MathsCBSE Class 9 · Introduction to Probability

Experimental and Theoretical Probability: Class 9

Updated 17 October 2026 · Fact-checked

Experimental probability is the fraction of trials in which an event happened: $\frac{\text{number of times it occurred}}{\text{total number of trials}}$. Theoretical probability, for equally likely outcomes, is $\frac{\text{favourable outcomes}}{\text{all possible outcomes}}$. Every probability lies between 0 and 1, and $P(\text{not } E) = 1 - P(E)$. With more trials the experimental value usually settles near the theoretical value.

Experimental probability comes from counting trials (124/200), theoretical from equally likely outcomes (3/6); both lie between 0 and 1.

Illustration drawn for QuestionBuddy.

Understand Experimental and Theoretical Probability

Probability measures how likely an event is, on a scale from $0$ (impossible) to $1$ (certain). An event with probability $\frac{1}{2}$ is as likely to happen as not.

Experimental probability. Repeat an experiment many times and record how often the event occurs. The relative frequency is
$ $P (E) = \frac{number of trials in which E occurs}{total number of trials} .$ $
If a coin is tossed $200$ times and shows heads $124$ times, the experimental probability of heads is $\frac{124}{200} = 0.62$ and of tails $0.38$ . A second set of $200$ tosses will probably give a slightly different number: experimental probabilities vary from experiment to experiment.

Theoretical probability. When all outcomes of an experiment are equally likely, the probability of an event is
$ $P (E) = \frac{number of outcomes favourable to E}{total number of possible outcomes} .$ $
For a fair die the outcomes $1$ to $6$ are equally likely, so $P (prime) = \frac{3}{6} = \frac{1}{2}$ (the primes are $2, 3, 5$ ). This definition needs equally likely outcomes: "it will rain or it will not rain" is two outcomes but not equally likely, so the probability is not $\frac{1}{2}$ by that argument.

Complement. $P (not E) = 1 - P (E)$ . In a bag of 6 yellow, 9 white and 5 black balls, $P (not black) = 1 - \frac{5}{20} = \frac{3}{4}$ . Often it is quicker to count the opposite event (for "at least one", count "none").

Combining outcomes. For events that cannot happen together, add the favourable counts: $P (white or black) = \frac{9 + 5}{20}$ . If events overlap, subtract the overlap once. Among numbers 1 to 30, 10 are prime and 4 are multiples of 7, but 7 is both, so the number that are prime *or* a multiple of 7 is $10 + 4 - 1 = 13$ .

Estimating from samples. A sample can estimate a probability for a population. If $7$ of $250$ bulbs sampled from a factory are defective, the estimated probability of a defective bulb is $\frac{7}{250} = 0.028$ , and in a lot of $10, 000$ bulbs we would expect about $280$ defective ones. This is an estimate, not a certainty, and a larger and more random sample gives a better estimate.

Law of large numbers. As the number of trials grows, the experimental probability tends to move close to the theoretical probability. In 10 tosses, 7 heads gives $0.7$ ; in 100 tosses, 56 heads gives $0.56$ ; in 1000 tosses, 508 heads gives $0.508$ . Small samples can mislead. Also, past results do not change the chances of the next trial: after five heads in a row a fair coin still has probability $\frac{1}{2}$ of heads next time. Thinking otherwise is the *gambler's fallacy*.

Experimental and Theoretical Probability: formulas to remember

Experimental probability: $P (E) = \frac{times E occurred}{total trials}$; Relative frequency; varies between experiments.
Theoretical probability: $P (E) = \frac{favourable outcomes}{all outcomes}$; Only for equally likely outcomes.
Range: $0 \leq P (E) \leq 1$; 0 is impossible, 1 is certain.
Complement: $P (not E) = 1 - P (E)$; Use for "at least one".
Sum of all outcomes: $P (E_{1}) + P (E_{2}) + \dots = 1$; For all single outcomes of an experiment.
Expected number: $expected = P (E) \times N$; $N$ = number of trials or size of the lot.

How to solve Experimental and Theoretical Probability questions

Decide whether the data is from trials (experimental) or from equally likely outcomes (theoretical).

1List or count all possible outcomes and check that they are equally likely.
2Count the outcomes that make the event happen.
3Write probability as favourable over total and reduce the fraction.
4For "not", "at least one" or "neither", count the opposite and subtract from 1.
5For "or" with overlap, add the counts and subtract the overlap once.
6For data tables, use frequency over total trials; for estimates, multiply the probability by the population size.
7Check that the answer lies between 0 and 1.

Experimental and Theoretical Probability in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Experimental probability

Suppose a weather record shows rain on 18 of the last 60 days of a monsoon month.

Estimate the probability of rain on a day, and the number of rainy days to expect in the next 30 days.

Try it first, then show the solutionSolution

Experimental probability $= \frac{18}{60} = 0.3$ .
Expected rainy days $= 0.3 \times 30 = 9$ .

Answer: The estimated probability of rain is 0.3, so about 9 rainy days are expected in 30 days.

Memory hook: Probability times number of days.

Tricks to remember Experimental and Theoretical Probability

Short memory aids. Say each one aloud once, then use it on a question the same day.

Favourable over total
Count what you want and divide by everything that can happen, provided all outcomes are equally likely.
Use it when: Theoretical probability.
One minus the opposite
For "at least one" or "not", find the opposite event and subtract from 1.
Use it when: Complement questions.
Coins have no memory
Past tosses do not change the next toss.
Use it when: Gambler's fallacy questions.

Experimental and Theoretical Probability: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Coin experiment · 3 marks · Short answerBoard level

A coin is tossed 200 times and heads appear 124 times. Find the experimental probability of heads and of tails, and compare the probability of heads with the theoretical value.

Try it first, then show the solutionSolution

$P (heads) = \frac{124}{200} = 0.62$ . (1)
Tails appeared $200 - 124 = 76$ times: $P (tails) = \frac{76}{200} = 0.38$ ; check $0.62 + 0.38 = 1$ . (1)
For a fair coin the theoretical $P (heads) = 0.5$ ; the experimental value is a little higher, which is possible with a finite number of tosses. (1)

Answer: $P (heads) = 0.62$ , $P (tails) = 0.38$ ; theoretical value $0.5$ .

Example 2: Balls in a bag · 3 marks · Short answerBoard level

A bag has 6 yellow, 9 white and 5 black balls. One ball is drawn at random. Find the probability that it is (i) yellow, (ii) not black, (iii) white or black.

Try it first, then show the solutionSolution

Total balls $= 6 + 9 + 5 = 20$ . (½)
(i) $P (yellow) = \frac{6}{20} = \frac{3}{10}$ . (½)
(ii) Not black $= 6 + 9 = 15$ balls: $\frac{15}{20} = \frac{3}{4}$ , or $1 - \frac{5}{20} = \frac{3}{4}$ . (1)
(iii) White or black $= 14$ balls: $\frac{14}{20} = \frac{7}{10}$ . (1)

Answer: (i) $\frac{3}{10}$ ; (ii) $\frac{3}{4}$ ; (iii) $\frac{7}{10}$ .

Example 3: Adding a ball changes the odds · 5 marks · Long answerHarder board level

A bag contains $x$ red and $y$ blue marbles. The probability of drawing a red marble is $\frac{2}{5}$ . When 5 more red marbles are added, it becomes $\frac{1}{2}$ . Find $x$ and $y$ .

Try it first, then show the solutionSolution

First condition: $\frac{x}{x + y} = \frac{2}{5}$ , so $5 x = 2 x + 2 y$ and $3 x = 2 y$ . (1½)
Second condition: $\frac{x + 5}{x + y + 5} = \frac{1}{2}$ , so $2 x + 10 = x + y + 5$ and $y = x + 5$ . (1½)
Substitute: $3 x = 2 (x + 5)$ , so $x = 10$ and $y = 15$ . (1)
Check: $\frac{10}{25} = \frac{2}{5}$ and $\frac{15}{30} = \frac{1}{2}$ . (1)

Answer: The bag has 10 red and 15 blue marbles.

Example 4: Estimating from a sample · 3 marks · Short answerHarder board level

A random sample of 250 bulbs from a factory contains 7 defective bulbs. Estimate the probability that a bulb is defective, the probability that it is not defective, and the number of defective bulbs to expect in a lot of 10,000.

Try it first, then show the solutionSolution

$P (defective) \approx \frac{7}{250} = 0.028$ . (1)
$P (not defective) \approx 1 - 0.028 = 0.972$ . (1)
Expected defective bulbs $= 0.028 \times 10000 = 280$ . (1)

Answer: About 0.028, 0.972 and 280 bulbs.

Example 5: Many tosses or few? · 4 marks · Case-basedHOTS / competency

Case: a student tosses a coin and records heads: 7 in 10 tosses, 56 in 100 tosses, 508 in 1000 tosses. (i) Find the experimental probability of heads in each case. (ii) What do you notice? (iii) Leela says "it came up heads 5 times in a row, so tails is due". Is she right? (iv) Does 7 heads in 10 tosses prove the coin is biased?

Try it first, then show the solutionSolution

(i) $0.7$ , $0.56$ and $0.508$ . (1)
(ii) The values come closer to $0.5$ as the number of tosses grows: the law of large numbers. (1)
(iii) No. Tosses are independent, so the next toss still has probability $\frac{1}{2}$ of tails; thinking otherwise is the gambler's fallacy. (1)
(iv) No. A fair coin can easily give 7 heads in 10 tosses; a small sample cannot prove bias, and the larger samples are close to $0.5$ . (1)

Answer: (i) 0.7, 0.56, 0.508; (ii) approaches 0.5; (iii) no; (iv) no, 10 tosses are too few.

Where students lose marks in Experimental and Theoretical Probability

The slips examiners see most often, and what to write instead.

Mistake: Assuming equally likely when they are not
Counting outcomes without checking the chances.
Wrong working
"Rain or no rain": $P (rain) = \frac{1}{2}$ .
Right working
The two outcomes are not equally likely; use data from past days.
Do this: Check the equal-likelihood condition before using favourable over total.
Usually costs: The conclusion.
Mistake: Counting the overlap twice
Adding both counts without subtracting the common outcome.
Wrong working
Prime or a multiple of 7 in 1 to 30: $10 + 4 = 14$ .
Right working
$10 + 4 - 1 = 13$ since 7 is in both.
Do this: List both sets and look for common members.
Usually costs: 2 marks.
Mistake: Probability bigger than 1
Dividing total by favourable.
Wrong working
$\frac{20}{5}$ for the probability of red.
Right working
$\frac{5}{20} = \frac{1}{4}$ .
Do this: Favourable on top; check the answer is at most 1.
Usually costs: The whole answer.

Exam tips

Always write the total number of outcomes first.
Give probabilities as reduced fractions unless decimals are asked for.
State the reason in "fair" or "unfair" questions: compare with the theoretical value.
Use the complement for "at least one" questions.

Practice questions on Experimental and Theoretical Probability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A number is chosen at random from 1 to 20. The probability that it is a multiple of 3 is
(a) $\frac{1}{4}$
(b) $\frac{3}{10}$
(c) $\frac{1}{3}$
(d) $\frac{7}{20}$
Hint
List 3, 6, 9, 12, 15, 18.
Answer
(b) $\frac{3}{10}$ . There are 6 multiples of 3 in 20 numbers: $\frac{6}{20} = \frac{3}{10}$ .
Q2 · 3 marks · Short answerBoard level
A fair die is rolled once. Find the probability of getting (i) a prime number, (ii) a number greater than 4, (iii) an even number greater than 2.
Hint
Write the favourable outcomes explicitly.
Answer
(i) ${2, 3, 5}$ : $\frac{3}{6} = \frac{1}{2}$ . (ii) ${5, 6}$ : $\frac{2}{6} = \frac{1}{3}$ . (iii) ${4, 6}$ : $\frac{2}{6} = \frac{1}{3}$ .
Q3 · 3 marks · Short answerHarder board level
A card is drawn at random from a standard pack of 52 cards. Find the probability that it is (i) a red face card, (ii) neither a heart nor a king.
Hint
There are 13 hearts and 4 kings, and the king of hearts is counted in both.
Answer
(i) There are 3 face cards (jack, queen, king) in each of the 2 red suits: $6$ red face cards, $\frac{6}{52} = \frac{3}{26}$ . (ii) Hearts or kings: $13 + 4 - 1 = 16$ cards. Neither: $52 - 16 = 36$ , so $\frac{36}{52} = \frac{9}{13}$ .
Q4 · 5 marks · Long answerHarder board level
Cards numbered 1 to 30 are placed in a box and one is drawn at random. Find the probability that the number is (i) a perfect square or a multiple of 7, (ii) neither of these, (iii) a prime number or a multiple of 7.
Hint
Squares: 1, 4, 9, 16, 25; multiples of 7: 7, 14, 21, 28. Check for overlap in each part.
Answer
(i) 5 squares and 4 multiples of 7, with no number in both: $\frac{9}{30} = \frac{3}{10}$ . (ii) $1 - \frac{3}{10} = \frac{7}{10}$ (21 cards). (iii) Primes up to 30: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$ (10 numbers); multiples of 7: 4 numbers; 7 is in both. So $10 + 4 - 1 = 13$ cards and the probability is $\frac{13}{30}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: an 8-sector spinner (sectors numbered 1 to 8, assumed fair) is spun 80 times. The results are: 1 came 9 times, 2: 11, 3: 8, 4: 12, 5: 10, 6: 9, 7: 11, 8: 10. (i) Find the experimental probability of an even number. (ii) Find the theoretical probability of an even number. (iii) Find the experimental probability of a number greater than 5 and compare with the theoretical value. (iv) Does the data suggest that the spinner is unfair?
Hint
Add the frequencies of the sectors you need.
Answer
(i) Even: $11 + 12 + 9 + 10 = 42$ , so $\frac{42}{80} = 0.525$ . (ii) $\frac{4}{8} = 0.5$ . (iii) Greater than 5 (6, 7, 8): $9 + 11 + 10 = 30$ , so $\frac{30}{80} = 0.375$ , the same as the theoretical $\frac{3}{8} = 0.375$ . (iv) The experimental values are close to the theoretical ones and the frequencies are spread fairly evenly, so there is no strong sign of unfairness.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Range of probability?Show answer
0 to 1.
2.Probability of a certain event?Show answer
1.
3. $P (not E)$ ?Show answer
$1 - P (E)$ .
4.Experimental probability formula?Show answer
Number of times the event occurred over total trials.
5.What is the gambler's fallacy?Show answer
Believing past results change the chance of the next independent trial.

Experimental and Theoretical Probability: frequently asked questions

Can probability be negative or more than 1?

No. It always lies between 0 and 1 inclusive.

Why does experimental probability differ from theoretical?

Because it comes from a limited number of random trials; with many more trials it tends to move closer.

What does probability 0.5 mean?

The event is as likely to happen as not; in many trials it happens about half the time.