Shunya / Probability & Statistics

Shunya · Maths lab

Probability & statistics

Build a sample space, tap together an event, and watch P(E) update live. Flip coins and roll dice thousands of times and watch chance settle onto the theory. Draw balls from a bag with and without replacement, drag data points to see the mean, median and mode react, work with grouped data and the ogive, then explore how binomial and normal distributions take shape.

Lessons

Pick your class, follow the steps in the lab, and check what you see against the working under the chart.

Class 9 — Probability from experiments (coins and dice)

Concept: when you cannot list all the outcomes in advance, you find a probability by experiment: $\text{experimental probability} = \dfrac{\text{number of trials in which the event happened}}{\text{total number of trials}}$ The more trials you do, the closer it gets to the true (theoretical) value.

Open Coins & dice, choose 1 coin and press ×10 a few times. The bars for Heads and Tails wobble a lot.
Press ×1,000. Now both bars sit close to 50%. Look at the graph on the right: the running frequency settles onto the dashed line.
Move Chance of heads to 0.8 to make a biased coin, run 1,000 trials and check that Heads comes up about 80% of the time.
Switch to 1 die. Follow the outcome 6 and run 1,000 trials: it happens about 1 time in 6 (0.1667).
Tick Auto-run and watch the wobble die down as the number of trials grows.

Takeaway: experimental probability is an estimate. It moves towards the theoretical value as the number of trials grows.

Class 9 — Mean, median and mode

Concept: the mean is the sum divided by the number of values, the median is the middle value of the sorted list, and the mode is the value that occurs most often.

Open Mean, median, mode with the Runs scored data. Read the mean (20.1), the median (21) and the mode (22) from the cards.
Read the working under the graph: the mean is 201 ÷ 10, and the median is the average of the 5th and 6th sorted values.
Drag the right-most dot far to the right. The mean follows it, but the median barely moves.
Press Two modes: some data has more than one mode. Press All equal: mean, median and mode are all the same.
Type your own numbers in the box (for example your test marks) and read their mean, median and mode.

Takeaway: the mean uses every value, so extreme values pull it. The median only cares about the middle, so it resists them.

Class 10 — Probability of an event (equally likely outcomes)

Concept: if all outcomes are equally likely, $P(E) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ It is always between 0 (impossible) and 1 (certain), and $P(\text{not } E) = 1 - P(E)$ .

Open the Probability mode, pick Die, and tap the preset Even number: E = {2, 4, 6}, n(E) = 3, n(S) = 6, so P(E) = 3/6 = 1/2.
Still in Probability, switch to Playing cards and try the preset A king: 4 kings out of 52 cards, so P(king) = 4/52 = 1/13. Then try A face card (J, Q, K) (12/52 = 3/13) and A red card (26/52 = 1/2) — the classic "a card is drawn from a well-shuffled deck" question, worked out live.
Open Coins & dice and choose 2 dice (sum). Follow the sum 7: read the working. There are 6 pairs that make 7 out of 36, so P = 6/36 = 1/6.
Follow the sum 2, then 12: only one pair each, so P = 1/36. The sums 6, 7, 8 are the most likely.
Run 1,000 trials and compare the bars (what happened) with the orange marks (the theory).
Open Bag of balls with 4 red, 3 blue and 2 green. P(red) = 4/9, and P(not red) = 1 − 4/9 = 5/9.

Takeaway: list the equally likely outcomes, count the favourable ones, and divide.

Class 10 — Mean, median and mode of grouped data

Concept: for data in classes, use the class mark $x = \tfrac{l + u}{2}$ . Then $\bar{x} = \dfrac{\sum fx}{\sum f}$ , and the median and mode use the formulas shown in the working.

Open Grouped data and press Marks in a test. Read the table under the graph: class, f, x, f·x and the cumulative frequency cf.
The mean is Σ f·x ÷ Σ f = 1350 ÷ 50 = 27. The median class is the first class whose cf is at least n/2 = 25.
Find the median with $l + \dfrac{n/2 - cf}{f}\times h$ and the mode with $l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\times h$ . Check them against the cards.
Drag a bar top up or down and watch all three numbers change.
Switch to Ogive: the median is where the height n/2 meets the curve, the same number the formula gave.

Takeaway: the formulas assume the values are spread evenly inside each class, which is why they give estimates.

Class 11 — Variance and standard deviation

Concept: the variance is the average squared distance from the mean: $\sigma^2 = \dfrac{\sum (x - \bar{x})^2}{n}$ and the standard deviation is $\sigma = \sqrt{\sigma^2}$ .

Open Mean, median, mode and tick show each distance from the mean. Long lines mean a large spread.
Tick show mean ± standard deviation: most values sit inside the band.
Press All equal: every distance is 0, so the variance and standard deviation are 0.
Press Symmetric and drag one dot away from the rest. Watch σ grow.
Read the working: each (x − mean)² is added, divided by n, then the square root is taken.

Takeaway: the standard deviation measures how far values typically sit from the mean, in the same units as the data.

Class 12 — Conditional probability and the multiplication rule

Concept: $P(A \cap B) = P(A)\,P(B \mid A)$ . If the first draw changes what is left in the bag, the second probability is conditional on the first.

Open Bag of balls with 4 red, 3 blue, 2 green and choose Without replacement. Focus on red.
P(1st red) = 4/9. After a red is taken, 3 red remain among 8 balls, so P(2nd red | 1st red) = 3/8.
Multiply: P(both red) = 4/9 × 3/8 = 1/6. Check it on the card, and against the simulation (press ×1,000).
Switch to With replacement: the second probability is 4/9 again, so P(both red) = 16/81. The draws are now independent.
Read P(same colour) by adding the branches RR, BB, GG, and P(different) = 1 − P(same).

Takeaway: without replacement the draws are dependent; with replacement they are independent.

Class 12 — Conditional probability with events, and independence

Concept: P(E | B) is the probability of E given that B has happened. Once B is known, only the outcomes in B are still possible, so the sample space shrinks to B: P(E | B) = n(E ∩ B) ÷ n(B).

Open Probability, choose Die, and press Even, given more than 3. B = {4, 5, 6} (dashed blue); E = {2, 4, 6}. Only 4 and 6 are in both, so P(E | B) = 2/3, while P(E) alone is 1/2.
Choose Playing cards and press A king, given a face card: there are 12 face cards and 4 of them are kings, so P(king | face card) = 4/12 = 1/3, although P(king) = 1/13. The reverse, P(face card | king), is 1: P(E | B) and P(B | E) are different questions.
Choose Two coins and press Both heads, given at least one head: the sample space shrinks from {HH, HT, TH, TT} to {HH, HT, TH}, so the answer is 1/3, not 1/2 (a classic trap). If instead you are told the first coin is heads, the answer is 1/2.
Independent events: on the cards, press An ace and a spade. P(ace | spade) = 1/13 = P(ace), so knowing the card is a spade tells you nothing about being an ace, and P(ace ∩ spade) = 1/52 = 4/52 × 13/52. For a die, “even” and “at most 4” are also independent (both 1/3).
Multiplication rule: P(E ∩ B) = P(B) · P(E | B). Addition rule: P(E ∪ B) = P(E) + P(B) − P(E ∩ B). The working panel applies both to your events.

Takeaway: P(E | B) = P(E ∩ B) ÷ P(B); events are independent exactly when P(E ∩ B) = P(E) × P(B), i.e. when P(E | B) = P(E).

Class 12 — Total probability and Bayes' theorem

Concept: if the sources H₁, H₂, … cover every case without overlap (a partition), then P(E) = Σ P(H_i) P(E | H_i) (total probability) and P(H_i | E) = P(H_i) P(E | H_i) ÷ P(E) (Bayes' theorem): it reverses a probability, from “how often does the cause give the evidence” to “how likely is the cause given the evidence”.

Open Bayes' theorem, scenario Medical test: 1% of people have a disease, the test is positive for 95% of the sick and, wrongly, for 5% of the healthy. In 10,000 people, 100 are sick (95 test positive) and 9,900 healthy (495 test positive).
P(positive) = 0.01 × 0.95 + 0.99 × 0.05 = 0.059. P(disease | positive) = 0.0095 ÷ 0.059 = 16.1%: a positive result on a good test still means probably healthy, because the healthy people are so many more (the base-rate trap). Press Common disease: 30% and it jumps to about 89%.
Press Given: test negative: P(disease | negative) is only 0.05%, so a negative result is very reassuring.
Three machines make 50%, 30% and 20% of the items with 1%, 2% and 3% defective. P(defective) = 0.005 + 0.006 + 0.006 = 0.017. A defective item came from A, B or C with probability 29.4%, 35.3% and 35.3%: machine A is the biggest producer but the least likely culprit.
Three urns (60%, 30% and 90% red, chosen equally likely): P(red) = 0.6. Given a red ball, urn III is the most likely (50%). Press Test 100 ×10 to simulate and watch the thin bar approach the theory.

Takeaway: add the branches to get P(E), then divide the branch you want by that total. Check that the posteriors add up to 1.

Class 12 — Random variables, mean and variance

Concept: a random variable X takes the value x with probability p, and the p's add up to 1. The mean (expected value) is E(X) = Σ x p, the variance is Var(X) = E(X²) − [E(X)]², and the standard deviation is σ = √Var.

Open Random variable (the fair die). E(X) = (1 + 2 + … + 6) ÷ 6 = 3.5 and E(X²) = 91/6 = 15.1667, so Var(X) = 15.1667 − 3.5² = 35/12 = 2.9167 and σ = 1.7078. The mean 3.5 is a long-run average: X can never equal it.
Sum of two dice: E(X) = 7 (the most likely sum, 6 ways out of 36) and Var(X) = 35/6 = 5.8333. Heads in 3 tosses: X = 0, 1, 2, 3 with probabilities 1/8, 3/8, 3/8, 1/8, so E(X) = 1.5 and Var(X) = 0.75, which is npq for n = 3, p = ½.
A game. Win ₹10 with probability ¼ or lose ₹4 with probability ¾: E(gain) = 10 × ¼ − 4 × ¾ = −0.5. You lose half a rupee per play on average. The “fair game” example (win ₹3 with probability 2/5, lose ₹2 with probability 3/5) has E = 0.
Change of variable. Move the sliders to Y = aX + b. For the die with a = 2 and b = 3: E(Y) = 2 × 3.5 + 3 = 10 and Var(Y) = 2² × 2.9167 = 11.667. Adding b slides the picture along the axis (the mean moves, the spread does not); multiplying by a stretches it and multiplies the variance by a².
Choose Your own distribution, set the weights, and watch the probabilities, the mean and the shaded band E ± σ update. The weights are scaled so the probabilities always add up to 1.

Takeaway: E(X) = Σ x p; Var(X) = E(X²) − [E(X)]²; E(aX + b) = aE(X) + b; Var(aX + b) = a² Var(X).

Class 12 — The Poisson distribution

Concept: for events that occur independently at an average rate λ per interval of time or space (calls, misprints, radioactive counts), the number X in one interval has P(X = k) = e^−λ λ^k ÷ k!, with mean and variance both equal to λ.

Open Poisson with λ = 3. P(X = 3) = e⁻³ × 27 ÷ 6 = 0.0498 × 4.5 = 0.2240. P(X ≤ 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472, so P(X ≥ 3) = 0.5768.
At least one event: P(X ≥ 1) = 1 − e^−λ = 1 − 0.0498 = 0.9502. For a switchboard that gets 2 calls a minute (λ = 2), the chance of no call in a minute is e⁻² = 0.1353 and of at least one is 0.8647.
The shape. P(k + 1) = P(k) × λ ÷ (k + 1), so the bars rise while k + 1 < λ and fall afterwards. For λ = 3 the two middle bars (k = 2 and k = 3) are equal, 0.2240, so there are two most likely values; for a non-integer λ there is one, ⌊λ⌋.
Mean = variance = λ. Press ×1,000 to simulate 1,000 intervals: the average of the counts lands close to 3 (and the dots close to the bars). A quick test: if a count’s mean and variance are far apart, it is probably not Poisson.
The limit of the binomial. Switch on compare with the binomial with n = 50 and p = λ ÷ n = 0.06: the biggest difference between a binomial bar and a Poisson bar is only 0.007. With n = 100 and p = 0.03, P(X = 2) is 0.2252 for the binomial against 0.2240 for Poisson. Use Poisson when n is large, p is small and np = λ is moderate.

Takeaway: P(X = k) = e^−λλ^k ÷ k!; E(X) = Var(X) = λ; Poisson(λ) is the limit of Binomial(n, λ ÷ n) as n → ∞.

Class 12 — The binomial distribution

Concept: repeat n independent trials, each with success probability p. The number of successes X has $P(X = k) = \binom{n}{k} p^k q^{n-k}$ with $q = 1 - p$ , mean $np$ and variance $npq$ .

Open Binomial and press 10 tosses, fair. The tallest bar is at k = 5.
Read the working for k = 5: C(10, 5) = 252, p⁵ = q⁵ = 1/32, so P(X = 5) = 252/1024 ≈ 0.2461.
Tick shade P(X ≤ k) and drag the orange dot: the cumulative probability grows as k grows.
Change p to 0.2: the distribution is skewed towards small k. The mean is np = n × p.
Press ×1,000 under the experiments and see the simulated dots land on the bars.

Takeaway: binomial means a fixed number of independent trials with the same chance of success each time.

Class 12 — The normal curve (a first look beyond the core syllabus)

Concept: many measurements (heights, marks, errors) pile up in a bell shape around the mean. The normal curve has $z = \dfrac{x - \mu}{\sigma}$ and the area under it between two values is a probability. This goes beyond the NCERT class 12 core, but it is used in statistics courses and competitive exams.

Open Normal curve with μ = 0 and σ = 1. Read P(−1 < X < 1) ≈ 0.6827 from the card.
Tick show the 68 – 95 – 99.7 rule: about 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations of the mean.
Increase σ: the bell gets wider and lower, but the area stays 1. Change μ: it slides sideways.
Drag a and b to choose any interval. The working shows z = (x − μ)/σ and P = Φ(z_b) − Φ(z_a).
Press Take 1,000 samples: the green bars follow the curve.

Takeaway: the shape depends only on μ and σ, and standardising with z turns every normal curve into the same one.

The maths behind it

What each mode covers

Coins & dice — flip one or two coins (fair or biased) or roll one or two dice up to 100,000 times. Bars show what happened, orange marks show the theory, and the running-frequency graph shows the law of large numbers. Class 9 and 10 probability.

Bag of balls — draw two balls from a bag you fill, with or without replacement. A tree of every outcome shows the exact fractions, and a simulation checks them. Conditional probability and the multiplication rule, class 12.

Bayes' theorem — a tree of 10,000 people (or items, or draws) for a medical test, three machines or three urns. Change the priors and likelihoods and see the total probability and the posterior P(cause | evidence) update, then check it by simulation. In Probability, a second event B gives conditional probability and the independence test. Class 12.

Random variable — a probability distribution as a bar chart (a die, two dice, coin tosses, a game, or your own weights) with its mean, variance and standard deviation, and a Y = aX + b change of variable. Class 12.

Poisson — the number of rare, independent events in an interval at an average rate λ, with the probabilities, the cumulative shading, a simulation and a comparison with the binomial it is the limit of. Class 12.

Mean, median, mode — type your own numbers or drag dots to see the mean, median, mode, range, quartiles, variance and standard deviation react, with a box plot. Class 9 and class 11 statistics.

Grouped data — a histogram whose bars you drag, with the mean, median and mode from the class formulas, a frequency polygon and the less-than ogive. Class 10 statistics.

Binomial — choose n and p, read P(X = k), the cumulative probability, the mean and variance, and compare simulated experiments and the normal curve. Class 12 probability distributions.

Normal curve — set the mean and standard deviation, drag two points to shade the area between them, see the 68–95–99.7 rule and draw random samples. A first look beyond the core class 12 syllabus.

Part of Shunya. See the syllabus map for the chapters these modes map to. Also try the Function & Graph Explorer. Random experiments use your browser's random number generator, so every run is different.

Formulas used in this lab

Experimental probability Class 9

Experimental probability: $P(E) = \dfrac{\text{trials in which } E \text{ happened}}{\text{total number of trials}}$
Expected number of times in n trials: $n \times P(E)$
Difference from the theoretical value: $\bigl|\,\text{experimental } P - \text{theoretical } P\,\bigr|$
Law of large numbers: $\text{experimental } P(E) \longrightarrow \text{theoretical } P(E) \text{ as } n \text{ grows}$

Probability of an event Class 10

Equally likely outcomes: $P(E) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
Range of a probability: $0 \le P(E) \le 1,\qquad P(\text{impossible}) = 0,\qquad P(\text{certain}) = 1$
Complement: $P(\text{not } E) = 1 - P(E)$
All outcomes together: $\sum P(\text{outcome}) = 1$
Two dice, sum s (out of 36 pairs): $P(\text{sum} = s) = \dfrac{6 - |s - 7|}{36},\qquad 2 \le s \le 12$
Two tosses of a coin with P(H) = p: $P(0\text{ H}) = q^2,\quad P(1\text{ H}) = 2pq,\quad P(2\text{ H}) = p^2,\quad q = 1 - p$

Mean, median and mode Class 9

Mean: $\bar{x} = \dfrac{\sum x}{n}$
Median, n odd (the middle value of the sorted list): $\left(\dfrac{n + 1}{2}\right)\text{th value}$
Median, n even: $\dfrac{\left(\frac{n}{2}\right)\text{th value} + \left(\frac{n}{2} + 1\right)\text{th value}}{2}$
Mode (there can be more than one, or none): $\text{most frequent value}$
Range: $\text{largest} - \text{smallest}$

Grouped data Class 10

Class mark and class width: $x = \dfrac{l + u}{2},\qquad h = u - l$
Mean (direct method): $\bar{x} = \dfrac{\sum f x}{\sum f}$
Median: $\text{Median} = l + \dfrac{\frac{n}{2} - cf}{f} \times h$
Mode: $\text{Mode} = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$
Empirical relation: $\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean}$
Median class (the first class with cumulative frequency at least n/2): $cf \ge \dfrac{n}{2}$

Spread: quartiles, variance and standard deviation Class 11

First quartile (median of the lower half): $Q_1 = \text{median}(\text{lower half})$
Third quartile (median of the upper half): $Q_3 = \text{median}(\text{upper half})$
Interquartile range: $\text{IQR} = Q_3 - Q_1$
Variance: $\sigma^2 = \dfrac{\sum (x - \bar{x})^2}{n}$
Standard deviation: $\sigma = \sqrt{\sigma^2}$

Conditional probability and independent events Class 12

Multiplication rule: $P(A \cap B) = P(A)\,P(B \mid A)$
Conditional probability: $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$
Independent events: $P(A \cap B) = P(A)\,P(B)$
Two draws with replacement: $P(\text{both red}) = \dfrac{r}{N}\times\dfrac{r}{N}$
Two draws without replacement: $P(\text{both red}) = \dfrac{r}{N}\times\dfrac{r - 1}{N - 1}$
Same colour, and its complement: $P(\text{same}) = P(RR) + P(BB) + P(GG),\qquad P(\text{different}) = 1 - P(\text{same})$

Total probability and Bayes' theorem Class 12

Multiplication rule: $P(E \cap B) = P(B)\,P(E \mid B) = P(E)\,P(B \mid E)$
Addition rule: $P(E \cup B) = P(E) + P(B) - P(E \cap B)$
Independent events: $P(E \cap B) = P(E)\,P(B), \text{ equivalently } P(E \mid B) = P(E)$
Partition: $H_1,\dots,H_n \text{ mutually exclusive and exhaustive: } P(H_1)+\dots+P(H_n) = 1$
Total probability: $P(E) = P(H_1)P(E \mid H_1) + \dots + P(H_n)P(E \mid H_n)$
Bayes' theorem: $P(H_i \mid E) = \dfrac{P(H_i)\,P(E \mid H_i)}{\sum_j P(H_j)\,P(E \mid H_j)}$

Binomial distribution Class 12

Probability of k successes in n trials: $P(X = k) = \binom{n}{k} p^k q^{n-k},\qquad q = 1 - p$
Binomial coefficient: $\binom{n}{k} = \dfrac{n!}{k!\,(n - k)!}$
At most k successes: $P(X \le k) = \sum_{i=0}^{k} P(X = i)$
At least k successes: $P(X \ge k) = 1 - P(X \le k - 1)$
Mean, variance and standard deviation: $\mu = np,\qquad \sigma^2 = npq,\qquad \sigma = \sqrt{npq}$
Most likely value: $\lfloor (n + 1)p \rfloor$
Normal curve fits well when: $np \ge 5 \ \text{ and } \ nq \ge 5$

Normal distribution Class 12

Density of the normal curve: $f(x) = \dfrac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu)^2}{2\sigma^2}}$
Standardising (z-score): $z = \dfrac{x - \mu}{\sigma}$
Probability between a and b: $P(a < X < b) = \Phi(z_b) - \Phi(z_a)$
The 68 – 95 – 99.7 rule: $P(|X - \mu| < \sigma) \approx 0.6827,\quad P(|X - \mu| < 2\sigma) \approx 0.9545,\quad P(|X - \mu| < 3\sigma) \approx 0.9973$
Total area under the curve: $\int_{-\infty}^{\infty} f(x)\,dx = 1$

Random variables and the Poisson distribution Class 12

Probability distribution: $p_i \ge 0,\qquad \sum_i p_i = 1$
Mean (expected value): $E(X) = \mu = \sum_i x_i\,p_i$
Variance and standard deviation: $\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \sum x_i^2 p_i - \mu^2,\qquad \sigma = \sqrt{\operatorname{Var}(X)}$
Linear change of variable: $E(aX + b) = aE(X) + b,\qquad \operatorname{Var}(aX + b) = a^2\operatorname{Var}(X)$
Poisson probability: $P(X = k) = \dfrac{e^{-\lambda}\lambda^k}{k!},\quad k = 0, 1, 2, \dots$
Mean and variance of Poisson: $E(X) = \operatorname{Var}(X) = \lambda$
Poisson as a limit of the binomial: $\binom nk p^k(1-p)^{n-k} \to \dfrac{e^{-\lambda}\lambda^k}{k!}\ \text{as } n\to\infty,\ np = \lambda$