The first activity.
Ten tosses of a coin.
Class 9 · Maths · Chapter 15 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Probability
How to use this page:
1. Read — Activity 1 · Activity 2 · Activity 3 · Activity 4 · Activity 5 · Exercise 15.1, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure shows one point of empirical probability on a line from 0 to 1.
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अनिश्चितता और क्रियाकलाप 1 · Textbook 15.1–15.2 · coin
Everyday words such as probably, doubt and a fifty-fifty chance are uncertainty. Probability gives that uncertainty a number. The subject began with games of chance and is now used in weather and science too.
Activity 1: toss a coin ten times, count heads and tails, then twenty times, then more. The fraction = number of heads / total tosses. As the trials grow, this fraction moves near 0.5.
Take any coin. Fill the table of ten tosses: total, heads, tails. Write both fractions. Then twenty tosses. Then increase again. Watch the fraction move near 1/2. This is not yet the name of the formula. It is a count.
Question: In 40 tosses, 22 are heads. Find the head fraction.
Formula: fraction = number of heads / total tosses.
Substitution: 22 / 40 = 0.55.
Unit: none. It is a ratio.
Do not leave 22/40 as 22. Show the division. The answer 0.55 is a ratio.
A 0.7 from ten tosses is not yet the rule. Increase the trials, then move near 0.5.
The first activity.
Ten tosses of a coin.
True — this is what Activity 1 shows.
22/40.
0.55.
Count ÷ count.
There is no unit.
Ten tosses, then twenty, then more. Fraction = heads / total. As the trials grow, the fraction moves near 0.5.
क्रियाकलाप 2 — समूह और संचयी भिन्न · Group coin · cumulative fraction
The printed heading has a spelling slip, but the order is still the second activity. Divide the class into groups of 2 or 3. Each group tosses a coin of the same kind 15 times. Write cumulative heads and tails on the board.
The next group adds its count to the earlier sum and finds the fraction on the new total. As the tosses grow, the columns move near 0.5. This is not the die activity. The die comes next.
Use a coin of one value in every group, so the tosses count as one coin. The first row is the fraction of its own 15 tosses. The second row is the fraction of the sum of both groups. That is what cumulative means.
Question: The first group has 6 heads and 9 tails. The second has 8 heads and 7 tails. Find the cumulative head fraction.
Formula: cumulative fraction = total heads / total tosses.
Substitution: heads = 6 + 8 = 14. Tosses = 30. 14 / 30 = 0.466….
Unit: none.
Do not find the second group’s fraction on its own 15 only. Add the earlier sum.
Do not call this Activity 3. The die has not come yet.
The group coin.
A coin 15 times.
True.
The sum.
14. The fraction is 14/30.
The cumulative fraction is 14/30, not only 8/15. The earlier group is added.
क्रियाकलाप 3 और 4 — पासा और दो सिक्के · Die 1/6 · two coins
Activity 3: throw a die 20 times, then 40. The fraction of each score from 1 to 6 moves near 1/6 when the trials are large. Add groups the way the coin groups were added. A die is a balanced cube with 1 to 6 on the faces.
Activity 4: toss two coins together ten times, then more. No head, one head, two heads. For a large count the fractions are about 0.25, 0.5 and 0.25.
Do not change the order. First one die. Six columns in the table. Then two coins together. Three columns: 0 heads, 1 head, 2 heads. Write their own fractions. Do not force 1/6 and 0.25 onto a small number of throws.
Question: In 60 throws the score 3 appeared twelve times. Find the fraction and compare it with 1/6.
Formula: fraction = (times 3 appeared) / (total throws). 1/6 ≈ 0.167.
Substitution: 12 / 60 = 0.2. This is near 0.167, not equal.
Unit: none.
Writing 12/60 = 0.2 as 1/6 is hasty. It is near, not the same.
The three fractions of two coins should add near 1. Leaving out one column spoils the sum.
Six faces.
1/6.
True.
12/60.
0.2.
0.25, 0.5, 0.25.
About 0.25.
50/200 = 0.25, 100/200 = 0.5, 50/200 = 0.25. The sum is 1. There is no unit.
False — the die is 3, two coins are 4.
परीक्षण, घटना और सूत्र · Activity 5 · empirical probability
A trial is an action that produces an outcome. One toss is a trial. One throw is a trial. One simultaneous toss of two coins is also a trial. The outcomes are head, tail, or 1 to 6. An event is a collection of some outcomes. The event of an even score contains 2, 4 and 6.
Activity 5 says: from the table of Activity 3 find the probability of the score 3, and see how it changes as the trials grow. The formula is P(E) = (trials in which E happened) / (total trials). For convenience the book just writes probability.
Open your own table from Activity 3. Count of score 3 ÷ total throws. Write the fraction on the first 20 throws, then on 40. The changing number is the empirical probability. Do not give it a new number. This is the fifth activity.
Question: In 1000 tosses there are 480 heads and 520 tails. Find both probabilities.
Formula: P(E) = (trials in which E happened) / (total trials).
Substitution: P(heads) = 480/1000 = 0.48. P(tails) = 520/1000 = 0.52.
Unit: none. The sum is 1.00.
Changing 480/1000 into a percent is not the first answer of this chapter. Keep 0.48 or the fraction.
An even score is not one outcome. It is an event of three outcomes: 2, 4, 6.
Divide by the total.
Favourable / total.
True — 2, 4 and 6.
480/1000.
0.48.
The die score 3.
Activity 3.
A trial is an action. An outcome is its result. An event is a collection of some outcomes. 0.48 + 0.52 = 1.
योग 1, और 0 तथा 1 · Impossible · certain · complementary event
When two events split every trial exactly once, their values add to 1. Heads and tails are like that. Probability runs from 0 to 1, both ends included. If no student has that weight, the probability is 0. If every student meets the condition, the probability is 1.
A value above 1 or below 0 is a calculation error. Writing 100 as a percent is not the answer of this formula.
Question: Of 200 students, 135 like the subject. Find the probabilities of like and dislike.
Formula: P = count / 200. Dislike = 1 − like, when together they are all the students.
Substitution: P(like) = 135/200 = 0.675. P(dislike) = 65/200 = 0.325. The sum is 1.
Unit: none.
Do not treat 65 as the total and write 65/135. The total is 200.
1.2 is not a probability. Check the division again.
10-second revision
Not even once.
0.
True.
Divide.
0.675.
Dislike = 1 − 0.675 = 0.325. The sum is 1. There is no unit.
अभ्यास 15.1 · Only Exercise 15.1 · questions 9 and 10 are activities
The only exercise of this chapter is 15.1. Later pages are an answers appendix, not new exercises. In each question find the total trials, then the favourable count. P = favourable / total.
In the family table, check that the three probabilities add to 1. In the 200 coin trials, take the row of two heads. In the opinion poll of 135 and 65, the total is 200.
1) The probability of not hitting a boundary. 2) Families with two children, number of girls, sum 1. 3) A student born in August. 4) Three coins, 200 times, two heads. 5) The income and vehicle table. 6) The marks table. 7) Likes or dislikes statistics. 8) An engineer’s distance, empirical. 9) Count vehicles at the gate. This is an activity, not a new number. 10) A three-digit number divisible by 3. This is an activity too. 11) A flour bag above 5 kg. 12) The sulphur dioxide interval. 13) Blood group AB.
Question: In 30 balls a boundary came 8 times. Find the probability that a boundary did not come.
Formula: P(not) = (total − favourable) / total.
Substitution: (30 − 8) / 30 = 22/30 = 0.733….
Unit: none.
Not hitting a boundary is not 8/30. It is 22/30. Read the wording of the question.
Do not turn appendix Exercise 1.1 into a question of this chapter. That is the answers section.
10-second revision
The remaining balls.
22/30.
True — they are unnumbered activities inside that same exercise.
22 ÷ 30.
About 0.733.
Before the summary.
15.1.
135/200 = 0.675 and 65/200 = 0.325. The sum is 1. There is no unit.
False — the value stays between 0 and 1.
Pick a type. The 30 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Divide by the total trials.
Favourable / total.
Activity 1.
0.5.
Six faces.
1/6.
Activity 4.
0.25.
Not even once.
0.
Every trial.
1.
Divide.
0.48.
Like.
0.675.
The rest.
22/30.
15 tosses.
The group coin.
Score 3.
Activity 3.
One, before the summary.
Only 15.1.
A ratio.
There is no unit.
False — between 0 and 1.
True.
False — 4 is two coins. The die is 3.
True — an even score is three outcomes.
False — they sit inside Exercise 15.1.
False — it adds and then finds the new fraction.
True.
False — a ratio has no unit.
Divide.
0.55.
Cumulative.
14.
The die.
0.2.
Heads.
0.48.
Tails.
0.52.
Like.
0.675.
Dislike.
0.325.
The remaining balls.
22.
1 is your coin, 2 is the group, 3 is the die, and 4 is two coins.
0 is impossible, 1 is certain, 1/6 is a die score, and 0.5 is the coin.
Assertion (A): P(E) divides favourable trials by the total.
Reason (R): Empirical probability is a ratio of counts.
Both are true and R is the reason.
Assertion (A): For large tosses, heads move near 0.5.
Reason (R): Each die score also moves near 0.5.
A is true. R is false — the die moves near 1/6.
Assertion (A): Probability lies between 0 and 1.
Reason (R): 1.5 is also allowed when the trials are large.
A is true. R is false.
Assertion (A): Question 9 is Activity 6.
Reason (R): Questions 9 and 10 are activities inside Exercise 15.1.
A is false. R is true.
Assertion (A): Heads 0.48 and tails 0.52 add to 1.
Reason (R): The two events split every toss exactly once.
Both are true and R is the reason.
0.5 is the coin. 1/6 and the even scores are the die. 0.25 is two coins.
Ten tosses are 1, the die is 3, the boundary is Exercise 15.1, and cumulative groups are 2.
P(E) = (trials in which E happened) / (total trials).
Impossible is 0. Certain is 1.
22/40 = 0.55. There is no unit.
An action with one or more outcomes.
2, 4 and 6.
0.48 and 0.52. The sum is 1. There is no unit.
0.675 and 0.325. The sum is 1.
(30−8)/30 = 22/30.
12/60 = 0.2 and 1/6 ≈ 0.167. They are near, not equal.
1 coin from ten onward. 2 cumulative group fraction. 3 die. 4 two coins. 5 the changing probability of score 3. P(E) = favourable / total.
0 is impossible, 1 is certain. A complement adds to 1. 0.48+0.52=1. 0.675+0.325=1. There is no unit.
1 is no boundary. 7 is like and dislike. 9 is vehicles at the gate. 10 is a three-digit number. They are activities inside the exercise, not Activity 6 or 7.
This model set is for practice. It is not a question from any year’s annual examination. Annual questions will be added only when a source page is available.
The remaining balls.
A boundary did not come.
The sum.
14/30.
Complement.
1.
False — those are answers of the whole book. The exercise of this chapter is only 15.1.
0.48 and 0.52. There is no unit. It is a ratio. The sum is 1.
These are competency-based practice questions. They are not copies of a CBSE paper.
Past days.
Empirical probability.
The remaining balls.
22/30.
Assertion (A): For large trials a coin moves near 0.5.
Reason (R): So every result of ten tosses will be exactly 0.5.
A is true. R is false — small trials still move.
False — a value above 1 does not occur in this chapter.
P(on time) = (50−20)/50 = 30/50 = 0.6. There is no unit.
18/40 = 0.45. It is an activity inside Exercise 15.1, not a new number.
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What you learned
| What | Keep this |
|---|---|
| Formula | P(E) = favourable / total |
| Coin | near 0.5 |
| Die | near 1/6 |
| Two coins | near 1/4, 1/2, 1/4 |
| Impossible | 0 |
| Certain | 1 |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.