NCERT Solutions Class 9 Maths Exercise 7.2 — The Mathematics of Maybe: Introduction to Probability

NCERT Class 9 Mathematics — Ganita Manjari Part I, The Mathematics of Maybe: Introduction to Probability, Exercise Set 7.2. All 6 questions solved.

Two ways of putting a number on a probability, and this exercise uses both.

$$\text{Experimental probability} = \frac{\text{number of times the event occurred}}{\text{total number of trials}}$$

$$\text{Theoretical probability} = \frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$$

The experimental one comes from data — counting what actually happened. The theoretical one comes from reasoning, and it requires the outcomes to be equally likely. Where that assumption fails — a paper cup, a drawing pin, a bag of sweets in unknown proportions — only the experimental route is available.

Key insight. Experimental probability is an estimate, and estimates from small samples wobble. Twelve rolls of a die can easily give three threes rather than two; six thousand rolls will not stray nearly so far from a sixth. More data means a better estimate, and that is the whole content of Question 6.

Question 1

A teacher mixes a large bag of sweets of different colours and randomly selects a sample of $30$ sweets: $10$ red, $8$ green, $7$ yellow, $5$ blue.

Solution.

(i) Calculate the probability that a randomly picked sweet from the sample is green

$$P(\text{green}) = \frac{8}{30} = \frac{4}{15} \approx 0.267$$

(ii) If there are $600$ sweets in total in the large bag, estimate how many are likely to be yellow

The sample is assumed to reflect the whole bag, so the proportion of yellow sweets should be about the same:

$$\frac{7}{30} \text{ of the sample} \quad\Longrightarrow\quad \frac{7}{30} \times 600 = 140$$

So about $140$ yellow sweets.

A quick sense check: $600$ is $20$ times $30$, so multiply each sample count by $20$ — $200$ red, $160$ green, $140$ yellow, $100$ blue, totalling $600$ ✓

This is an estimate, not a fact. A different sample of $30$ would give slightly different numbers, and the true count could be anywhere near $140$.

Question 2

A survey of $40$ students gives: $14$ Science Club, $11$ Arts Club, $9$ Sports Club, $6$ Debate Club. Assume there are $800$ students in the whole school.

Solution.

(i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?

$$P(\text{Arts}) = \frac{11}{40} = 0.275$$

(ii) Estimate how many students in the whole school are likely to prefer the Sports Club

$$\frac{9}{40} \times 800 = 9 \times 20 = 180 \text{ students}$$

Check the whole school by the same scaling ($800 \div 40 = 20$): $280$ Science, $220$ Arts, $180$ Sports, $120$ Debate — totalling $800$ ✓

The estimate is only as good as the sample is representative. If the $40$ students had all been surveyed outside the science lab, the figures would be badly skewed — which is why the question says the sample was random.

Question 3

Toss a coin $20$ times and record the result each time.

Solution. This is an experiment to carry out, so the numbers depend on your own tosses. Suppose the result were $12$ heads and $8$ tails — a plausible outcome — and note where the answer does not depend on it.

(i) How many times did you get heads?

Count them. With the illustrative data, $12$.

(ii) How many times did you get tails?

$$20 – 12 = 8$$

The two counts must add to $20$, which is the check to run on your tally.

(iii) Calculate the experimental probability of getting heads

$$P(\text{heads}) = \frac{\text{heads}}{20} = \frac{12}{20} = \frac{3}{5} = 0.6$$

Note this is not $0.5$, even for a perfectly fair coin. Twenty tosses is far too few to land on the theoretical value exactly, and results between about $6$ and $14$ heads are entirely ordinary.

(iv) If you toss the coin once more, what is the probability of getting tails?

$$P(\text{tails}) = \frac{1}{2}$$

This does not depend on the previous twenty tosses at all. The coin has no memory: it is not “due” a tail after a run of heads, and the next toss is a fresh, fair, two-outcome experiment.

Here the theoretical probability is the right one to quote, because the coin is fair and its two outcomes are equally likely. The experimental figure of $0.6$ from part (iii) is an estimate of that same $\frac12$, made from too little data.

Question 4

Toss a paper cup into the air $100$ times. After each toss record whether the cup lands on its bottom, upside down on its top, or on its side. Assign probabilities to the outcomes by using experimental probability.

Solution. The method. Keep a tally with three rows — bottom, top, side — and make a mark for each toss. Then

$$P(\text{outcome}) = \frac{\text{number of times it landed that way}}{100}$$

A representative set of results. A paper cup is wider at the top than at the bottom, so it lands on its side far more often than either end. Typical counts might be:

The three probabilities must add to $1$, since one of the three must happen. That is the check on your tally.

Why this experiment and not a coin. The three outcomes are not equally likely — and, more importantly, there is no way to reason out how likely each is. The shape of the cup, its weight and the height of the toss all matter in ways no formula captures.

So the theoretical route is unavailable, and the experimental one is the only route there is. That is exactly why the chapter uses a paper cup here: it shows what experimental probability is for. Real questions — whether a drug works, whether a machine will fail — are nearly all of this kind.

Question 5

What is the probability of getting an even number when rolling a fair $6$-sided die?

Solution. Here the outcomes are equally likely, so theoretical probability applies.

Possible outcomes: $1, 2, 3, 4, 5, 6$ — six of them.

Favourable outcomes (even numbers): $2, 4, 6$ — three of them.

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

Question 6

Suppose you roll a $6$-sided die $12$ times and get a ‘$3$’ three times.

Solution.

(i) What is the experimental probability of rolling a ‘$3$’?

$$\frac{3}{12} = \frac{1}{4} = 0.25$$

(ii) What is the theoretical probability of rolling a ‘$3$’?

One face in six, all equally likely:

$$\frac{1}{6} \approx 0.167$$

(iii) Why might these probabilities be different? What would you expect if you roll the die $60$, $600$ or $6000$ times?

Why they differ. Twelve rolls is a very small number. With a fair die the expected number of threes in $12$ rolls is $12 \times \frac16 = 2$, and getting $3$ instead of $2$ is an entirely ordinary fluctuation — a single extra three. It is no evidence at all that the die is loaded.

What more trials would do. As the number of rolls grows, the experimental probability settles down towards the theoretical one. This is the law of large numbers.

The count drifts further from the expected number in absolute terms — $1$ off, then $4$, then $12$ — but the proportion gets steadily closer to $\frac16 \approx 0.167$. That is the whole point: errors grow more slowly than the number of trials, so dividing by the number of trials shrinks them away.

Common mistakes

• Question 1(ii), giving the sample count. $7$ is how many yellow sweets were in the sample of $30$. The question asks about all $600$, so the proportion must be scaled up.
• Questions 1 and 2, treating the estimate as exact. A sample gives an estimate. “About $140$” is the honest answer; “exactly $140$” is not.
• Question 3(iii), answering $0.5$ without doing the experiment. That is the theoretical probability. The experimental one is whatever your own tosses gave, and it will usually differ.
• Question 3(iv), thinking the coin is “due” a tail. After ten heads in a row the next toss is still $\frac12$. This is the gambler’s fallacy, and coins have no memory.
• Question 4, assuming the three outcomes are equally likely. They are not — a cup lands on its side far more often than on either end — which is precisely why the experiment is worth doing.
• Question 4, probabilities that do not add to $1$. One of the three landings must happen, so the three counts must total $100$ and the three probabilities must total $1$.
• Question 6(iii), concluding the die is loaded. Three threes out of twelve is normal variation, not evidence of bias. Many more rolls would be needed before a claim like that could be made.

Practise next

• Exercise Set 7.3 — sample spaces and events, the vocabulary that makes theoretical probability precise.
• The End-of-Chapter Exercises, Questions 2 and 7 — a relative-frequency question and a tyre survey, both of which are Questions 1 and 2 here with different data.