
- Estimate for a population = probability from the sample × population size.
- Larger, more representative samples give more reliable estimates.
- Theoretical probability = favourable outcomes / possible outcomes, for equally likely outcomes.
- A fair (unbiased) coin or die is symmetrical; no outcome is favoured.
- P(4 on a die) = 1/6 ≈ 0.167; P(B from PROBABILITY) = 2/11.
- Law of Large Numbers: more trials bring experimental probability closer to theoretical.
- Gambler’s Fallacy: past independent trials do not change the next one.
- Sample space S lists every outcome once; n(S) is the sample size.
- An event is a subset of S.
- Coin 2, die 6, two coins 4, three coins 8, coin and die 12, two dice 36 outcomes.
- Each full path of a tree diagram is one outcome.
- With replacement: n × n; without replacement: n × (n − 1).
- P(not E) = 1 − P(E).
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, it belongs to Unit VI, Statistics and Probability, which carries 10 marks out of the 80-mark theory paper. That 10 is shared between the Statistics chapter and this chapter; the course structure does not split it further. The unit is allotted 24 teaching periods. The remaining 20 marks of the year are internal assessment: pen-paper tests and multiple assessment (10), portfolio (5) and lab practical (5).
Its listed competencies: describe likelihood on the probability scale, estimate empirical probability from data, apply theoretical probability, use tree diagrams and tables, and model random experiments by simulation.
The CBSE Class IX question paper design weights questions by thinking skill: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating. The 1, 2, 3 and 5-mark grouping above is a practice ladder; the question types in a given school paper are set by that paper.
This chapter is the foundation for probability in Class 10, where the same counting of sample spaces and favourable outcomes is used.
Class 9 Maths · Chapter 7 – swipe through all 9 cards to understand the whole chapter.
Random experiments
You know every possible outcome but cannot predict the result of one trial.
A coin toss to decide who bats first is fair because nobody can predict it.
- Can be repeated, like tossing a coin or rolling a die
- All outcomes are known: die gives 1 to 6
- Which day follows Sunday is not random
From 0 to 1
Probability measures likelihood on a scale from impossible to certain.
A probability below 0 or above 1 means a mistake.
- 0 impossible, 0.5 even chance, 1 certain
- Below 0.5 less likely, above 0.5 more likely
- Number 2 to 10 from 52 cards: 36/52, more likely
Experimental probability
Estimate probability from what actually happened in trials.
Also called relative frequency; the frequency is the count itself.
- Die rolled 50 times, 4 came up 8 times: 8/50 = 0.16
- The only method for a paper cup landing
- Results change from one set of trials to the next
Survey data and samples
Use a sample to estimate for a whole population.
Larger, more representative samples give better estimates.
- 20 of 50 like mango: P = 0.4
- School of 1500: about 0.4 × 1500 = 600 mangoes
- Include students from different classes
Theoretical probability
Reason from equally likely outcomes, with no experiment needed.
Only valid for equally likely outcomes, such as a fair coin or die.
- P(4 on a die) = 1/6 ≈ 0.167
- P(B from PROBABILITY) = 2/11
- 3 red, 7 blue marbles: P(red) = 3/10, not 1/2
Law of Large Numbers
More trials bring experimental probability closer to theoretical.
Small samples can differ a lot from theory.
- 12 rolls, three 3s: 0.25 against 1/6
- 600 or 6000 rolls settle near 1/6
- It describes the long run, not the next trial
Gambler’s Fallacy
Past independent trials do not change the next one.
Each coin flip or die roll is an independent event.
- Six heads in a row: P(tails next) is still 1/2
- Three 6s in Snakes and Ladders: next 6 still 1/6
- A streak does not make the opposite result due
Sample space and events
List every outcome once; an event is a subset of that list.
Match the detail of S to the question being asked.
- Two coins: S = {HH, HT, TH, TT}, n(S) = 4
- At least one head: E = {HH, HT, TH}, P = 3/4
- Die, greater than 4: E = {5, 6}, P = 1/3
Multi-step experiments
Each full path of a tree is one outcome of the whole experiment.
Without replacement, the second step has one item fewer.
- Coin and die: 2 × 6 = 12 outcomes; two dice: 36
- 4 balls with replacement 16, without 12
- 4 red, 5 blue, no replacement: P(red then blue) = 5/18
📝 Practice Introduction to Probability - 10 questions
Concept-check questions · with answers & explanations
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Why C: Even chance means probability 1/2. For P(purple) = (purple cards)/6 = 1/2, the deck needs 3 purple cards, so purple and green are equally likely.
Why not A: With 0 purple cards, picking purple is impossible, probability 0.
Why not B: With 6 purple cards, picking purple is certain, probability 1.
Why not D: With 2 purple cards, P(purple) = 2/6 = 1/3, which is less likely.
Remember: Even chance = half the outcomes favourable.
Why C: S = {H1, …, H6, T1, …, T6}, so n(S) = 2 × 6 = 12. The primes on a die are 2, 3 and 5, so the event is {H2, H3, H5}, which has 3 elements.
Why not A: 8 comes from adding 2 + 6; each coin result pairs with each die result, so there are 2 × 6 outcomes.
Why not B: 4 counts 1 as a prime; 1 is not prime, so the primes on a die are 2, 3 and 5.
Why not D: 6 counts every head outcome, but only the prime faces belong to the event.
Remember: Coin and die together: 12 outcomes; 1 is not a prime.
Why C: Step 1 (combine the classes in the range): 210 + 325 = 535 tyres lasted between 4000 and 14000 km. Step 2 (relative frequency): 535/1000 = 0.535.
Why not A: 0.325 uses only the 9001 to 14000 km group and misses the 4001 to 9000 km group.
Why not B: 0.555 wrongly adds the 20 tyres that lasted less than 4000 km.
Why not D: 0.445 is the probability of lasting more than 14000 km.
Remember: Add every class inside the range, then divide by the total.
Why C: Step 1 (what decides evenness): a number is even when its last digit is even. Step 2 (equally likely last digit): by symmetry each of 1, 2, 3, 4 is equally likely to be last, and 2 of them (2 and 4) are even, so P(even) = 2/4 = 1/2. Checking by counting: 24 arrangements in all, 12 end in 2 or 4, and 12/24 = 1/2.
Why not A: 1/4 counts only numbers ending in 2 (or only in 4), but both 2 and 4 make the number even.
Why not B: 3/4 has no basis; only 2 of the 4 possible last digits are even.
Why not D: 2/3 wrongly takes 3 as the number of choices for the last digit.
Remember: Evenness depends only on the last digit.
Why C: Step 1 (first branch): P(red first) = 4/9. Step 2 (second branch, without replacement): 8 balls remain and all 5 blue are still there, so P(blue second) = 5/8. Step 3 (multiply along the path): 4/9 × 5/8 = 20/72 = 5/18. Check by counting: 9 × 8 = 72 equally likely ordered pairs, of which 4 × 5 = 20 are red then blue.
Why not A: 20/81 uses 9 balls for the second draw, as if the first ball were put back.
Why not B: 4/9 is only the probability of red on the first draw.
Why not D: 1/2 has no basis in the counts; red and blue are not equally likely.
Remember: Without replacement, the second branch has one ball fewer.
Why A: No face shows a number greater than 6, so there are 0 favourable outcomes and the event is impossible. Impossible events have probability 0. R states the reason A holds.
Why not B: R gives exactly why the event is impossible, and an impossible event has probability 0, so R does explain A.
Why not C: R is true: a standard die shows only 1 to 6.
Why not D: A is true: an impossible event has probability 0.
Remember: No way for it to happen means probability 0.
Why C: With three coins the number of heads can be 0 (TTT), 1, 2 or 3 (HHH). Nothing else is possible, so S = {0, 1, 2, 3}.
Why not A: It misses 0, which happens when all three coins show tails.
Why not B: It misses 3, which happens when all three coins show heads.
Why not D: It includes 4, which is impossible with only three coins.
Remember: Every possible outcome in, every impossible one out.
Why B: Step 1 (probability from the sample): P(yellow) = 7/30. Step 2 (scale to the population): expected yellow = (7/30) × 600 = 7 × 20 = 140.
Why not A: 150 is 600/4, assuming the four colours are equally common, which the sample does not show.
Why not C: 7 is the number of yellow sweets in the sample, not the estimate for the bag.
Why not D: 70 comes from 7 × 10, but 600 is 20 times 30, not 10 times.
Remember: Sample probability × population size = estimated count.
Why A: Step 1 (count cards): P, E, A, C, E are 5 equally likely cards. Step 2 (favourable for ‘not E’): P, A and C, so 3 cards. P(not E) = 3/5.
Why not B: 2/5 is the probability that it IS an E.
Why not C: 3/4 uses the 4 different letters P, E, A, C, but there are 5 cards.
Why not D: 4/5 treats only one card as E, but PEACE has two Es.
Remember: Not E = total minus the E cards.
Why A: Step 1 (with replacement): the second draw again has 4 choices, so 4 × 4 = 16 outcomes, including (1, 1). Step 2 (without replacement): the second draw has only 3 choices, so 4 × 3 = 12 outcomes, and repeats like (1, 1) cannot occur.
Why not B: Without replacement the first ball is missing, so the second draw has only 3 choices.
Why not C: The two cases are swapped; putting the ball back gives more outcomes, not fewer.
Why not D: 8 comes from adding 4 + 4 instead of multiplying, and 6 ignores order, but (1, 2) and (2, 1) are different outcomes.
Remember: Put back: same choices again; not put back: one fewer.
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Frequently Asked Questions
Experimental probability comes from actual data: the number of times an event occurred divided by the total number of trials. Theoretical probability comes from reasoning: the number of favourable outcomes divided by the number of possible outcomes, and it needs all outcomes to be equally likely. A die rolled 50 times showing 4 eight times gives an experimental value of 0.16, while the theoretical value is 1/6, about 0.167.
Probability is measured on a scale from 0 to 1. The favourable outcomes are part of all possible outcomes, and the times an event occurs can never exceed the number of trials, so the fraction can never be larger than 1. A value of 1 means the event is certain. An answer above 1, or a negative answer, means a counting mistake.
The sample space S is the list of all possible outcomes of a random experiment, written in curly brackets and separated by commas. It must include every possible outcome and must not list any outcome twice. The number of elements is the sample size n(S). For two coins, S = {HH, HT, TH, TT} and n(S) = 4, because HT and TH are different outcomes.
As the number of trials increases, the experimental probability tends to get closer to the theoretical probability. With 12 rolls of a die you may see a 3 three times, giving 0.25 against the theoretical 1/6. With 600 or 6000 rolls the proportion of 3s usually settles close to 1/6. The law describes the long run and says nothing certain about the next trial.
It is the false belief that after a streak, the opposite result becomes more likely. A fair coin that shows heads six times in a row still has probability 1/2 of tails on the next flip, because the coin has no memory and each flip is independent. The same holds for dice in Snakes and Ladders: after three 6s, the chance of another 6 is still 1/6.
First find the probability from the sample, then multiply by the population size. If 20 of 50 students in a sample like mango, the probability is 0.4, so in a school of 1500 students about 0.4 × 1500 = 600 would prefer mango. The estimate is more reliable when the sample is larger and represents the whole population, for example by including different classes.
Use a tree diagram for a multi-step experiment, such as tossing a coin twice or drawing two balls one after another. Each step splits into branches for its possible outcomes, and every full path from start to end is one outcome of the whole experiment. Two coin tosses give 4 paths: HH, HT, TH, TT. Without replacement, the second step has one item fewer, which changes the branches.
