Class 9 Maths Ch 7: Probability Notes

1The probability scale

The probability of an event is a number from 0 to 1 that measures how likely it is. 0 means impossible (rolling a 7 on a normal die), 1 means certain (the sun rising tomorrow), and $\dfrac{1}{2}$ means even chance (a fair coin landing heads).

Probabilities can be written as fractions, decimals or percentages: $\dfrac{1}{4} = 0.25 = 25\%$ . A probability can never be negative or greater than 1 — if your answer is, something has gone wrong.

Probability describes the long run, not a guarantee. A fair coin that has shown heads five times in a row is still equally likely to show heads or tails next.

Key formulas & definitions

$0 \le P(E) \le 1$
Impossible: 0 · Even chance: $\dfrac{1}{2}$ · Certain: 1

Common mistakes

The 'gambler's fallacy': thinking tails is 'due' after many heads.
Writing a probability like $\dfrac{7}{5}$ — impossible, since favourable outcomes cannot exceed the total.

2Experimental and theoretical probability

Experimental (empirical) probability comes from what actually happened: $P(E) = \dfrac{\text{number of trials in which } E \text{ happened}}{\text{total number of trials}}$ . It is also called the relative frequency.

Theoretical probability comes from reasoning about a model where all outcomes are equally likely: $P(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ .

With few trials the two can differ a lot; as the number of trials grows, the experimental value usually settles close to the theoretical one. If it does not, the model's assumption (for example, 'the die is fair') may be wrong — that is how data can test fairness.

Key formulas & definitions

Experimental: $\dfrac{\text{times event occurred}}{\text{total trials}}$
Theoretical: $\dfrac{\text{favourable outcomes}}{\text{total equally likely outcomes}}$

Worked example 1

A die is rolled 300 times and a six comes up 72 times. Find the experimental probability of a six and compare it with the theoretical value. Is the die likely to be fair?

Experimental: $\dfrac{72}{300} = 0.24$ .
Theoretical for a fair die: $\dfrac{1}{6} \approx 0.167$ , i.e. about 50 sixes in 300 rolls.
72 is well above 50 for this many rolls, so the die may be biased towards six (more rolls would confirm).

Answer: 0.24 vs 0.167 — the die looks suspicious.

Worked example 2

In a survey of 250 students, 90 come to school by bus, 110 walk and the rest cycle. Estimate the probability that a randomly chosen student cycles.

Cyclists $= 250 - 90 - 110 = 50$ .
$P(\text{cycle}) = \dfrac{50}{250} = \dfrac{1}{5} = 0.2$ .

Answer: 0.2.

Common mistakes

Using the favourable/total formula when outcomes are not equally likely (e.g. 'win, lose, draw' is not automatically $\dfrac{1}{3}$ each).
Expecting exactly 50 heads in 100 tosses.

3Sample space, events and complements

An experiment is any action with uncertain results. Each possible result is an outcome, and the list of all outcomes is the sample space S. An event is a set of outcomes we are interested in.

Tossing two coins: $S = \{HH, HT, TH, TT\}$ — four outcomes, not three, because HT and TH are different. Rolling two dice gives $6 \times 6 = 36$ outcomes, best shown in a $6 \times 6$ table.

The complement of E ('not E') contains every outcome not in E. Since one of them must happen, $P(E) + P(\text{not } E) = 1$ . This is often the quickest route: 'at least one' = 1 − 'none'.

Key formulas & definitions

$P(\text{not } E) = 1 - P(E)$
Two coins: 4 outcomes · Two dice: 36 outcomes · One card from a pack: 52.

Worked example 1

Two dice are rolled. Find the probability that the sum is 8, and that the sum is not 8.

Pairs with sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.
$P(\text{sum} = 8) = \dfrac{5}{36}$ .
$P(\text{sum} \neq 8) = 1 - \dfrac{5}{36} = \dfrac{31}{36}$ .

Answer: $\dfrac{5}{36}$ and $\dfrac{31}{36}$ .

Trick / memory aid

For two dice, sums 2 to 12 occur 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 times — rising to 6 ways at sum 7, then falling.

Common mistakes

Listing two coins as {2 heads, 1 head, 0 heads} and giving each $\dfrac{1}{3}$ . 'One head' is twice as likely ( $\dfrac{2}{4}$ ).
Counting (4,4) twice.

4Tree diagrams

A tree diagram shows a multi-stage experiment: the first set of branches for stage 1, and from the end of each, branches for stage 2. Write the probability on every branch; branches from one point add to 1.

Each complete path from the start to an end is one combined outcome. Multiply along a path to get its probability. If several paths lead to the event you want, add those paths.

Trees make it easy to handle 'without replacement' problems, where the second-stage probabilities change because something was removed.

Key formulas & definitions

Along a path: multiply · Between paths: add
All end probabilities together add to 1.

Worked example 1

A bag has 3 red and 2 blue balls. Two balls are drawn one after another without replacement. Find the probability that both are the same colour.

First draw: $P(R) = \dfrac{3}{5}$ , $P(B) = \dfrac{2}{5}$ .
After a red: $P(R) = \dfrac{2}{4}$ . After a blue: $P(B) = \dfrac{1}{4}$ .
$P(RR) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}$ ; $P(BB) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20}$ .
Same colour $= \dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}$ .

Answer: $\dfrac{2}{5}$ .

Worked example 2

A school bus is late with probability 0.2 on any day, independently. Find the probability it is late on at least one of two days.

On time both days: $0.8 \times 0.8 = 0.64$ .
At least one late $= 1 - 0.64 = 0.36$ .

Answer: 0.36.

Common mistakes

Adding along a path instead of multiplying.
Keeping the same denominator on the second draw when there is no replacement.

Important questions

Try each question on paper first, then tap it to check the answer.

Basic Recall and direct use

A card is drawn from a well-shuffled pack of 52. Find P(a king) and P(a red card).Answer

Kings: 4, so

\dfrac{4}{52} = \dfrac{1}{13}

. Red cards: 26, so

\dfrac{26}{52} = \dfrac{1}{2}

If the probability of winning a game is 0.35, what is the probability of not winning?Answer

1 - 0.35 = 0.65

Write the sample space when a coin is tossed and a die is rolled together. How many outcomes are there?Answer

\{H1, H2, \dots, H6, T1, \dots, T6\}

2 \times 6 = 12

outcomes.

Medium Two or three steps

Three coins are tossed. Find the probability of getting at least two heads.Answer

8 outcomes. At least two heads: HHT, HTH, THH, HHH = 4. Probability

= \dfrac{4}{8} = \dfrac{1}{2}

A cricket player hit a boundary in 18 of his last 120 balls. Estimate the probability that he does not hit a boundary on the next ball.Answer

P(\text{boundary}) = \dfrac{18}{120} = 0.15

, so

P(\text{no boundary}) = 0.85

A number is chosen from 1 to 30. Find the probability that it is a prime number.Answer

Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 = 10. Probability

= \dfrac{10}{30} = \dfrac{1}{3}

Advanced Reasoning and HOTS

Two dice are thrown. Find the probability that the product of the numbers is even.Answer

Product is odd only when both are odd:

3 \times 3 = 9

outcomes.

P(\text{odd}) = \dfrac{9}{36} = \dfrac{1}{4}

, so

P(\text{even}) = \dfrac{3}{4}

A box has 4 good and 2 defective bulbs. Two bulbs are taken out one by one without replacement. Draw a tree and find the probability that exactly one is defective.Answer

Paths GD and DG.

P(GD) = \dfrac{4}{6} \times \dfrac{2}{5} = \dfrac{8}{30}

;

P(DG) = \dfrac{2}{6} \times \dfrac{4}{5} = \dfrac{8}{30}

. Total

= \dfrac{16}{30} = \dfrac{8}{15}

Self-check quiz

Five quick questions. Tap an option to see if you are right.

Which of these cannot be a probability?

Two coins are tossed. P(exactly one head) is

P(E) = 0.08

, then

P(\text{not } E)

In a tree diagram, the probability of a single path is found by

The probability of getting a sum of 7 with two dice is

FAQ

What is the difference between experimental and theoretical probability?

Experimental probability is measured from actual trials or data. Theoretical probability is calculated from equally likely outcomes. With many trials, they usually come close.

Why are there 4 outcomes, not 3, when tossing two coins?

Think of the coins as a ₹1 and a ₹2 coin. 'Heads on ₹1, tails on ₹2' and 'tails on ₹1, heads on ₹2' are different outcomes, so one head can happen in two ways.

How do I solve 'at least one' questions quickly?

Use the complement: P(at least one) = 1 − P(none). It avoids listing many cases.

Book references

NCERT Ganita Manjari Part I, Chapter 7: The Mathematics of Maybe: Introduction to ProbabilityRead the chapter and do every exercise set first.
Mathematics for Class 9 — R.D. Sharma (Dhanpat Rai)Extra graded practice; its chapter numbers follow the old syllabus, so match by topic.
Mathematics for Class 9 — R.S. Aggarwal (Bharati Bhawan)Good for drill on basics; match by topic.
Exemplar Problems in Mathematics, Class IX — NCERTHarder, concept-testing questions (written for the old book).

These notes, examples and questions are written by PrepRung in our own words to help you study the NCERT Ganita Manjari book. NCERT and CBSE are not affiliated with PrepRung. Always study from your textbook too.

Class 9 Maths Chapter 7: The Mathematics of Maybe: Introduction to Probability — Notes, Examples & Practice

1The probability scale

2Experimental and theoretical probability

3Sample space, events and complements

4Tree diagrams

Important questions

Basic Recall and direct use

Medium Two or three steps

Advanced Reasoning and HOTS

Self-check quiz

FAQ

Book references