NCERT Solutions Class 9 Maths Exercise 7.3 — The Mathematics of Maybe: Introduction to Probability

NCERT Class 9 Mathematics — Ganita Manjari Part I, The Mathematics of Maybe: Introduction to Probability, Exercise Set 7.3. All 3 questions solved.

This exercise asks for no arithmetic at all. It asks you to list — carefully, completely, and in a way that leaves nothing out and nothing counted twice.

• A sample space $S$ is the set of all possible outcomes of a random experiment.
• An event is any subset of the sample space — one outcome, several, or none.
• When an experiment has two stages, with $m$ outcomes at the first and $n$ at the second, the sample space has $m \times n$ outcomes. That is the multiplication principle, and it is the check to run on every list you write.

Key insight. A sample space is not unique — it depends on what you decide to record. Twelve balls in a box can be listed as twelve distinct objects or as two colours, and both are legitimate sample spaces for the same experiment. But only the first has equally likely outcomes, and the formula $P = \frac{\text{favourable}}{\text{possible}}$ is valid only when they are. Choosing the sample space is therefore a mathematical decision, not a clerical one.

Question 1

When a single $6$-sided die is rolled, what is the total number of possible outcomes in the sample space?

Solution. The die has six faces, and exactly one of them finishes uppermost:

$$S = \{1, 2, 3, 4, 5, 6\}$$

so the sample space has $6$ outcomes.

Because the die is fair, these six are equally likely, and each has probability $\frac16$. That is what makes this sample space useful — a list of outcomes that were not equally likely would be a correct list but a useless one for computing probabilities.

Question 2

For the following experiments write down the sample space $S$.

Question 2 (i)

Rolling a die and tossing a coin together.

Solution. Each outcome has to record both results, so write it as an ordered pair: the die face first, the coin second.

The die contributes $6$ outcomes and the coin $2$, so by the multiplication principle the sample space has $6 \times 2 = 12$ outcomes:

$$S = \{(1, H), (2, H), (3, H), (4, H), (5, H), (6, H),$$ $$\qquad (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)\}$$

A tidy way to be sure nothing is missed is to run through the die faces once with $H$ and once with $T$ — six and six. All twelve are equally likely, each with probability $\frac{1}{12}$.

Question 2 (ii)

Choosing a random integer between $-5$ and $+5$.

Solution. “Between” ordinarily means strictly between, so the endpoints $-5$ and $+5$ are excluded:

$$S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}, \qquad n(S) = 9$$

Do not forget $0$ — it is an integer, it lies between $-5$ and $5$, and leaving it out is the commonest slip here. Counting: four negative, four positive, and zero makes nine.

The book does not say whether the endpoints are meant to count. If they are included the sample space is

$$S = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}, \qquad n(S) = 11$$

Either is defensible provided you state which reading you have taken. The strict one is the standard meaning of “between”, and that is the answer given here.

Question 2 (iii)

A box containing $5$ green and $7$ red balls. One ball is drawn at random.

Solution. This one has two defensible sample spaces, and the difference between them is the real content of the question.

If you record only the colour, there are two possible outcomes:

$$S = \{\text{Green}, \text{Red}\}, \qquad n(S) = 2$$

If you treat the balls as distinguishable — label them $G_1, \dots, G_5$ and $R_1, \dots, R_7$ — there are twelve:

$$S = \{G_1, G_2, G_3, G_4, G_5, R_1, R_2, R_3, R_4, R_5, R_6, R_7\}, \qquad n(S) = 12$$

Which should you use? It depends on what you are going to do with it. The colour list is shorter and answers the question as asked. But its two outcomes are not equally likely — red is more likely than green — so you cannot put $n(S) = 2$ into $P = \frac{\text{favourable}}{\text{possible}}$ and write $P(\text{green}) = \frac12$. That would be wrong.

The twelve-ball list does have equally likely outcomes, and it gives the right answers at once:

$$P(\text{green}) = \frac{5}{12}, \qquad P(\text{red}) = \frac{7}{12}$$

which add to $1$, as they must.

Question 3

In a village fair, there are $3$ popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.

Solution.

(i) List the sample space of all possible snack and drink combinations a person could choose at the fair

Each choice is a snack and a drink, so again write ordered pairs. Three snacks and two drinks give $3 \times 2 = 6$ combinations:

$$S = \{(\text{Samosa}, \text{Chai}), (\text{Samosa}, \text{Lassi}),$$ $$\qquad (\text{Pakora}, \text{Chai}), (\text{Pakora}, \text{Lassi}),$$ $$\qquad (\text{Bhaji}, \text{Chai}), (\text{Bhaji}, \text{Lassi})\}$$

Listing them snack by snack — each snack paired with each drink in turn — is what guarantees the list is complete.

(ii) List the event ‘Selecting Samosa as a snack’

An event is a subset of the sample space, so pick out those outcomes whose snack is Samosa:

$$E = \{(\text{Samosa}, \text{Chai}), (\text{Samosa}, \text{Lassi})\}$$

Note what the event must not be. It is not “$\{\text{Samosa}\}$”: the outcomes of this experiment are pairs, so the elements of any event are pairs too. The drink is unconstrained by this event, which is exactly why both of the Samosa pairs belong to it.

If every combination were equally likely, this event would have probability

$$P(E) = \frac{2}{6} = \frac{1}{3}$$

which is just the probability of choosing Samosa from three snacks — a useful sanity check that the event was picked out correctly.

Common mistakes

• Question 2(i), writing $S = \{1,2,3,4,5,6,H,T\}$. That is $8$ outcomes, and it is wrong: the die and the coin both land, so every outcome records both. Eight would be the answer if you rolled the die or tossed the coin, which is a different experiment.
• Question 2(ii), forgetting $0$. It is an integer and it lies between $-5$ and $5$. Leaving it out gives $8$ outcomes instead of $9$.
• Question 2(iii), writing $P(\text{green}) = \frac12$. The colour sample space has two outcomes, but they are not equally likely, so dividing by $2$ is not allowed. This is the central trap of the chapter.
• Question 3, listing outcomes as single items. “Samosa, Pakora, Bhaji, Chai, Lassi” is a list of choices, not of outcomes. An outcome of this experiment is one snack paired with one drink.
• Question 3(ii), confusing an event with an outcome. An event is a set. Writing $E = (\text{Samosa}, \text{Chai})$ names one outcome and omits the other half of the event.
• Miscounting a two-stage sample space. Adding when you should multiply — $3 + 2 = 5$ rather than $3 \times 2 = 6$ — is the quickest way to produce an incomplete list. Always cross-check a list against $m \times n$ before you trust it.

Practise next

• Exercise Set 7.4 — the same listing job done with tree diagrams, which make it much harder to miss an outcome.
• The End-of-Chapter Exercises, Questions 4, 13, 14 and 15 — four more sample spaces, one of which asks you to say why three offered lists are not sample spaces.