Chapter 14 of 16

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JEE Main Mathematics★ ~2 questions (8 marks) in JEE Main, usually one on variance or standard deviation and one on conditional probability, Bayes' theorem or the binomial distribution

Statistics and Probability

Every question here reduces a lot of information to one number, and the marks go to noticing what that number leaves out. A mean hides the spread, and a test's accuracy is not the chance you are ill.

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Timed practiceMCQs from this chapter with instant scoring

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JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Compute mean, median and mode for grouped and ungrouped data, and read what their ordering says about the distribution
2Compute mean deviation, variance and standard deviation, using the mean-of-squares shortcut rather than deviations
3Predict the effect of shifting and scaling on mean and variance, and combine the variances of two groups correctly
4Apply the addition theorem and distinguish mutually exclusive events from independent ones
5Use conditional probability by shrinking the sample space, and apply Bayes' theorem with the prior included
6Find the mean and variance of a random variable, and apply the binomial distribution to Bernoulli trials

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Why this chapter matters in JEE Main

A test that is 99 per cent accurate in both directions, for a disease affecting one person in ten thousand, returns a positive result. Almost everyone answers that the chance of being ill is 99 per cent. Count a million people instead: 100 have the disease and 99 of them test positive, while 999,900 are healthy and one per cent of that enormous group still gives 9,999 false positives. The chance of being ill is 99 out of 10,098, which is about one per cent, a hundredfold error. The accuracy figure describes the probability of a positive test given the disease, and the question asked for the probability of the disease given a positive test. Those are different numbers, and Bayes' theorem converts one into the other. The statistics half fails the same way: two classes can both average fifty out of a hundred while one has every mark identical and the other spread from zero to a hundred. Every question here reduces a lot of information to one number, and the marks go to noticing what that number leaves out.

Before you start — revise these

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Combinations, for counting favourable outcomes, from Permutations and Combinations

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The binomial coefficient and its counting meaning, from Binomial Theorem

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Inclusion-exclusion for two sets, from Sets, Relations and Functions

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Comfort with fractions, percentages and simple algebraic manipulation of sums

Statistics and Probability

A test for a rare disease is $99$ per cent accurate, in both directions: it catches $99$ per cent of those who have it and clears $99$ per cent of those who do not. The disease affects one person in ten thousand.

You test positive. What is the chance you have it?

The number in the question is $99$ per cent, so that is the answer everyone gives.

Take a million people and count.

Group	How many	Test positive
Have the disease	$100$	$99$
Do not	$999, 900$	$9, 999$
Total positives		$10, 098$

$P (disease ∣ positive) = \frac{99}{10 , 098} = 0.0098 \approx 1 per cent$

Not $99$ per cent. About one per cent, a hundredfold error.

The healthy group is ten thousand times larger, so even a one per cent error rate applied to it produces a hundred times more false positives than there are true ones. The test's accuracy was never the answer, because it describes $P (positive ∣ disease)$ and the question asked for $P (disease ∣ positive)$ . Those are different numbers, and Bayes' theorem is the machinery for turning one into the other.

The statistics half of the chapter fails in the same way. Two classes both average $50$ out of $100$ .

${50, 50, 50, 50, 50} and {0, 25, 50, 75, 100}$

Same mean, and nothing else in common: the first has standard deviation $0$ , the second $35.4$ . One number described both and distinguished neither.

Every question in this chapter reduces a lot of information to one number, and the marks go to noticing what that number leaves out.

The number	What it hides
the mean	the spread, entirely
$P (A ∣ B)$	it is not $P (B ∣ A)$
"mutually exclusive"	this is not the same as independent

1. Central Tendency

Measure	Definition	Best when
Mean	$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$	data is symmetric, no extreme values
Median	the middle value in order	outliers are present
Mode	the most frequent value	data is categorical

For grouped data the median and mode need interpolation inside the modal or median class.

$median = ℓ + \frac{\frac{n}{2} - c}{f} \times h, mode = ℓ + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$

Here $ℓ$ is the lower boundary of the relevant class, $h$ its width, $c$ the cumulative frequency before it, and $f_{0}, f_{1}, f_{2}$ the frequencies of the previous, modal and next classes.

Trap. The mean is dragged by a single extreme value and the median is not. A dataset of nine salaries near thirty thousand and one of ten million has a mean that describes nobody in it.

Illustration 1

For the distribution below, find the mean, median and mode.

Class	$0$ – $10$	$10$ – $20$	$20$ – $30$	$30$ – $40$	$40$ – $50$
Frequency	$5$	$8$	$15$	$16$	$6$

Mean uses class midpoints $5, 15, 25, 35, 45$ and $n = 50$ .

$\overset{x}{ˉ} = \frac{25 + 120 + 375 + 560 + 270}{50} = \frac{1350}{50} = 27$

Median needs the cumulative frequencies $5, 13, 28, 44, 50$ . Half of $50$ is $25$ , which first exceeds a cumulative total inside the class $20$ – $30$ .

$median = 20 + \frac{25 - 13}{15} \times 10 = 20 + 8 = 28$

Mode sits in the class with the highest frequency, $30$ – $40$ , with $f_{0} = 15$ and $f_{2} = 6$ on either side.

$mode = 30 + \frac{16 - 15}{32 - 15 - 6} \times 10 = 30 + \frac{10}{11} \approx 30.9$

The three answers differ, and their ordering carries information: mean below median below mode signals a distribution with a longer tail to the left. A question asking which measure best represents the data is asking you to read exactly that.

2. Measures of Dispersion

$mean deviation = \frac{\sum f _{i} ∣ x _{i} - x ˉ ∣}{\sum f _{i}}, σ^{2} = \frac{\sum f _{i} ( x _{i} - x ˉ ) ^{2}}{\sum f _{i}}$

Squaring rather than taking a modulus is what makes the variance algebraically tractable, and it produces the shortcut that does almost all the work.

$σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\frac{\sum x _{i}}{n})^{2} = \overline{x^{2}} - \overset{x}{ˉ}^{2}$

Mean of the squares minus the square of the mean. Standard deviation is $σ$ , in the same units as the data; variance is $σ^{2}$ , in squared units.

Illustration 2

Find the variance and standard deviation of $4, 8, 11, 17, 20$ , both directly and by the shortcut.

$\sum x = 60 \Rightarrow \overset{x}{ˉ} = 12, \sum x^{2} = 16 + 64 + 121 + 289 + 400 = 890$

$σ^{2} = \frac{890}{5} - 1 2^{2} = 178 - 144 = 34$

Check directly. The deviations are $- 8, - 4, - 1, 5, 8$ , whose squares total $64 + 16 + 1 + 25 + 64 = 170$ .

$σ^{2} = \frac{170}{5} = 34 ✓, σ = 34 \approx 5.83$

The shortcut needed no deviations at all, which matters when the mean is not a whole number: subtracting $12.37$ from every value is where arithmetic errors come from, and the shortcut avoids it entirely.

3. Transformations and Combined Data

Change to every value	Effect on mean	Effect on variance
add $a$	increases by $a$	unchanged
multiply by $k$	multiplied by $k$	multiplied by $k^{2}$

Shifting the whole dataset moves it without stretching it, so the spread cannot change. Scaling stretches every deviation by $k$ , and variance is built from squared deviations.

For two groups combined, the variance picks up the gap between the group means as well as the variances within them.

$σ^{2} = \frac{n _{1} ( σ _{1}^{2} + d _{1}^{2} ) + n _{2} ( σ _{2}^{2} + d _{2}^{2} )}{n _{1} + n _{2}}, d_{i} = \overset{x}{ˉ}_{i} - \overset{x}{ˉ}_{combined}$

Illustration 3

The data of Illustration 1 has mean $12$ and variance $34$ . Find the mean and variance after adding $5$ to each value, and after multiplying each by $3$ .

Adding shifts the whole set without stretching it.

$new mean = 17, new variance = 34$

Multiplying stretches every deviation threefold, and variance squares deviations.

$new mean = 36, new variance = 34 \times 9 = 306$

The standard deviation, being the square root, is only tripled: $306 = 334$ . That is the practical reason to keep standard deviation and variance apart in your head — one scales by $k$ , the other by $k^{2}$ .

Illustration 4

Group A has $5$ values with mean $12$ and variance $34$ . Group B has $5$ values with mean $20$ and variance $16$ . Find the variance of the combined data.

The combined mean comes first, because the gaps are measured from it.

$\overset{x}{ˉ} = \frac{5 ( 12 ) + 5 ( 20 )}{10} = 16, d_{1} = - 4, d_{2} = 4$

$σ^{2} = \frac{5 ( 34 + 16 ) + 5 ( 16 + 16 )}{10} = \frac{250 + 160}{10} = 41$

Notice that $41$ exceeds both group variances, $34$ and $16$ . Combining groups whose means differ always increases the spread, because the separation between the means becomes part of the variation in the pooled data. Averaging the two variances to get $25$ is the standard error, and it is always too small.

4. Probability: Events and the Axioms

$P (A) = \frac{favourable outcomes}{total outcomes}, 0 \leq P (A) \leq 1$

That counting definition needs the outcomes to be equally likely, which is a real assumption and not a formality.

$P (not A) = 1 - P (A)$

The complement rule is the most under-used line in the chapter. "At least one" is almost always faster as one minus the probability of none.

5. Addition Theorem, and Two Words That Are Not Synonyms

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

The subtraction is inclusion-exclusion again, correcting for the overlap counted twice.

Term	Condition	Meaning
Mutually exclusive	$P (A \cap B) = 0$	they cannot both happen
Independent	$P (A \cap B) = P (A) P (B)$	one happening tells you nothing about the other

Illustration 5

Show that two events with non-zero probability cannot be both mutually exclusive and independent.

Suppose they were. Exclusivity gives $P (A \cap B) = 0$ , and independence gives $P (A \cap B) = P (A) P (B)$ .

$0 = P (A) P (B)$

But both probabilities were assumed non-zero, so their product is non-zero. Contradiction.

The intuition is sharper than the algebra. Mutually exclusive events are maximally dependent: learning that $A$ happened tells you $B$ certainly did not, which is the largest possible amount of information one event can carry about another. Independence means learning $A$ tells you nothing at all. The two are opposite ends of the same scale, not two names for the same idea.

Illustration 6

A card is drawn from a standard pack. Find the probability that it is a king or a heart.

The two events overlap in exactly one card, the king of hearts, and that card would otherwise be counted twice.

$P (K \cup H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

Are they independent? Test the definition rather than guessing.

$P (K) P (H) = \frac{4}{52} \cdot \frac{13}{52} = \frac{1}{52} = P (K \cap H)$

They are, which surprises most people. Being a heart really does not change the chance of being a king, because each suit contains exactly one king. Change the pack by removing one non-heart king and independence breaks immediately.

6. Conditional Probability

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) \neq = 0$

Conditioning shrinks the sample space to $B$ and re-measures $A$ inside it. That is the entire content of the definition, and reading it that way solves most questions without algebra.

$P (A \cap B) = P (B) P (A ∣ B) = P (A) P (B ∣ A)$

Illustration 7

A family has two children. Given that at least one is a boy, what is the probability that both are?

The reflex answer is $\frac{1}{2}$ , reasoning that the other child is equally likely to be either.

List the sample space, in birth order, and then shrink it.

${B B, B G, GB, GG}$

"At least one boy" removes $GG$ and leaves three equally likely outcomes.

$P (B B ∣ at least one boy) = \frac{1}{3}$

The reflex fails because "at least one is a boy" does not identify which child, so it eliminates one outcome rather than fixing one child's sex. Contrast the different question "the elder is a boy", which leaves ${B B, B G}$ and does give $\frac{1}{2}$ .

Trap. Read exactly what the condition specifies. "At least one", "the first one" and "a randomly chosen one" produce three different answers to the same-sounding question.

7. Bayes' Theorem

$P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum _{j} P ( E _{j} ) P ( A ∣ E _{j} )}$

The denominator is the total probability of $A$ , assembled from every way it could happen. The theorem reverses the conditioning: it converts $P (evidence ∣ cause)$ , which experiments give you, into $P (cause ∣ evidence)$ , which is what you actually want to know.

The prior $P (E_{i})$ is what people leave out. In the opening example the prior was one in ten thousand, and ignoring it moved the answer by a factor of a hundred.

Illustration 8

Bag A holds $3$ red and $4$ black balls; bag B holds $5$ red and $6$ black. A bag is chosen at random and a red ball is drawn. What is the probability it came from bag B?

Set out the priors and the conditionals before combining anything.

$P (A) = P (B) = \frac{1}{2}, P (R ∣ A) = \frac{3}{7}, P (R ∣ B) = \frac{5}{11}$

The denominator is every route to a red ball.

$P (R) = \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{5}{11} = \frac{3}{14} + \frac{5}{22} = \frac{33 + 35}{154} = \frac{68}{154} = \frac{34}{77}$

$P (B ∣ R) = \frac{\frac{1}{2} \cdot \frac{5}{11}}{\frac{34}{77}} = \frac{5}{22} \cdot \frac{77}{34} = \frac{35}{68} \approx 0.515$

Slightly above a half, because bag B has the marginally higher proportion of red: $5/11 \approx 0.455$ against $3/7 \approx 0.429$ . The evidence shifted the answer from the prior of $0.5$ only a little, which is what weak evidence should do.

8. Random Variables, Mean and Variance

A random variable assigns a number to each outcome. Its probability distribution lists the values with their probabilities, which must sum to $1$ .

$μ = E (X) = \sum x_{i} p_{i}, σ^{2} = \sum x_{i}^{2} p_{i} - μ^{2}$

The variance formula is the same "mean of the squares minus square of the mean" as in the statistics half, and that is not a coincidence: a probability distribution is a dataset with the frequencies replaced by probabilities.

Illustration 9

Find the mean and variance of the score on one throw of a fair die.

Each value has probability $\frac{1}{6}$ , so both sums are straightforward.

$μ = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$

$E (X^{2}) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6}$

$σ^{2} = \frac{91}{6} - (\frac{7}{2})^{2} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.92$

The mean of $3.5$ is a value the die can never show, which is worth noticing: an expected value is a long-run average, not a prediction of any single throw.

9. Bernoulli Trials and the Binomial Distribution

Bernoulli trials are repeated, independent, with two outcomes and a constant probability of success.

$P (X = r) = (r n) p^{r} q^{n - r}, q = 1 - p$

The binomial coefficient counts the orderings in which $r$ successes can occur, which is the Binomial Theorem's counting argument reappearing in a new setting.

$mean = n p, variance = n pq$

Since $q < 1$ , the variance is always less than the mean for a binomial distribution. That is a fast check on any answer.

Illustration 10

A fair die is thrown $5$ times. Find the probability of exactly $3$ sixes, and the mean and variance of the number of sixes.

The trials are independent with constant $p = \frac{1}{6}$ , so the binomial applies.

$P (X = 3) = (3 5) (\frac{1}{6})^{3} (\frac{5}{6})^{2} = 10 \cdot \frac{1}{216} \cdot \frac{25}{36} = \frac{250}{7776} \approx 0.032$

$mean = 5 \cdot \frac{1}{6} = \frac{5}{6} \approx 0.83, variance = 5 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{25}{36} \approx 0.69$

The variance is below the mean, as it must be. The mean under $1$ also matches intuition: five throws will most often produce no six at all.

Illustration 11

A die is thrown $5$ times. Find the probability of at least one six.

Listing the cases for one, two, three, four and five sixes means five binomial terms. The complement needs one.

$P (at least one) = 1 - P (none) = 1 - (\frac{5}{6})^{5} = 1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.598$

Just under $60$ per cent, which is worth remembering as a corrective: five throws is not enough to make a six likely-but-certain, and the intuition that six throws should "guarantee" one is wrong too, giving only about $66.5$ per cent.

Trap. "At least one" almost always means take the complement. Adding the individual cases gives the same answer and takes five times as long, with five times as many chances to slip.

10. A Note on Syllabus Emphasis

The unit is titled Measures of Dispersion and Probability, and the statistics half is exactly that: means, medians and modes of grouped and ungrouped data, and mean deviation, variance and standard deviation for both.

Where the marks are	What it looks like
variance and standard deviation	one computation, often with a transformation attached
conditional probability and Bayes	the most examined idea in the unit
binomial distribution	mean, variance, and one specified probability

Probability carries most of the weight, and within it Bayes' theorem carries most again. The statistics half is short, formulaic and worth securing completely, because the questions are nearly identical from year to year.

Summary

Every question here reduces a lot of information to one number, and the marks go to noticing what that number leaves out.

A $99$ per cent accurate test for a one-in-ten-thousand disease gives a positive result that means about a one per cent chance, because the healthy group is vastly larger. $P (A ∣ B)$ is not $P (B ∣ A)$ .

Two datasets can share a mean and share nothing else. The mean hides the spread completely.

For grouped data, the median and mode are interpolated inside their classes, and mean below median below mode signals a left tail.

Variance is the mean of the squares minus the square of the mean, which avoids computing any deviations.

Adding a constant leaves the variance unchanged; multiplying by $k$ multiplies it by $k^{2}$ , while the standard deviation is multiplied by $k$ .

Combining groups with different means always increases the variance, because the gap between the means becomes part of the spread.

Mutually exclusive and independent are opposite ends of one scale, and events with non-zero probability can never be both.

Conditioning shrinks the sample space and re-measures inside it, which is why "at least one is a boy" gives $\frac{1}{3}$ and "the elder is a boy" gives $\frac{1}{2}$ .

Bayes' theorem reverses the conditioning, and the prior is what people leave out.

An expected value is a long-run average and need not be an attainable value, as the die's $3.5$ shows.

For a binomial distribution the mean is $n p$ and the variance $n pq$ , so the variance is always the smaller of the two.

"At least one" almost always means one minus the probability of none.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The organising principle

every question reduces a lot of information to one number; the marks are in what that number leaves out

The mean hides the spread entirely. P(A given B) is not P(B given A). Mutually exclusive is not independent. Three different single numbers, three different omissions.

Central tendency for grouped data

median = l + [(n/2 - c)/f] h; mode = l + [(f1 - f0)/(2f1 - f0 - f2)] h

Interpolation inside the median or modal class. Mean below median below mode signals a longer tail to the left, and the reverse ordering a longer tail to the right.

Variance, the working form

sigma squared = mean of the squares minus the square of the mean

Avoids computing any deviations, which matters most when the mean is not a whole number. Standard deviation is in the data's own units, variance in squared units.

Mean deviation

the average of the absolute deviations from the mean or median

Uses a modulus rather than a square, which makes it easy to state and awkward to manipulate. That is precisely why variance squares instead.

Effect of transformations

adding a: mean shifts, variance unchanged; multiplying by k: mean times k, variance times k squared

Shifting moves the data without stretching it. Standard deviation is multiplied by k, not k squared, which is the reason to keep the two quantities separate in your head.

Combined variance

weight each group's variance plus its squared distance from the combined mean, then average

Combining groups with different means ALWAYS increases the spread, since the gap between the means becomes part of the variation. Averaging the two variances is always too small.

Addition theorem

P(A or B) = P(A) + P(B) - P(A and B)

Inclusion-exclusion again, correcting for the overlap counted twice. For a card being a king or a heart, the king of hearts is the single overlap.

Mutually exclusive against independent

exclusive means P(A and B) = 0; independent means P(A and B) = P(A)P(B)

Events with non-zero probability can never be both. Exclusivity is the strongest possible dependence, since learning A happened tells you B certainly did not.

Conditional probability

P(A given B) = P(A and B) / P(B)

Conditioning shrinks the sample space to B and re-measures A inside it. Reading it that way answers most questions without algebra.

Multiplication theorem

P(A and B) = P(B) P(A given B) = P(A) P(B given A)

For independent events this collapses to the product of the two probabilities, which is the definition of independence rather than a separate rule.

Bayes' theorem

P(cause given evidence) is the prior times the likelihood, over the total probability of the evidence

It reverses the conditioning. The prior is what people leave out, and leaving it out moved the disease-test answer by a factor of a hundred.

Mean and variance of a random variable

mean is the sum of x times p; variance is the sum of x squared times p, minus the mean squared

The same mean-of-squares structure as the statistics half, because a distribution is a dataset with frequencies replaced by probabilities. An expected value need not be attainable: a die's mean is 3.5.

Binomial distribution

P (X = r) = C (n, r) p^{r} q^{n - r}, w i t hm e ann p an d v a r ian ce n pq

Requires independent trials with constant p. Since q is below 1, the variance is ALWAYS less than the mean, which is a one-second check on any answer.

JEE Main test series

Ready to be tested on this?

Statistics and Probability shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~2Est. Qs

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Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Answering P(A given B) with the value of P(B given A)

✓ They are different numbers whenever the two events have different probabilities. A 99 per cent accurate test for a one-in-ten-thousand disease gives P(positive given ill) of 0.99 and P(ill given positive) of about 0.01. Write down which way the conditioning runs before computing anything, and use Bayes to reverse it.

Why it happens: Only one conditional probability is stated in the question, so it gets used as though it were the answer.

WATCH OUT

✗ Ignoring the prior in a Bayes question

✓ The prior multiplies the likelihood in every term of the numerator and denominator. A rare cause with a strong likelihood can still lose to a common cause with a weak one, which is exactly what happens with rare diseases. Set out priors and likelihoods in two labelled rows before combining them.

Why it happens: The likelihoods are the numbers the question emphasises, and the base rate often appears as an easily overlooked aside.

WATCH OUT

✗ Treating mutually exclusive and independent as the same thing

✓ They are opposites. Exclusive means P(A and B) = 0, so learning A happened tells you B did not, the maximum possible dependence. Independent means P(A and B) = P(A)P(B), so learning A tells you nothing. Events with non-zero probability cannot be both, and the one-line proof is that a product of non-zero numbers is not zero.

Why it happens: Both describe a relationship between two events and both sound like a kind of separateness.

WATCH OUT

✗ Averaging two group variances when combining data

✓ Each group's contribution is its variance PLUS the square of its distance from the combined mean. For groups of five with means 12 and 20 and variances 34 and 16, the combined variance is 41, not the average of 25. Combining groups with different means always increases the spread.

Why it happens: Averaging the means works, so averaging the variances looks like the same operation.

WATCH OUT

✗ Multiplying the variance by k rather than k squared under scaling

✓ Variance is built from squared deviations, so scaling every value by k scales it by k squared. Standard deviation, being the square root, scales by k. Adding a constant changes the mean but leaves both untouched, because shifting does not stretch.

Why it happens: The mean scales by k, and both are called measures of the data.

WATCH OUT

✗ Summing the individual cases for an at-least-one probability

✓ The complement is one term instead of many. For at least one six in five throws, compute one minus five-sixths to the fifth, giving about 0.598. Adding the five binomial terms gives the same number after five times the work and five times the opportunities to slip.

Why it happens: The cases are each computable, so the direct route feels safe.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Statistics and Probability?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Every answer here is one number, and the marks are in what it leaves out
•P(A given B) is not P(B given A); Bayes converts one to the other
•The prior is what people leave out, and it moved the disease answer a hundredfold
•Two datasets can share a mean and share nothing else
•Variance is the mean of the squares minus the square of the mean
•Adding leaves variance unchanged; multiplying by k multiplies it by k squared
•Combining groups with different means always increases the variance
•Mutually exclusive and independent are opposites, never both for non-zero probabilities
•Conditioning shrinks the sample space and re-measures inside it
•At least one is a boy gives one third; the elder is a boy gives one half
•An expected value is a long-run average and need not be attainable
•Binomial mean is np and variance npq, so the variance is always the smaller

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: 8

Question style	Marks each	Typical count
Central tendency and grouped data	2	1
Dispersion, variance and standard deviation	2	1
Probability: addition, multiplication and conditional	2	1
Bayes' theorem and the binomial distribution	2	1

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Write down which way the conditioning runs before touching a probability question. Underline the word given, and if the question asks for the reverse of what is supplied, you are in a Bayes question whatever it looks like.
Set out priors and likelihoods in two labelled rows before combining them. Most Bayes errors are bookkeeping rather than conceptual, and the layout prevents them.
Use the mean-of-squares shortcut for every variance. It is faster, and it removes the decimal deviations that cause most arithmetic slips when the mean is not a whole number.
Check any binomial answer against the rule that the variance is less than the mean. It costs a second and catches a swapped np and npq immediately.
Reach for the complement whenever you read at least one. One subtraction replaces a sum of several terms, and the saved time is usually the difference between finishing the section and not.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

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Spam filters and fraud detection are Bayes' theorem applied at scale, updating a prior probability with each piece of evidence, and their failure mode is precisely a mis-estimated base rate

Quality control uses the binomial distribution to set acc…

Quality control uses the binomial distribution to set acceptance limits, since a batch is accepted or rejected on the number of defectives in a sample, and the variance being smaller than the mean is what makes such sampling workable

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Because the question asks for a different conditional probability from the one the accuracy figure describes. Ninety-nine per cent accuracy means that given you have the disease, the test almost certainly says so. What you want to know is the reverse: given the test says so, how likely is it that you have the disease. The two differ enormously when the disease is rare, because the healthy group is so much larger that even a small error rate on it produces far more false positives than there are true cases. Counting a million people makes it concrete: 99 true positives against 9,999 false ones, so a positive result carries about a one per cent chance. Bayes' theorem is exactly the machinery for that reversal, and the prevalence figure is the prior it needs.

Test the definition and nothing else: compute P(A)P(B) and compare it with P(A and B). If they are equal the events are independent, and if not they are not. Intuition is unreliable here, because independence is a numerical coincidence rather than a physical separation. Drawing a card that is a king and drawing one that is a heart turn out to be independent, which surprises most people, because each suit contains exactly one king. Remove one non-heart king from the pack and independence breaks immediately, with no change to the physical setup. Equally, do not use mutual exclusivity as evidence: exclusive events with non-zero probability are always dependent, and strongly so.

Because the pooled data now varies for two reasons: the spread within each group, and the gap between the two group means. The combined variance formula makes this explicit, weighting each group by its own variance plus the square of its distance from the combined mean. Only if the two means happen to be identical do those distance terms vanish, and then the combined variance really is a weighted average of the two. For groups of five with means 12 and 20 and variances 34 and 16, the answer is 41, comfortably above both. This is also why an average of averages is a dangerous quantity: it discards exactly the information that the variance formula is careful to keep.

Whenever the phrase at least one appears, and often for at most one as well. The opposite of at least one is none, which is a single easy case, so the complement replaces a whole sum with one subtraction. For at least one six in five throws of a die, the direct route needs five binomial terms while the complement needs one: one minus five-sixths to the fifth power, giving about 0.598. The same reasoning applies to at least one defective item, at least one head, and at least one success in any Bernoulli setting. For at most one the complement is at least two, which may or may not be shorter, so count the cases on each side before committing.

Structurally almost nothing, which is why the two halves of this chapter share their formulas. A dataset lists values with frequencies and a distribution lists values with probabilities, and dividing the frequencies by the total converts one into the other. That is why the mean of a random variable is the sum of value times probability, matching the grouped-data mean with frequencies replaced, and why the variance is again the mean of the squares minus the square of the mean. The one genuine difference is interpretation: a dataset's mean is a summary of numbers you actually observed, while an expected value is a long-run average that no single trial need produce. A fair die has expected score 3.5, a value it can never show.