EN हिं

NCERT Solutions for Class 9th Maths Chapter 7 Exercise Set 7.2 — Measuring Probability Objectively

Book page 165–166 Updated on2026-09-08

Q1.
A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour: 10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets. (i) Calculate the probability that a randomly picked sweet from the sample is green. (ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.
Answer

Check the sample first: 10 + 8 + 7 + 5 = 30 sweets. ✓

(i) Every sweet in the sample is equally likely to be picked, so count green ones against all of them.

P(green) = 8/30
= 4/15 ≈ 0.267 or 26.7%

(ii) The sample gives the yellow proportion; apply it to the whole bag.

P(yellow) = 7/30
Expected yellow in 600 = 600 × 7/30
= 20 × 7
= 140 yellow sweets
Why it happens: this is statistical probability — the estimate comes from data, not from symmetry. The sweets are not equally likely by colour (there is no reason a bag should hold equal numbers), so we cannot argue P(yellow) = 1/4 from "there are four colours". What is equally likely is the pick of one sweet from the 30 in the sample, and that is what licenses the fraction 7/30. Scaling to 600 then assumes the sample is representative of the whole bag.
Tip: 140 is an estimate, not a count. A second sample of 30 might give 6 or 9 yellow and shift the estimate to 120 or 180. A larger sample would pin it down more tightly — that is exactly why the chapter recommends bigger, more representative samples.
Q2.
A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are: 14 students: Science Club | 11 students: Arts Club | 9 students: Sports Club | 6 students: Debate Club. Assume there are 800 students in the whole school. (i) What is the probability that a randomly chosen student from the sample prefers the Arts Club? (ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.
Answer

Check the sample: 14 + 11 + 9 + 6 = 40 students. ✓

(i)

P(Arts Club) = 11/40
= 0.275 or 27.5%

(ii) The Sports Club share in the sample is 9 out of 40. Apply that share to all 800 students.

P(Sports Club) = 9/40 = 0.225
Estimated number in school = 800 × 9/40
= 20 × 9
= 180 students
ClubSample countRelative frequencyEstimate out of 800
Science1414/40 = 0.35280
Arts1111/40 = 0.275220
Sports99/40 = 0.225180
Debate66/40 = 0.15120
Total401.00800
Why it happens: the four relative frequencies add to 1, exactly as the probabilities of a full sample space must. That is a useful check: if your fractions do not add to 1, some response has been missed or double-counted. Multiplying each by 800 therefore splits the school into four groups that add back to 800.
Tip: the estimate is only as good as the sample. If those 40 students were all picked from one class, or all from the school choir, the "random sample" claim fails and the 180 means very little. A sample must be drawn from across the whole school for the scaling step to be honest.
Q3.
Toss a coin 20 times and record the result each time (heads or tails). (i) How many times did you get heads? (ii) How many times did you get tails? (iii) Calculate the experimental probability of getting heads. (iv) If you toss the coin once more, what is the probability of getting tails?
Answer

This one you must actually perform — the answer is your own data. Here is a worked set of results so you can see the method; your counts will differ, and that is the point.

Toss1–56–1011–1516–20
ResultH T H H TT H T H HH T T H TH H T H T

(i) Heads = 11    (ii) Tails = 9    (11 + 9 = 20 ✓)

(iii)

Experimental P(heads) = number of heads / total tosses
= 11/20
= 0.55 or 55%

(iv) This part is not answered from your data. The 21st toss is a fresh, independent trial of a fair coin.

P(tails on the next toss) = 1/2 = 0.5
Why it happens: parts (iii) and (iv) ask two different questions, and mixing them is the classic error here. Part (iii) is experimental probability — it summarises 20 trials that have already happened, so 0.55 is a fact about your data. Part (iv) is theoretical probability — it is about a trial that has not happened yet, and a fair coin does not remember that heads led 11–9. Answering (iv) with 9/20 would be the Gambler's Fallacy in reverse.
Try This: pool the class's results. If 30 students each toss 20 times, you have 600 tosses. The combined proportion of heads will sit far closer to 0.5 than most individual results did — the Law of Large Numbers, visible in one lesson.
Q4.
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 (See Fig. 7.5). Assign probabilities to the outcomes by using experimental probability.
Answer

Here the sample space is S = {bottom, top, side}, with n(S) = 3 — but you cannot write 1/3 for each. A paper cup is not symmetric, so the three outcomes are not equally likely. The only honest way to get their probabilities is to toss and count.

A typical set of 100 tosses, with the tally turned into probabilities:

Landing positionTally (out of 100)Relative frequencyExperimental probability
Bottom (upright)2222/1000.22
Top (upside down)1818/1000.18
Side6060/1000.60
Total100100/1001.00
Experimental P(side) = 60/100 = 0.60
Experimental P(bottom) = 22/100 = 0.22
Experimental P(top) = 18/100 = 0.18
Check: 0.60 + 0.22 + 0.18 = 1 ✓
Why it happens: this experiment is in the chapter precisely to break the habit of dividing by the number of outcomes. The counting formula P = n(A)/n(S) is valid only when every outcome is equally likely, and the cup gives no reason to believe that: its curved wall has far more area to rest on than either rim, so "side" dominates. With no symmetry to argue from, evidence is the only route — which is the whole distinction between theoretical and experimental probability.
Check it yourself: record your results in blocks of 20 and plot the running proportion of "side". It will jump about early on and steady down as the tosses accumulate. Use a cup of a different shape — a wider, shorter one — and the three probabilities change, proving they are properties of that particular cup, not of the number 3.
Q5.
What is the probability of getting an even number when rolling a fair 6-sided die?
Answer
S = {1, 2, 3, 4, 5, 6}, so n(S) = 6
E = 'an even number' = {2, 4, 6}, so n(E) = 3
P(E) = n(E)/n(S) = 3/6
= 1/2 = 0.5 or 50%
Why it happens: the die is fair, so all six faces are equally likely and the counting formula applies without further thought. Three of the six faces carry even numbers, so exactly half the outcomes are favourable. The same argument gives P(odd) = 3/6 = 1/2, and P(even) + P(odd) = 1, as it must, since every roll is one or the other.
Tip: the answer is 1/2 only because evens and odds are equally represented on this die. On a die numbered 1, 1, 2, 3, 4, 5 the sample space of faces still has six equally likely members, but only two show an even number, so P(even) = 2/6 = 1/3. Always count faces, not the numbers written on them.
Q6.
Suppose you roll a 6-sided die 12 times and get a ‘3’ three times. (i) What is the experimental probability of rolling a ‘3’? (ii) What is the theoretical probability of rolling a ‘3’? (iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?
Answer

(i) From the data:

Experimental P(3) = number of 3s / number of rolls
= 3/12
= 1/4 = 0.25 or 25%

(ii) From the symmetry of a fair die:

S = {1, 2, 3, 4, 5, 6}, n(S) = 6; favourable = {3}, n = 1
Theoretical P(3) = 1/6 = 0.1666… ≈ 0.167 or 16.7%

(iii) They differ because 12 rolls is a very short run. In 12 rolls the expected number of 3s is 12 × 1/6 = 2, and getting 3 instead of 2 is an ordinary fluctuation — one extra 3 shifts the fraction from 0.167 to 0.25 because the denominator is so small.

Number of rollsExpected 3s (n × 1/6)One extra 3 changes P byWhat to expect
1220.083Wide swings; 0.25 or 0.08 are unremarkable
60100.017Usually within about 0.10 to 0.23
6001000.0017Usually close to 0.15–0.19
600010000.00017Very close to 0.167
Why it happens: the experimental probability is a fraction whose denominator is the number of trials. As that denominator grows, a single unexpected result moves the fraction less and less, so the value settles. This is the Law of Large Numbers: as trials increase, the experimental probability tends towards the theoretical value of 1/6. Note what it does not promise — it never says the counts must catch up, only that the ratio steadies.
Tip: if you rolled 6000 times and still got about 0.25, the sensible conclusion would not be "the Law of Large Numbers has failed" but "this die is not fair". That is how experimental probability is used to test an assumption.
Was this helpful? Report an error