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NCERT Solutions for Class 9th Maths Chapter 7 Exercise Set 7.3 — Elements of Probability: Sample Spaces and Events

Book page 167–168 Updated on2026-09-08

Q1.
When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Answer
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
Why it happens: a die has six faces and exactly one of them ends up on top, so there are six possible results and no more. The list satisfies both requirements of a sample space — every result appears, and none appears twice — and since the die is fair, all six are equally likely, each with probability 1/6.
Tip: n(S) counts outcomes, not the answers to some question about them. "Even or odd" is a way of grouping these six outcomes into two events; it does not turn the sample space into a two-element set.
Q2.
For the following experiments write down the sample space S. (i) Rolling a die and tossing a coin together. (ii) Choosing a random integer between – 5 and + 5. (iii) A box containing 5 green and 7 red balls. One ball is drawn at random.
Answer

(i) Die and coin together. Each of the 6 die faces can pair with each of the 2 coin faces.

S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H),
     (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
n(S) = 6 × 2 = 12

All 12 are equally likely, each with probability 1/12.

(ii) A random integer between −5 and +5. Read "between" strictly, so the two ends are not included:

S = {−4, −3, −2, −1, 0, 1, 2, 3, 4}
n(S) = 9

If your teacher intends the endpoints to be included, then S = {−5, −4, …, 4, 5} and n(S) = 11. Say which reading you are using — that is part of the answer.

(iii) One ball from 5 green and 7 red. Two correct sample spaces, and they are not interchangeable:

Sample spacen(S)Equally likely?Use it for
S = {Green, Red}2No — P(G) = 5/12, P(R) = 7/12Naming what can happen
S = {G1, …, G5, R1, …, R7}12Yes — 1/12 eachCalculating probabilities
Why it happens: this is the single most important idea in the section. The list {Green, Red} is a perfectly valid sample space — it is complete and has no repeats — but its two members are not equally likely, because there are more red balls than green. So the counting formula must not be used on it: P(green) is 5/12, not 1/2. Labelling the balls G1…G5, R1…R7 produces 12 outcomes that are equally likely, because the draw is at random and every ball is as easy to pick as every other. Only then does n(A)/n(S) give the right answer: 5/12.
Tip: whenever you are about to divide by n(S), stop and ask "would I bet the same amount on each of these?" If the answer is no, refine the list until it is yes.
Q3.
In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi. (i) List the sample space of all possible snack and drink combinations a person could choose at the fair. (ii) List the event ‘Selecting Samosa as a snack.’
Answer

(i) Pair every snack with every drink:

S = {(Samosa, Chai), (Samosa, Lassi),
     (Pakora, Chai), (Pakora, Lassi),
     (Bhaji, Chai), (Bhaji, Lassi)}
n(S) = 3 × 2 = 6
ChaiLassi
Samosa(Samosa, Chai)(Samosa, Lassi)
Pakora(Pakora, Chai)(Pakora, Lassi)
Bhaji(Bhaji, Chai)(Bhaji, Lassi)

(ii) The event is the set of outcomes in which the snack is a samosa — the shaded first row:

E = {(Samosa, Chai), (Samosa, Lassi)}
n(E) = 2

If a villager chooses snack and drink completely at random, then P(E) = 2/6 = 1/3.

Why it happens: the grid shows why 3 × 2 = 6 and not 3 + 2 = 5. The two choices are made together, so each snack must be offered every drink; the sample space is one cell of the grid, not one row or one column. The event 'Samosa as a snack' is an entire row, which is a reminder that an event is a subset of S and may hold more than one outcome — here two.
Tip: the same 3 × 2 grid is the sample space for a tree diagram with 3 first branches and 2 second branches — the next section's Fig. 7.6 is exactly this picture drawn sideways.
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