GATE DA Probability and Statistics Formula Sheet with Traps

Q: What is the variance of an exponential distribution?

For Exponential(λ) the mean is 1/λ and the variance is 1/λ². The tail is P(X > x) = e^(-λ x) and the median is (ln 2)/λ. A common slip is to give 1/λ as the variance. The exponential is also memoryless, a property it shares only with the geometric distribution among the standard distributions.

Q: What is the difference between independent and mutually exclusive events?

Independent events satisfy P(A ∩ B) = P(A) P(B), so knowing one tells you nothing about the other. Mutually exclusive events satisfy P(A ∩ B) = 0, so one happening rules out the other. Two events that both have non-zero probability can never be both independent and mutually exclusive, which is a frequent multiple-select trap.

Q: Which value of z is used for a 95 per cent confidence interval?

A two-sided 95 per cent interval uses z = 1.96, because Phi(1.96) = 0.975. A one-sided 5 per cent test uses 1.645 and a 99 per cent two-sided interval uses 2.576. When the population standard deviation is unknown, replace z with the t value for n - 1 degrees of freedom and use the sample standard deviation s.

Q: How many degrees of freedom does a chi-squared test have?

A goodness-of-fit test with k categories has k - 1 degrees of freedom, minus one more for every parameter you estimate from the data. A test of independence on an r × c table has (r - 1)(c - 1). A test for one variance uses n - 1. The expected count in a cell is the row total times the column total, divided by the grand total.

By MD ANISH AHAMADUpdated 4 Oct 20267 min read

GATE DA Probability and Statistics Formula Sheet with Traps

The probability and statistics formulas you need for GATE DA fit on a few tables: counting and probability rules, expectation and variance, a dozen distributions, and the sampling results behind confidence intervals and tests. This sheet gives each formula with its condition, a one-line example of my own and the trap that costs marks. Every formula is checked against the last-minute sheet in the GATE DA 2027 book.

In this guide

Key takeaways
The terms this sheet uses
Counting and probability rules
Expectation, variance and correlation
Distributions: PMF, mean and variance
Sampling and the central limit theorem
Confidence intervals and tests
Using the sheet in the exam
Quick revision

Key takeaways

Write the conditional the question asks for, $P(A \mid B)$ or $P(B \mid A)$ , before you substitute anything.
Variance scales by the square: $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$ .
The sample variance divides by $n - 1$ ; the population form divides by $n$ .
Learn each distribution as a set of three: its PMF or PDF, its mean and its variance.
Interval and test questions turn on one choice: $z$ when $\sigma$ is known, $t$ with $n - 1$ degrees of freedom when it is not.
No counted paper (2024 to 2026) has examined confidence intervals or the $z$ , $t$ and chi-squared tests, but the syllabus names them, so prepare them.

The terms this sheet uses

A random variable is a number attached to each outcome of an experiment. A discrete one has a probability mass function (PMF), $P(X = k)$ . A continuous one has a probability density function (PDF), $f(x)$ , whose area over an interval is a probability.

The cumulative distribution function (CDF) is $F(x) = P(X \le x)$ . It never decreases and runs from 0 to 1. A sample is the data you observe; the population is what it was drawn from. $\bar{x}$ is the sample mean, $s$ the sample standard deviation, and $\mu$ and $\sigma$ their population counterparts. $\ln$ is the natural logarithm.

Counting and probability rules

Formula	Watch out for
$\displaystyle {}^nP_r = \frac{n!}{(n - r)!}$ , $\displaystyle \binom{n}{r} = \frac{n!}{r!\,(n - r)!}$	Order matters only for permutations
Arrangements with repeats: $\displaystyle \frac{n!}{n_1!\,n_2!\cdots n_k!}$	Divide once for each repeated group
Non-negative solutions of $x_1 + \cdots + x_k = n$ : $\binom{n + k - 1}{k - 1}$ ; positive: $\binom{n - 1}{k - 1}$	Read "non-negative" against "positive"
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$	Subtract the overlap once
$P(\text{at least one}) = 1 - P(\text{none})$	The fastest route for "at least"
$\displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , $P(B) > 0$	Not symmetric in $A$ and $B$
$P(B) = \sum_i P(B \mid A_i)\,P(A_i)$	The $A_i$ must partition the space
$\displaystyle P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{\sum_j P(B \mid A_j)\,P(A_j)}$	The denominator is total probability

Example: $x_1 + x_2 + x_3 + x_4 = 6$ has $\binom{9}{3} = 84$ non-negative solutions and $\binom{5}{3} = 10$ positive ones.

Bayes in one line: a condition affects 2 per cent of people, a test detects it 90 per cent of the time and gives a false positive 5 per cent of the time. Then $\displaystyle P(\text{ill} \mid +) = \frac{0.018}{0.018 + 0.049} \approx 0.269$ .

Trap: A test that is 90 per cent accurate does not make a positive result 90 per cent reliable. The small prior drags the posterior down, as the example shows.

Expectation, variance and correlation

Formula	Watch out for
$E[aX + bY + c] = aE[X] + bE[Y] + c$	No independence needed
$\operatorname{Var}(X) = E[X^2] - (E[X])^2$	Square of the mean, not mean of the square
$\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$	The shift $b$ disappears
$\operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X, Y)$	Drop the last term only if uncorrelated
$\operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y]$	$E[XY] = E[X]E[Y]$ needs independence or zero covariance
$\displaystyle \rho = \frac{\operatorname{Cov}(X, Y)}{\sigma_X \sigma_Y}$ , $-1 \le \rho \le 1$	Unchanged by positive rescaling
$E[X] = E\big[E[X \mid Y]\big]$	The law of total expectation
$\operatorname{Var}(X) = E[\operatorname{Var}(X \mid Y)] + \operatorname{Var}(E[X \mid Y])$	Both terms, every time
$\displaystyle s^2 = \frac{1}{n - 1}\sum (x_i - \bar{x})^2$	Population form divides by $n$
New mean after one point: $\displaystyle \frac{n\bar{x} + x_{\text{new}}}{n + 1}$	Recompute the total, then divide

Examples: if $\operatorname{Var}(X) = 5$ , then $\operatorname{Var}(2X - 3) = 20$ . If $\operatorname{Cov}(X, Y) = 6$ , $\sigma_X = 2$ and $\sigma_Y = 5$ , then $\rho = 0.6$ . Four values with mean 10 plus a new value 20 give a mean of $\displaystyle \frac{40 + 20}{5} = 12$ .

The median of a continuous variable is the $m$ with $F(m) = 0.5$ . For data, it is the middle sorted value, or the mean of the two middle values when $n$ is even. The mode is the most frequent value, or the maximiser of a density.

Distributions: PMF, mean and variance

Distribution	PMF or PDF	Mean	Variance
$\text{Bernoulli}(p)$	$P(1) = p$	$p$	$p(1 - p)$
$\text{Binomial}(n, p)$	$\binom{n}{k}p^k(1 - p)^{n - k}$	$np$	$np(1 - p)$
Uniform on $\{1, \ldots, n\}$	$\displaystyle \frac{1}{n}$	$\displaystyle \frac{n + 1}{2}$	$\displaystyle \frac{n^2 - 1}{12}$
Geometric, $k \ge 1$	$(1 - p)^{k - 1}p$	$\displaystyle \frac{1}{p}$	$\displaystyle \frac{1 - p}{p^2}$
$\text{Poisson}(\lambda)$	$\displaystyle \frac{e^{-\lambda}\lambda^k}{k!}$	$\lambda$	$\lambda$
$\text{Uniform}(a, b)$	$\displaystyle \frac{1}{b - a}$	$\displaystyle \frac{a + b}{2}$	$\displaystyle \frac{(b - a)^2}{12}$
$\text{Exponential}(\lambda)$	$\lambda e^{-\lambda x}$ , $x \ge 0$	$\displaystyle \frac{1}{\lambda}$	$\displaystyle \frac{1}{\lambda^2}$
$N(\mu, \sigma^2)$	$\displaystyle \frac{1}{\sigma\sqrt{2\pi}}e^{-(x - \mu)^2/(2\sigma^2)}$	$\mu$	$\sigma^2$
$\chi^2_k$	sum of $k$ squared $N(0, 1)$	$k$	$2k$
$t_\nu$	symmetric, heavier tails	$0$ ( $\nu > 1$ )	$\displaystyle \frac{\nu}{\nu - 2}$ ( $\nu > 2$ )

Examples: $\text{Binomial}(10, 0.3)$ has mean 3 and variance 2.1. For $\text{Exponential}(0.5)$ , $P(X > 4) = e^{-2} \approx 0.135$ and the median is $\displaystyle \frac{\ln 2}{0.5} \approx 1.386$ .

Fact	Watch out for
$P(X > s + t \mid X > s) = P(X > t)$	Only exponential and geometric are memoryless
$\text{Binomial}(n, p) \approx \text{Poisson}(np)$	Needs large $n$ and small $p$
$aX + b \sim N(a\mu + b,\, a^2\sigma^2)$ ; $\displaystyle Z = \frac{X - \mu}{\sigma}$	Divide by $\sigma$ , not $\sigma^2$
$\Phi(1.645) = 0.95$ , $\Phi(1.96) = 0.975$ , $\Phi(2.576) = 0.995$ ; $\Phi(-z) = 1 - \Phi(z)$	Two-sided tails split $\alpha$ in half
$P(a < X \le b) = F(b) - F(a)$	A discrete CDF jumps at each value

Remember: Independent Poissons add with their rates added. Independent normals add with means added and variances added, never standard deviations.

This sheet covers one section. The book's last-minute sheet does the same for all seven technical sections, each result with its condition beside it. It sits in the GATE DA 2027 book alongside 907 questions with worked solutions and 10 full mock tests.

Sampling and the central limit theorem

The central limit theorem says a sum or mean of many independent draws is close to normal, whatever the original shape, provided the variance is finite.

Formula	Watch out for
$S_n \approx N(n\mu,\, n\sigma^2)$ ; $\displaystyle \bar{X} \approx N\!\left(\mu, \frac{\sigma^2}{n}\right)$	Finite variance and independence required
Standard error of a mean: $\displaystyle \frac{\sigma}{\sqrt{n}}$ ; of a proportion: $\displaystyle \sqrt{\frac{p(1 - p)}{n}}$	$\sqrt{n}$ , not $n$
$\displaystyle \frac{(n - 1)s^2}{\sigma^2} \sim \chi^2_{n - 1}$ ; $\displaystyle \frac{\bar{x} - \mu}{s/\sqrt{n}} \sim t_{n - 1}$	Normal samples only

Example: the sum of 100 independent $\text{Bernoulli}(0.2)$ variables has mean 20, variance 16 and standard deviation 4.

Confidence intervals and tests

Formula	Watch out for
$\displaystyle \bar{x} \pm z_{\alpha/2}\,\frac{\sigma}{\sqrt{n}}$	$\sigma$ known
$\displaystyle \bar{x} \pm t_{\alpha/2,\, n - 1}\,\frac{s}{\sqrt{n}}$	$\sigma$ unknown
$\displaystyle \hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$	Uses the sample proportion
Width halves when $n$ is multiplied by 4	Not by 2
$\displaystyle z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$ ; reject at 5 per cent two-sided if $\lvert z \rvert > 1.96$	One-sided uses 1.645
$\displaystyle t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ , $n - 1$ degrees of freedom	A paired test is a one-sample test on differences
Pooled two-sample: $\displaystyle s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$	$n_1 + n_2 - 2$ degrees of freedom
$\displaystyle \chi^2 = \sum \frac{(O - E)^2}{E}$	Degrees of freedom $k - 1 - (\text{parameters estimated})$
Independence: $\displaystyle E = \frac{\text{row total} \times \text{column total}}{\text{grand total}}$	Degrees of freedom $(r - 1)(c - 1)$
Variance test: $\displaystyle \chi^2 = \frac{(n - 1)s^2}{\sigma_0^2}$	$n - 1$ degrees of freedom

Examples: with $\bar{x} = 80$ , $\sigma = 10$ and $n = 25$ , the 95 per cent interval is $80 \pm 3.92$ . With $\bar{x} = 52$ , $\mu_0 = 50$ , $s = 6$ and $n = 9$ , $t = 1$ on 8 degrees of freedom. A $2 \times 3$ table has 2 degrees of freedom.

A Type I error rejects a true null hypothesis, with probability $\alpha$ . A Type II error keeps a false one, with probability $\beta$ ; the power is $1 - \beta$ . The p-value is the probability, under the null, of a statistic at least as extreme as the one observed. Reject when $p < \alpha$ .

In one line: Pick $z$ or $t$ by whether $\sigma$ is known, count the degrees of freedom, then compare with the critical value.

Using the sheet in the exam

These questions often end in a decimal answer. Practise the arithmetic on the GATE virtual calculator, and round only the final value. Before you guess on a probability MCQ, read the MCQ, MSQ and NAT marking scheme, common to every GATE paper, because only MCQs carry negative marks.

If you know the CS syllabus, GATE CS vs GATE DA shows that statistical inference belongs to DA alone. The General Aptitude guide covers the simpler probability questions in that section.

More formula sheets: all of GATE DA · DBMS and Algorithms · Linear Algebra and Calculus · Machine Learning

Quick revision

Bayes: posterior equals likelihood times prior, over total probability.
$\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$ , and $s^2$ divides by $n - 1$ .
Independence implies zero correlation; zero correlation does not imply independence.
Exponential: mean $\displaystyle \frac{1}{\lambda}$ , variance $\displaystyle \frac{1}{\lambda^2}$ , median $\displaystyle \frac{\ln 2}{\lambda}$ .
Poisson mean and variance are both $\lambda$ ; $\chi^2_k$ has mean $k$ and variance $2k$ .
$\displaystyle \bar{X} \approx N\!\left(\mu, \frac{\sigma^2}{n}\right)$ ; quadruple $n$ to halve the interval width.
1.96 for 95 per cent two-sided, 1.645 one-sided, 2.576 for 99 per cent.
Chi-squared independence has $(r - 1)(c - 1)$ degrees of freedom.

Frequently asked questions

What is the variance of an exponential distribution?

For $\text{Exponential}(\lambda)$ the mean is $\displaystyle \frac{1}{\lambda}$ and the variance is $\displaystyle \frac{1}{\lambda^2}$ . The tail is $P(X > x) = e^{-\lambda x}$ and the median is $\displaystyle \frac{\ln 2}{\lambda}$ . A common slip is to give $\displaystyle \frac{1}{\lambda}$ as the variance. The exponential is also memoryless, a property it shares only with the geometric distribution among the standard distributions.

What is the difference between independent and mutually exclusive events?

Independent events satisfy $P(A \cap B) = P(A)\,P(B)$ , so knowing one tells you nothing about the other. Mutually exclusive events satisfy $P(A \cap B) = 0$ , so one happening rules out the other. Two events that both have non-zero probability can never be both independent and mutually exclusive, which is a frequent multiple-select trap.

Which value of z is used for a 95 per cent confidence interval?

A two-sided 95 per cent interval uses $z = 1.96$ , because $\Phi(1.96) = 0.975$ . A one-sided 5 per cent test uses 1.645 and a 99 per cent two-sided interval uses 2.576. When the population standard deviation is unknown, replace $z$ with the $t$ value for $n - 1$ degrees of freedom and use the sample standard deviation $s$ .

How many degrees of freedom does a chi-squared test have?

A goodness-of-fit test with $k$ categories has $k - 1$ degrees of freedom, minus one more for every parameter you estimate from the data. A test of independence on an $r \times c$ table has $(r - 1)(c - 1)$ . A test for one variance uses $n - 1$ . The expected count in a cell is the row total times the column total, divided by the grand total.

Does zero correlation mean two variables are independent?

No. Independence implies zero correlation, but zero correlation does not imply independence. Correlation measures only linear association. If $X$ is uniform on $(-1, 1)$ and $Y = X^2$ , then $Y$ is completely determined by $X$ , yet their correlation is zero. So a zero correlation only rules out a linear relationship, never every kind of dependence.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.