MathsISC Class 12 · Probability

Probability Distribution and Mean of a Random Variable

Q: What is a random variable in simple words?

It is a number attached to each outcome of an experiment, such as the number of heads in some tosses or the larger of two chosen numbers. It is called random because its value depends on the outcome.

Q: Is variance in the ISC 2027 syllabus?

The 2027 syllabus lists the probability distribution and the mean of a random variable. This page therefore concentrates on building distributions and finding the mean.

Q: Can E(X) be a value X never takes?

Yes. The mean is a long-run average, so it can fall between the possible values, like 1.5 heads in three tosses.

Q: How do I find k in a probability distribution?

Add all the probabilities, set the total equal to 1 and solve. If you get two values, reject any that makes a probability negative or greater than 1.

Q: How many marks does this topic carry in ISC?

It was a 6-mark question in 2024 (a "larger number" distribution) and a 1-mark expected-value item in 2025, within the 9 marks for Probability.

Updated 15 October 2026 · Fact-checked

A random variable X assigns a number to each outcome of an experiment. Its probability distribution lists every value $x_i$ with its probability $p_i$, where each $p_i \ge 0$ and $\sum p_i = 1$. The mean, or expected value, is $E(X) = \sum x_i p_i$: a probability-weighted average, not the plain average of the values.

The mean E(X) = Σ x·p is the balance point of the bars: heavier probabilities pull it closer, and here the symmetry puts it at 1.

Illustration drawn for QuestionBuddy.

Understand Probability Distribution and Mean of a Random Variable

Random variable. An experiment has outcomes; a random variable $X$ is a rule that gives each outcome a real number. Toss two coins and let $X$ be the number of heads: the outcome HT gives $X = 1$ . Different outcomes can give the same value. In ISC Class 12 the random variables are discrete: they take a list of separate values.

Probability distribution. For each value $x_{i}$ , add the probabilities of all outcomes that give it: $p_{i} = P (X = x_{i})$ . Present the result as a table:

$X$	$x_{1}$	$x_{2}$	…	$x_{n}$
$P (X)$	$p_{1}$	$p_{2}$	…	$p_{n}$

Two conditions make a table a valid distribution: **every $p_{i} \geq 0$ and $\sum p_{i} = 1$ **. Questions with an unknown $k$ use the second condition to find $k$ , then the first to reject a root that would make some probability negative or bigger than 1.

Building a distribution. The 2024 ISC paper built one from a classroom activity: two numbers were chosen from 2, 3, 4, 5 without replacement and $X$ was the larger number. The method is always the same:
1. List the sample space (ordered or unordered, as the experiment demands) and check that the outcomes are equally likely.
2. Read off the value of $X$ for each outcome.
3. Count how many outcomes give each value and divide by the total.
4. Check that the probabilities add to 1.

Probabilities of ranges. $P (X \leq a)$ , $P (X > a)$ or $P (a < X < b)$ are found by adding the relevant entries. Watch the strict and non-strict inequalities: $P (X < 5)$ leaves out $X = 5$ .

Mean (expectation). The mean of $X$ is
$μ = E (X) = \sum_{i} x_{i} p_{i} .$
It is the average value of $X$ over a very large number of repetitions. Each value is weighted by how often it occurs, which is why it is not the simple average of the $x_{i}$ , and not the average of the $p_{i}$ either. The ISC 2025 examiners noted candidates who took "just the average of P(X) instead of the sum of X·P(X)".

Useful facts. The mean need not be a value $X$ can take (the mean number of heads in 3 tosses is 1.5). If every value is multiplied by $a$ and increased by $b$ , the mean becomes $a E (X) + b$ . For a game, let $X$ be the net gain (prize minus stake); the game is fair when $E (X) = 0$ , favourable to the player when $E (X) > 0$ .

Link to statistics. $E (X) = \sum x_{i} p_{i}$ is the same as the mean of a frequency distribution $\frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , with relative frequencies $\frac{f _{i}}{N}$ replaced by probabilities.

Probability Distribution and Mean of a Random Variable: formulas to remember

Validity: $p_{i} \geq 0, \sum p_{i} = 1$; Use the sum to find an unknown constant; use $p_{i} \geq 0$ to reject a root.
Probability of a range: $P (a \leq X \leq b) = \sum_{a \leq x_{i} \leq b} p_{i}$; Check < against ≤.
Mean (expected value): $E (X) = \sum x_{i} p_{i}$; A weighted average.
Linear change: $E (a X + b) = a E (X) + b$; Rescaling a prize or a unit.
Fair game: $E (net gain) = 0$; Net gain = prize − cost to play.

How to solve Probability Distribution and Mean of a Random Variable questions

A distribution question usually has three parts: build or complete the table, find probabilities of ranges, and find the mean.

1Define $X$ in words and list all the values it can take.
2Write the sample space, with its size, and check that outcomes are equally likely.
3Count the outcomes for each value of $X$ and write $P (X = x_{i})$ as a fraction over the same denominator.
4Check that the probabilities add to 1 (or use this to find $k$ , then reject any root that gives a negative probability).
5Add the required entries for range probabilities.
6Make a row $x_{i} p_{i}$ and add it to get $E (X)$ ; interpret the answer in context.

Probability Distribution and Mean of a Random Variable in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Expected value as a long-run average

Suppose a phone repair shop finds that the number of repairs X a walk-in customer needs is 1, 2 or 3 with probabilities 0.6, 0.3 and 0.1, and each repair takes 20 minutes.

Find the expected number of repairs per customer and the expected repair time per customer.

Try it first, then show the solutionSolution

$E (X) = 1 (0.6) + 2 (0.3) + 3 (0.1) = 1.5$ .
Time $T = 20 X$ minutes, so $E (T) = 20 E (X) = 30$ minutes.

Answer: 1.5 repairs and 30 minutes per customer, on average.

Memory hook: Mean = each value × how often it happens, all added.

Is a game fair?

Imagine a fete stall where you pay ₹30 to roll one fair die. You receive ₹120 for a six, ₹30 for a four or five, and nothing otherwise.

Find the expected net gain per game. Who does the game favour?

Try it first, then show the solutionSolution

Net gain: ₹90 with probability $\frac{1}{6}$ ; ₹0 with probability $\frac{2}{6}$ ; −₹30 with probability $\frac{3}{6}$ .
$E = 90 \cdot \frac{1}{6} + 0 - 30 \cdot \frac{3}{6} = 15 - 15 = 0$ .

Answer: Expected net gain ₹0: the game is fair, favouring neither side.

Memory hook: Fair means the expected NET gain is zero; always subtract the stake.

Tricks to remember Probability Distribution and Mean of a Random Variable

Short memory aids. Say each one aloud once, then use it on a question the same day.

Two rules of a distribution
"None negative, total one."
Use it when: Checking a table and finding $k$ .
Multiply across, add down
Write an extra row $x P$ under the table; its total is $E (X)$ .
Use it when: Every mean calculation.
Mean is not the middle
The mean leans towards the values with the biggest probabilities.
Use it when: Sanity-checking $E (X)$ .
Net, not prize
For a game, X is what you end up with minus what you paid.
Use it when: Fair-game questions.

Probability Distribution and Mean of a Random Variable: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mean from a table · 1 mark · MCQBoard level

X takes the values −20, 0 and 40 with probabilities $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ . Then $E (X)$ is:
(a) 5
(b) $\frac{20}{3}$
(c) 10
(d) 0

Try it first, then show the solutionSolution

$E (X) = - 20 \cdot \frac{1}{4} + 0 + 40 \cdot \frac{1}{4} = - 5 + 10 = 5$ . (Option (b) is the simple average of the values.)

Answer: (a) 5.

Example 2: Finding k and rejecting a root · 3 marks · Short answerHarder board level

The probability distribution of X is $P (X = 0) = 2 k^{2}$ , $P (X = 1) = k$ , $P (X = 2) = 3 k$ , $P (X = 3) = k^{2} + 2 k$ . Find k, $P (X \geq 2)$ and $E (X)$ .

Try it first, then show the solutionSolution

Sum: $3 k^{2} + 6 k = 1$ , so $3 k^{2} + 6 k - 1 = 0$ and $k = \frac{- 6 \pm 48}{6} = - 1 \pm \frac{2}{3}$ . (1 mark)
Probabilities cannot be negative, so $k = \frac{2}{3} - 1 = \frac{2 3 - 3}{3} \approx 0.1547$ . (1 mark)
$P (X \geq 2) = 3 k + k^{2} + 2 k = k^{2} + 5 k \approx 0.797$ ; $E (X) = k + 6 k + 3 (k^{2} + 2 k) = 3 k^{2} + 13 k \approx 2.083$ . (1 mark)

Answer: $k = \frac{2 3 - 3}{3} \approx 0.155$ ; $P (X \geq 2) \approx 0.797$ ; $E (X) \approx 2.08$ .

Example 3: Building a distribution by counting · 6 marks · Long answerHarder board level

Two numbers are chosen at random, one after the other without replacement, from 1, 2, 3, 4, 5, 6. Let X be the smaller of the two. Find the probability distribution of X, $P (X < 3)$ and $E (X)$ .

Try it first, then show the solutionSolution

Ordered pairs with distinct numbers: $6 \times 5 = 30$ , all equally likely. (1 mark)
Smaller number $m$ with the other larger: $6 - m$ choices for the other number, and 2 orders, so $2 (6 - m)$ pairs. (1 mark)
X = 1, 2, 3, 4, 5 with counts 10, 8, 6, 4, 2: probabilities $\frac{1}{3}, \frac{4}{15}, \frac{1}{5}, \frac{2}{15}, \frac{1}{15}$ ; they add to 1. (2 marks)
$P (X < 3) = \frac{10 + 8}{30} = \frac{3}{5}$ . (1 mark)
$E (X) = \frac{1 ( 10 ) + 2 ( 8 ) + 3 ( 6 ) + 4 ( 4 ) + 5 ( 2 )}{30} = \frac{70}{30} = \frac{7}{3}$ . (1 mark)

Answer: X: 1, 2, 3, 4, 5 with P: $\frac{1}{3}, \frac{4}{15}, \frac{1}{5}, \frac{2}{15}, \frac{1}{15}$ ; $P (X < 3) = \frac{3}{5}$ ; $E (X) = \frac{7}{3}$ .

Example 4: Draws without replacement · 4 marks · Long answerHarder board level

A drawer has 4 working and 2 dead batteries. Three batteries are taken at random together. Let X be the number of dead batteries taken. Find the distribution of X and its mean.

Try it first, then show the solutionSolution

Total selections: $(3 6) = 20$ . X can be 0, 1 or 2. (1 mark)
$P (0) = \frac{( 3 4 )}{20} = \frac{1}{5}$ ; $P (1) = \frac{( 1 2 ) ( 2 4 )}{20} = \frac{3}{5}$ ; $P (2) = \frac{( 2 2 ) ( 1 4 )}{20} = \frac{1}{5}$ . (2 marks)
$E (X) = 0 + \frac{3}{5} + \frac{2}{5} = 1$ . (1 mark)

Answer: P(0) = P(2) = $\frac{1}{5}$ , P(1) = $\frac{3}{5}$ ; $E (X) = 1$ .

Example 5: Expected number of trials · 4 marks · Case-basedHOTS / competency

A learner driver is allowed at most three attempts at a test. Suppose the chance of passing is 0.5 on the first attempt, 0.6 on the second and 0.8 on the third (each given the earlier attempts failed). Let X be the number of attempts taken (stop at a pass or after three attempts). Find the distribution of X, $E (X)$ and the probability that the learner never passes.

Try it first, then show the solutionSolution

$P (X = 1) = 0.5$ ; $P (X = 2) = 0.5 \times 0.6 = 0.3$ ; $P (X = 3) = 0.5 \times 0.4 = 0.2$ (the third attempt happens whatever its result).
Check: $0.5 + 0.3 + 0.2 = 1$ .
$E (X) = 0.5 + 0.6 + 0.6 = 1.7$ attempts.
Never passes: $0.5 \times 0.4 \times 0.2 = 0.04$ .

Answer: X = 1, 2, 3 with P = 0.5, 0.3, 0.2; E(X) = 1.7; P(never passes) = 0.04.

Where students lose marks in Probability Distribution and Mean of a Random Variable

The slips examiners see most often, and what to write instead.

Mistake: Averaging instead of weighting
The ISC 2025 examiners noted candidates who took the average of P(X) instead of $\sum X \cdot P (X)$ .
Wrong working
$E (X) = \frac{p _{1} + p _{2} + p _{3}}{3}$ .
Right working
$E (X) = x_{1} p_{1} + x_{2} p_{2} + x_{3} p_{3}$ .
Do this: Always write the $x P$ row and add it.
Usually costs: 1 mark.
Mistake: Not checking the probabilities add to 1
A miscounted sample space goes unnoticed and every later part is wrong.
Do this: Add the row before finding the mean.
Usually costs: 2 to 3 marks.
Mistake: Keeping a root that gives a negative probability
Quadratics in k give two roots; only one is valid.
Do this: Substitute both and reject any that makes some $p_{i} < 0$ or $> 1$ .
Usually costs: 1 mark.
Mistake: Strict versus non-strict inequality
$P (X < 5)$ and $P (X \leq 5)$ differ by $P (X = 5)$ .
Do this: Circle the inequality sign before adding entries.
Usually costs: 1 mark.

Exam tips

Present the distribution as a two-row table; examiners look for it.
Keep a common denominator in the table so that checking the sum and finding the mean are easy.
State the sample space size and whether pairs are ordered.
Interpret $E (X)$ in context ("on average 1.7 attempts"), especially in case-study questions.

Practice questions on Probability Distribution and Mean of a Random Variable

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these can be a probability distribution of X = 0, 1, 2?
(a) 0.5, 0.6, −0.1
(b) 0.3, 0.3, 0.3
(c) 0.2, 0.5, 0.3
(d) 0.4, 0.4, 0.4
Hint
None negative, total one.
Answer
(c).
Q2 · 2 marks · Short answerBoard level
X takes values 1, 2, 3 with $P (X = x) = \frac{x + 1}{k}$ . Find k and $E (X)$ .
Hint
$\frac{2 + 3 + 4}{k} = 1$ .
Answer
$k = 9$ ; $E (X) = \frac{2 + 6 + 12}{9} = \frac{20}{9}$ .
Q3 · 3 marks · Short answerBoard level
A pair of fair dice is thrown. Let X be the number of sixes. Find the distribution of X and its mean.
Hint
$P (six) = \frac{1}{6}$ on each die.
Answer
P(0) = $\frac{25}{36}$ , P(1) = $\frac{10}{36}$ , P(2) = $\frac{1}{36}$ ; $E (X) = \frac{12}{36} = \frac{1}{3}$ .
Q4 · 3 marks · Short answerHarder board level
Two cards are drawn together from a pack of 52. Let X be the number of kings drawn. Find $E (X)$ .
Hint
Use $(2 52) = 1326$ .
Answer
P(0) = $\frac{1128}{1326}$ , P(1) = $\frac{192}{1326}$ , P(2) = $\frac{6}{1326}$ ; $E (X) = \frac{204}{1326} = \frac{2}{13}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If X takes values 1, 2, 3 with probabilities 0.2, 0.3, 0.5, then $E (X) = 2$ .
Reason (R): The mean of a random variable is the average of its values.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) Both false.
Hint
Compute $\sum x p$ .
Answer
(d): $E (X) = 0.2 + 0.6 + 1.5 = 2.3$ , and the mean is a weighted average, not the plain average.
Q6 · 4 marks · Long answerHOTS / competency
Suppose a coin is tossed until a head appears or until it has been tossed 4 times. Let X be the number of tosses. Find the distribution of X and $E (X)$ .
Hint
X = 4 includes TTTH and TTTT.
Answer
P(1) = $\frac{1}{2}$ , P(2) = $\frac{1}{4}$ , P(3) = $\frac{1}{8}$ , P(4) = $\frac{1}{8}$ ; $E (X) = \frac{15}{8}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose an insurer sells a ₹2000-a-year gadget policy. In a year a claim of ₹15 000 is made with probability 0.05, a claim of ₹5000 with probability 0.1, and no claim otherwise.
(i) Write the distribution of the insurer's net gain per policy. (2 marks)
(ii) Find the expected net gain. (1 mark)
(iii) What premium would make the policy fair? (1 mark)
Hint
Net gain = premium − claim.
Answer
(i) −₹13 000 (0.05), −₹3000 (0.1), ₹2000 (0.85). (ii) $- 650 - 300 + 1700 = 750$ , that is ₹750. (iii) Expected claim ₹1250, so a premium of ₹1250.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What two conditions must a probability distribution satisfy?Show answer
Each $p_{i} \geq 0$ and $\sum p_{i} = 1$ .
2.Define the mean of a discrete random variable.Show answer
$E (X) = \sum x_{i} p_{i}$ .
3.X = 0, 1 with P = 0.3, 0.7. E(X)?Show answer
0.7.
4.If E(X) = 4, what is E(3X − 2)?Show answer
10.
5.When is a game fair?Show answer
When the expected net gain of the player is zero.

Probability Distribution and Mean of a Random Variable: frequently asked questions

What is a random variable in simple words?

It is a number attached to each outcome of an experiment, such as the number of heads in some tosses or the larger of two chosen numbers. It is called random because its value depends on the outcome.

Is variance in the ISC 2027 syllabus?

The 2027 syllabus lists the probability distribution and the mean of a random variable. This page therefore concentrates on building distributions and finding the mean.

Can E(X) be a value X never takes?

Yes. The mean is a long-run average, so it can fall between the possible values, like 1.5 heads in three tosses.

How do I find k in a probability distribution?

Add all the probabilities, set the total equal to 1 and solve. If you get two values, reject any that makes a probability negative or greater than 1.

How many marks does this topic carry in ISC?

It was a 6-mark question in 2024 (a "larger number" distribution) and a 1-mark expected-value item in 2025, within the 9 marks for Probability.