Random Variables and Probability Distributions

Why this section exists. The rationalized NCERT trimmed random variables, probability distributions, Bernoulli trials and the binomial distribution from this chapter — but JEE Main kept all of them in its syllabus, and they appear nearly every session. If you are preparing only for boards you may treat this section as enrichment; JEE aspirants should treat it as core.

Random variables

A random variable $X$ is a real-valued function whose domain is the sample space of a random experiment — a rule that attaches a number to every outcome.

Toss a coin twice: $S = \{HH, HT, TH, TT\}$. If $X$ = number of heads, then $$X(HH) = 2, \quad X(HT) = 1, \quad X(TH) = 1, \quad X(TT) = 0$$

More than one random variable can live on the same sample space: $Y$ = (heads minus tails) gives $Y(HH) = 2$, $Y(HT) = Y(TH) = 0$, $Y(TT) = -2$. The experiment is random; the function is deterministic — the randomness of $X$ comes entirely from which outcome occurs.

Probability distribution of a random variable

The probability distribution of $X$ lists its possible values with their probabilities:

where $p_i = P(X = x_i)$ must satisfy the two conditions:

$$\textbf{(1)}\ \ p_i \geq 0 \ \text{for all } i \qquad \qquad \textbf{(2)}\ \ \sum_{i=1}^{n} p_i = 1$$

Any table meeting both conditions is a valid distribution; violating either (a negative entry, or a total other than 1) disqualifies it. "Find $k$" problems use condition (2) to pin down an unknown constant.

Mean (expectation) and variance

The mean or expected value of $X$ is the probability-weighted average of its values:

$$E(X) = \mu = \sum_{i=1}^{n} x_i\,p_i$$

The variance measures spread around the mean:

$$\operatorname{Var}(X) = \sigma^2 = E(X^2) - \big(E(X)\big)^2 = \sum x_i^2\,p_i - \left(\sum x_i\,p_i\right)^2$$

and the standard deviation is $\sigma = \sqrt{\operatorname{Var}(X)}$. The $E(X^2) - \mu^2$ form is almost always the faster route in exams.

Common mistakes to avoid. (1) $E(X^2) \neq \big(E(X)\big)^2$ — square the values before weighting. (2) The mean need not be one of the values of $X$ (a mean of $\frac{1}{3}$ sixes is fine). (3) Always check $\sum p_i = 1$ before computing anything — a mis-copied table poisons every later step.

Bernoulli Trials and the Binomial Distribution

Bernoulli trials

Trials of an experiment are Bernoulli trials if all four conditions hold:

(1) the number of trials $n$ is finite; (2) the trials are independent; (3) each trial has exactly two outcomes — success or failure; (4) the probability of success $p$ is the same in every trial.

Tossing a coin 10 times, or drawing balls with replacement, are Bernoulli trials. Drawing without replacement is not — the success probability changes from draw to draw, violating condition (4). This distinction is a favourite JEE trap.

The binomial distribution

Run $n$ Bernoulli trials with success probability $p$ (and $q = 1 - p$). If $X$ counts the successes, then

$$P(X = r) = \binom{n}{r} p^r q^{n-r}, \qquad r = 0, 1, 2, \ldots, n$$

We write $X \sim B(n, p)$. The three factors tell the story: $p^r$ for the $r$ successes, $q^{n-r}$ for the failures, and $\binom{n}{r}$ for the number of orders in which they can occur. The probabilities are the successive terms of the expansion $(q + p)^n$, which is why they add to 1.

Mean and variance of the binomial (memorise)

$$E(X) = np, \qquad \operatorname{Var}(X) = npq, \qquad \sigma = \sqrt{npq}$$

Two instant consequences that JEE loves:

• $\dfrac{\operatorname{Var}(X)}{E(X)} = q$, so mean and variance of a binomial determine $q$, then $p = 1 - q$, then $n$. Given "mean 4, variance 2": $q = \frac{2}{4} = \frac{1}{2}$, $p = \frac{1}{2}$, $n = 8$.
• Since $0 < q < 1$, variance $<$ mean always. A claimed binomial with variance exceeding its mean is impossible — an instant elimination in MCQs.

The standard question types

Type 1 — exactly $r$: direct formula. Type 2 — at least / at most $r$: sum the tail, or use the complement (for "at least 1", $P = 1 - q^n$). Type 3 — parameter hunting: given mean and variance (or a ratio of probabilities), find $n$ and $p$, then answer a probability question. Type 4 — validity checks: test the four Bernoulli conditions or the two distribution conditions.

JEE-Pattern Worked Examples

Example 1 — Building a distribution from scratch

A fair coin is tossed twice and $X$ is the number of heads. Write the probability distribution of $X$ and find $E(X)$.

Step 1 — probabilities from the sample space $\{HH, HT, TH, TT\}$: $$P(X = 0) = \frac{1}{4}, \qquad P(X = 1) = \frac{2}{4} = \frac{1}{2}, \qquad P(X = 2) = \frac{1}{4}$$

Step 2 — check: $\frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1$ ✓

Step 3 — expectation: $$E(X) = 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} = 1$$

Answer: $E(X) = 1$ — one head expected in two fair tosses, as intuition demands.

Example 2 — Number of sixes in two throws

Two dice are thrown and $X$ is the number of sixes. Find the distribution of $X$ and its mean.

Step 1 — distribution: $$P(X = 0) = \left(\frac{5}{6}\right)^2 = \frac{25}{36}, \qquad P(X = 1) = 2 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{10}{36}, \qquad P(X = 2) = \frac{1}{36}$$

Step 2 — mean: $$E(X) = 0 \cdot \frac{25}{36} + 1 \cdot \frac{10}{36} + 2 \cdot \frac{1}{36} = \frac{12}{36} = \frac{1}{3}$$

Answer: $E(X) = \dfrac{1}{3}$ — matching the binomial shortcut $np = 2 \times \frac{1}{6}$.

Example 3 — Finding the unknown constant

A random variable $X$ takes values $0, 1, 2$ with probabilities $k$, $2k$, $3k$. Find $k$, $P(X \geq 1)$ and $E(X)$.

Step 1 — condition (2): $k + 2k + 3k = 6k = 1$, so $k = \frac{1}{6}$.

Step 2 — tail probability: $P(X \geq 1) = 2k + 3k = 5k = \frac{5}{6}$.

Step 3 — mean: $E(X) = 0 \cdot k + 1 \cdot 2k + 2 \cdot 3k = 8k = \frac{8}{6} = \frac{4}{3}$.

Answer: $k = \dfrac{1}{6}$, $P(X \geq 1) = \dfrac{5}{6}$, $E(X) = \dfrac{4}{3}$.

Example 4 — Aces without replacement (not binomial!)

Two cards are drawn without replacement from a deck of 52. $X$ is the number of aces. Find the distribution of $X$ and $E(X)$.

Step 1 — the trials are dependent (no replacement), so use the multiplication theorem, not the binomial formula: $$P(X = 0) = \frac{48}{52} \cdot \frac{47}{51} = \frac{188}{221}, \qquad P(X = 1) = 2 \cdot \frac{4}{52} \cdot \frac{48}{51} = \frac{32}{221}, \qquad P(X = 2) = \frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$$

Step 2 — check: $\frac{188 + 32 + 1}{221} = 1$ ✓

Step 3 — mean: $$E(X) = 1 \cdot \frac{32}{221} + 2 \cdot \frac{1}{221} = \frac{34}{221} = \frac{2}{13}$$

Answer: $E(X) = \dfrac{2}{13}$. (Elegant check: each card is an ace with probability $\frac{1}{13}$, and expectations add even without independence: $E(X) = 2 \times \frac{1}{13}$.)

Example 5 — Mean and variance from a table

$X$ takes values $1, 2, 3, 4$ with probabilities $0.1, 0.2, 0.3, 0.4$. Find the mean, variance and standard deviation.

Step 1 — mean: $E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3$.

Step 2 — second moment: $E(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10$.

Step 3 — variance: $\operatorname{Var}(X) = 10 - 3^2 = 1$, so $\sigma = 1$.

Answer: mean $3$, variance $1$, standard deviation $1$.

Example 6 — Binomial with a fair coin

A fair coin is tossed 6 times. Find the probability of (i) exactly 4 heads, (ii) at least 5 heads.

Step 1 — set up: $X \sim B\left(6, \frac{1}{2}\right)$, and every outcome string has probability $\frac{1}{64}$.

Step 2 — (i): $$P(X = 4) = \binom{6}{4} \left(\frac{1}{2}\right)^6 = \frac{15}{64}$$

Step 3 — (ii): $$P(X \geq 5) = \binom{6}{5}\frac{1}{64} + \binom{6}{6}\frac{1}{64} = \frac{6 + 1}{64} = \frac{7}{64}$$

Answer: (i) $\dfrac{15}{64}$, (ii) $\dfrac{7}{64}$.

Example 7 — Exactly two sixes in four throws

A die is thrown 4 times. Find the probability of getting exactly two sixes.

Step 1 — set up: $X \sim B\left(4, \frac{1}{6}\right)$, $q = \frac{5}{6}$.

Step 2 — formula: $$P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 = 6 \cdot \frac{25}{1296} = \frac{150}{1296} = \frac{25}{216}$$

Answer: $\dfrac{25}{216}$.

Example 8 — Parameter hunting from mean and variance

A binomial distribution has mean 4 and variance 2. Find $n$, $p$ and $P(X = 1)$.

Step 1 — divide variance by mean: $\dfrac{npq}{np} = q = \dfrac{2}{4} = \dfrac{1}{2}$, so $p = \dfrac{1}{2}$.

Step 2 — recover $n$: $np = 4$ gives $n = 8$.

Step 3 — the required probability: $$P(X = 1) = \binom{8}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^7 = \frac{8}{256} = \frac{1}{32}$$

Answer: $n = 8$, $p = \dfrac{1}{2}$, $P(X = 1) = \dfrac{1}{32}$.