TN 12th MathsLearn · Visualize · Practice
📊

Chapter 11

Probability Distributions

Random variables, distributions, expectation and special discrete distributions.

1. Random variables

The outcome of a random experiment is not always a number. Toss a coin and you get HH or TT; draw balls from an urn and you get colours. To do arithmetic — averaging, spreading, computing probabilities — we attach a real number to each outcome. That rule is a random variable.

DefinitionRandom variable

A random variable XX is a function X:SRX : S \to \mathbb{R} that assigns a real number X(ω)X(\omega) to each outcome ω\omega of the sample space SS, in such a way that the inverse image of any interval is an event whose probability is defined.

We write random variables with capitals (X,Y,ZX, Y, Z) and their possible values with small letters (x,y,zx, y, z). The set of all values XX can take is its range.

Random variables come in two flavours. A discrete random variable takes isolated, countable values such as 0,1,2,0, 1, 2, \dots (things you count). A continuous random variable can take any value in an interval of the real line (things you measure).

Discrete vs continuous
  • Discrete (count): number of heads, number of defective items, score on two dice — you can list the values.
  • Continuous (measure): height, weight, waiting time, bulb lifetime — the values fill an interval and cannot be listed one by one.
ExampleValues and inverse images
Two coins are tossed once and XX is the number of tails. List the values of XX and the number of outcomes mapping to each.
Key idea: A random variable turns messy outcomes into plain numbers, so we can average them, measure their spread, and compute probabilities with algebra and calculus.

2. Discrete distributions and the pmf

For a discrete random variable we describe its behaviour by listing every value together with its probability. That list is the probability distribution, and the function giving those probabilities is the probability mass function (pmf).

DefinitionProbability mass function

For a discrete random variable XX, the pmf is f(x)=P(X=x)f(x) = P(X = x) — the probability that XX equals the value xx.

A list of numbers is a valid pmf only if it passes two tests:

Conditions for a pmf
f(x)0andxf(x)=1f(x) \ge 0 \quad \text{and} \quad \sum_{x} f(x) = 1
Note: Every probability is between 0 and 1, and because some value must occur, all the probabilities add to exactly 1. This 'sums to 1' rule is your tool for finding a missing probability.
ExampleFind a missing probability
XX takes values 0,1,2,30, 1, 2, 3 with f(x)=0.1, 0.3, k, 0.2f(x) = 0.1,\ 0.3,\ k,\ 0.2. Find kk and confirm it is a valid pmf.
ExampleFind the constant when probabilities are algebraic
A pmf on x=1,2,3,4,5,6x = 1,2,3,4,5,6 is f(x)=c2, 2c2, 3c2, 4c2, c, 2cf(x) = c^2,\ 2c^2,\ 3c^2,\ 4c^2,\ c,\ 2c. Find cc.

3. Cumulative distribution function (discrete)

Often we want the probability of being at or below a value rather than exactly equal to it. The cumulative distribution function (cdf) accumulates probability from the left.

DefinitionCumulative distribution function

The cdf is F(x)=P(Xx)F(x) = P(X \le x) — the total probability of all values up to and including xx.

Key properties of F
  • Discrete: F(x)=txf(t)F(x) = \sum_{t \le x} f(t) — add the probabilities of all values x\le x.
  • FF is a non-decreasing step function; it jumps by f(x)f(x) at each value xx.
  • F()=0F(-\infty) = 0 and F()=1F(\infty) = 1.
  • Interval probability by subtraction: P(a<Xb)=F(b)F(a)P(a < X \le b) = F(b) - F(a).
ExampleBuild the cdf from a pmf
X:0,1,2,3X: 0, 1, 2, 3 with f(x):0.1,0.3,0.4,0.2f(x): 0.1, 0.3, 0.4, 0.2. Find F(x)F(x) and use it to get P(X2)P(X \le 2) and P(1<X3)P(1 < X \le 3).

4. Continuous distributions and the pdf

For a continuous random variable the probability of hitting one exact value is 0 — there are infinitely many possibilities crammed into every interval. Instead we use a probability density function f(x)f(x) and read probability as area under its curve.

DefinitionProbability density function

A function f(x)f(x) is a pdf for a continuous random variable XX if f(x)0f(x) \ge 0 everywhere and the total area under it is 11.

Conditions for a pdf
f(x)0andf(x)dx=1f(x) \ge 0 \quad \text{and} \quad \int_{-\infty}^{\infty} f(x)\,dx = 1

Probability over an interval is the area under the curve on that interval:

P(aXb)=abf(x)dxP(a \le X \le b) = \int_{a}^{b} f(x)\,dx
Key idea: Discrete probabilities are heights you add up; continuous probabilities are areas you integrate. In both cases the grand total is 1.
ExampleFind the constant, then an interval probability
f(x)=kx2f(x) = kx^2 on 0x30 \le x \le 3 and 00 elsewhere. Find kk, then P(1<X<2)P(1 < X < 2).
Watch out: For a continuous $X$, $f(x)$ is a density, not a probability — it can exceed 1. Only areas (integrals) give probabilities, and $P(X = a) = 0$ for any single point $a$.

5. Cumulative distribution function (continuous)

The cdf is defined the same way for a continuous variable, F(x)=P(Xx)F(x) = P(X \le x), but now it is an area — the integral of the density from the far left up to xx.

cdf and its density
F(x)=xf(t)dt,f(x)=F(x)F(x) = \int_{-\infty}^{x} f(t)\,dt, \qquad f(x) = F'(x)
Two-way street
  • Given the pdf, integrate to get the cdf.
  • Given the cdf, differentiate to get the pdf: the density is the slope of the cdf.
  • FF is continuous and non-decreasing, with F()=0F(-\infty)=0, F()=1F(\infty)=1.
  • Interval probability: P(a<Xb)=F(b)F(a)P(a < X \le b) = F(b) - F(a).
ExampleUniform density: find the cdf and a probability
f(x)=14f(x) = \dfrac{1}{4} for 1x51 \le x \le 5 and 00 elsewhere. Find F(x)F(x) and P(X<3)P(X < 3).

6. Mathematical expectation (mean)

The expectation (mean) E(X)E(X) is the long-run average value of the random variable — a probability-weighted balance point. It generalises the ordinary average: instead of dividing by nn, we weight each value by its probability.

Expectation
E(X)=xxf(x)(discrete),E(X)=xf(x)dx(continuous)E(X) = \sum_{x} x\,f(x) \quad \text{(discrete)}, \qquad E(X) = \int_{-\infty}^{\infty} x\,f(x)\,dx \quad \text{(continuous)}

More generally, the expected value of any function g(X)g(X) weights gg by the probabilities. Taking g(X)=X2g(X) = X^2 gives the second moment E(X2)E(X^2), which we need for the variance:

E(X2)=xx2f(x)(discrete),E(X2)=x2f(x)dx(continuous)E(X^2) = \sum_{x} x^2\,f(x) \quad \text{(discrete)}, \qquad E(X^2) = \int x^2\,f(x)\,dx \quad \text{(continuous)}
ExampleExpectation from a table
X:4,1,2,5,6X: -4, 1, 2, 5, 6 with f(x):110,110,310,310,210f(x): \tfrac{1}{10}, \tfrac{1}{10}, \tfrac{3}{10}, \tfrac{3}{10}, \tfrac{2}{10}. Find E(X)E(X).
Note: $E(X)$ need not be a value $X$ can actually take. It is the balance point of the distribution, not a 'typical' outcome.

7. Variance and standard deviation

The mean tells you where a distribution sits; the variance tells you how widely it spreads around that mean. A small variance means values cluster tightly about the mean; a large variance means they scatter.

Variance and standard deviation
Var(X)=E(X2)[E(X)]2,σ=Var(X)\text{Var}(X) = E(X^2) - [E(X)]^2, \qquad \sigma = \sqrt{\text{Var}(X)}
Watch out: $E(X^2)$ is not the same as $[E(X)]^2$. Square each value first, weight by probability to get $E(X^2)$, then subtract the square of the mean. Mixing these two up is the single most common exam error here.

A useful shortcut: for constants a,ba, b,  Var(aX+b)=a2Var(X)\ \text{Var}(aX + b) = a^2\,\text{Var}(X) (adding a constant shifts but does not spread).

ExampleFull mean and variance from a table
For x=1,,6x = 1,\dots,6 the pmf is f(x)=125,225,325,425,525,1025f(x) = \tfrac{1}{25}, \tfrac{2}{25}, \tfrac{3}{25}, \tfrac{4}{25}, \tfrac{5}{25}, \tfrac{10}{25}. Find the mean and variance.
ExampleQuick variance from moments
A random variable has E(X)=2E(X) = 2 and E(X2)=5E(X^2) = 5. Find Var(X)\text{Var}(X) and σ\sigma.

8. Mean and variance by integration (continuous)

For a continuous random variable the recipe is identical, with integrals replacing sums. Find E(X)E(X) and E(X2)E(X^2) by integrating xf(x)x f(x) and x2f(x)x^2 f(x), then use the same variance formula.

E(X)=xf(x)dx,E(X2)=x2f(x)dx,Var(X)=E(X2)[E(X)]2E(X) = \int x f(x)\,dx, \qquad E(X^2) = \int x^2 f(x)\,dx, \qquad \text{Var}(X) = E(X^2) - [E(X)]^2
ExampleMean and variance of a density
f(x)=x29f(x) = \dfrac{x^2}{9} on 0x30 \le x \le 3 (and 00 elsewhere). Find E(X)E(X), E(X2)E(X^2) and Var(X)\text{Var}(X).

9. Bernoulli and binomial distributions

A Bernoulli trial is a single experiment with just two outcomes — success (prob pp) or failure (prob q=1pq = 1 - p). Scoring X=1X = 1 for success and 00 for failure gives the Bernoulli distribution, with mean pp and variance pqpq.

Repeat a Bernoulli trial nn independent times and count the successes: that count follows a binomial distribution, written XB(n,p)X \sim B(n, p).

Conditions for a binomial experiment
  • A fixed number nn of trials, with nn finite.
  • Each trial has exactly two outcomes: success or failure.
  • The success probability pp is the same on every trial.
  • The trials are independent of one another.
DefinitionBinomial probability

If XX is the number of successes in nn trials, then for x=0,1,,nx = 0, 1, \dots, n:

P(X=x)=(nx)pxqnx,q=1pP(X = x) = \binom{n}{x} p^{x} q^{\,n-x}, \qquad q = 1 - p

Here (nx)\binom{n}{x} counts the ways to place xx successes among nn trials, pxp^x is the chance those succeed, and qnxq^{n-x} the chance the rest fail. The mean and variance have neat closed forms:

Mean and variance of a binomial
E(X)=np,Var(X)=npq,σ=npqE(X) = np, \qquad \text{Var}(X) = npq, \qquad \sigma = \sqrt{npq}
Note: Because $q < 1$, the variance $npq$ is always smaller than the mean $np$. If a problem ever gives you a binomial with variance larger than the mean, recheck your reading — it is impossible.
ExampleExactly 2 heads in 5 tosses
A fair coin is tossed 55 times. Find the probability of exactly 22 heads.
ExampleAt least one correct: multiple-choice guessing
A student guesses on 1010 multiple-choice questions, each with success probability p=14p = \tfrac14. Find the probability of at least one correct answer.
ExampleMean and variance of a binomial
A die is rolled 1818 times; a success is rolling a six (p=16p = \tfrac16). Find the mean and variance of the number of sixes.
ExampleRecover n and p from the mean and variance
A binomial variate XX has mean 22 and variance 1.51.5. Find nn and pp, then P(X=0)P(X = 0).

10. Choosing the right model (and other distributions)

The binomial is the central discrete model for Class 12, but it is one of a family. When trials are rare but numerous (large nn, small pp, with npnp moderate), the Poisson distribution approximates the binomial. When measurements cluster symmetrically about a mean, the continuous normal distribution — the bell curve — takes over.

Key idea: Recognise the model first — is it a fixed count of independent yes/no trials? If so reach for the binomial, then the np, npqnp,\ npq shortcuts. If it is a measurement over an interval, you are in continuous territory: set up an integral.
Watch out: Exam pitfalls to avoid: (1) forgetting to check probabilities are non-negative when picking a constant; (2) writing $[E(X)]^2$ where $E(X^2)$ is needed; (3) using the binomial formula when trials are not independent or $p$ changes (e.g. drawing without replacement); (4) treating a continuous density value as a probability.