Probability and Random Events Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Probability and Random Events notes for XAT: 10 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Random Experiments and Sample Space

Random Experiment

An action with well-defined, uncertain outcomes.

Sample Space ( $S$ )

The set of all possible outcomes.

Event ( $E$ )

A subset of the sample space.

Example: Tossing two coins

$S = {H H, H T, T H, T T}$

Event $E$ (getting exactly one head) $= {H T, T H}$

Classical Definition of Probability

P (E) = \frac{Number of favorable outcomes n ( E )}{Total number of outcomes n ( S )}

Condition: All outcomes in the sample space must be equally likely.

Example: Rolling a fair six-sided die. Find P(prime number).

Total outcomes

n (S) = 6

Favorable outcomes (primes: 2, 3, 5)

n (E) = 3

P (E) = \frac{3}{6} = \frac{1}{2}

Basic Random Selection from a Group

Problem: A bag contains 3 red, 4 blue, and 5 green balls. One ball is drawn at random. Find the probability that it is blue.

Step 1: Total Items

T o t a l = 3 + 4 + 5 = 12

Step 2: Favorable Items

B l u e = 4

Step 3: Probability

P (B l u e) = \frac{4}{12} = \frac{1}{3}

Complementary Events in Selection

P (E) + P (not E) = 1

P (E) = 1 - P (not E)

Application

Highly recommended for "at least one" scenarios in multiple selections. Calculating the exact combinations for "at least one" is tedious, while calculating "none" requires only one combination.

Probability and Random Events: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $S$ be the set of all permutations of the digits ${1, 2, 3, 4, 5, 6}$ . A permutation is chosen uniformly at random from $S$ .

Let $A$ be the event that the digit 1 appears before the digit 2.

Let $B$ be the event that the digit 3 appears before the digit 4.

Let $C$ be the event that the digit 5 appears before the digit 6.

What is the probability that exactly two of these events occur?

A.
1/4
B.
3/8
C.
1/2
D.
5/8

Correct Answer:

Step-by-Step Solution

Key idea: Symmetry and Independence of relative orderings.

Step 1: Analyze individual probabilities.

In any random permutation, for any distinct pair of digits (e.g., 1 and 2), the probability that one appears before the other is $1/2$ .

So, $P (A) = 1/2$ , $P (B) = 1/2$ , $P (C) = 1/2$ .

Step 2: Analyze independence.

The relative ordering of disjoint pairs (1,2), (3,4), and (5,6) are mutually independent events.

Why? The positions of 1 and 2 are symmetric with respect to 3 and 4. There is no bias introduced by the relative order of 3 and 4 on the relative order of 1 and 2.

Thus, $A, B, C$ are independent events.

Step 3: Define the target event.

We want exactly two events to occur.

Possible scenarios:

$A \cap B \cap C^{c}$
$A \cap B^{c} \cap C$
$A^{c} \cap B \cap C$

Step 4: Calculate probability for one scenario.

$P (A \cap B \cap C^{c}) = P (A) P (B) P (C^{c})$ (due to independence)

$= (1/2) \times (1/2) \times (1/2) = 1/8$ .

Step 5: Sum the probabilities.

Since the scenarios are mutually exclusive:

$P (Exactly 2) = 1/8 + 1/8 + 1/8 = 3/8$ .

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

A box contains 4 Red, 4 Green, and 4 Blue balls. Three balls are drawn simultaneously at random.

Let $X$ be the number of different colors among the drawn balls.

What is the expected value of $X$ ?

Correct Answer:

2.2

Step-by-Step Solution

Key idea: Linearity of Expectation with Indicator Variables.

Step 1: Define Indicator Variables.

Let $I_{R}$ be 1 if at least one Red ball is drawn, 0 otherwise.

Let $I_{G}$ be 1 if at least one Green ball is drawn, 0 otherwise.

Let $I_{B}$ be 1 if at least one Blue ball is drawn, 0 otherwise.

Then $X = I_{R} + I_{G} + I_{B}$ .

By Linearity of Expectation:

$E [X] = E [I_{R}] + E [I_{G}] + E [I_{B}]$ .

Step 2: Calculate $E [I_{R}]$ .

$E [I_{R}] = P (I_{R} = 1) = P (At least one Red)$ .

$P (At least one Red) = 1 - P (No Red)$ .

Total balls = 12. Draw 3.

Total ways = $(3 12) = \frac{12 \times 11 \times 10}{6} = 220$ .

Ways to draw No Red (only from 4G, 4B = 8 balls):

$(3 8) = \frac{8 \times 7 \times 6}{6} = 56$ .

$P (No Red) = \frac{56}{220}$ .

$P (At least one Red) = 1 - \frac{56}{220} = \frac{164}{220}$ .

Step 3: Symmetry.

By symmetry, $E [I_{G}] = E [I_{B}] = E [I_{R}] = \frac{164}{220}$ .

Step 4: Sum.

$E [X] = 3 \times \frac{164}{220}$ .

Simplify fraction:

$164/220$ . Divide by 4: $41/55$ .

$E [X] = 3 \times \frac{41}{55} = \frac{123}{55}$ .

Calculate decimal:

$123 \div 55 \approx 2.236$ .

Wait, let's re-calculate $123/55$ .

$55 \times 2 = 110$ . Remainder 13.

$130/55 = 2.36...$

$123/55 = 2.23636...$

Is the answer a clean decimal?

Let's check the calculation of $(3 12)$ .

$12 \times 11 \times 10/6 = 2 \times 11 \times 10 = 220$ . Correct.

$(3 8) = 56$ . Correct.

$1 - 56/220 = 164/220$ . Correct.

$164/4 = 41$ . $220/4 = 55$ . Correct.

$3 \times 41 = 123$ .

$123/55$ .

The question asks for NAT. Usually 2 decimal places.

Answer: 2.24.

Let's double check if "Expected Value" implies a fraction or decimal. NAT usually accepts decimals.

Let's try to verify with direct counting.

Possible values for X: 1, 2, 3.

P(X=1): All same color.

3 Reds: $(3 4) = 4$ .

3 Greens: 4.

3 Blues: 4.

Total 12.

$P (X = 1) = 12/220$ .

P(X=3): All different colors (1R, 1G, 1B).

$4 \times 4 \times 4 = 64$ .

$P (X = 3) = 64/220$ .

P(X=2): 1 - P(X=1) - P(X=3) = $1 - 12/220 - 64/220 = 144/220$ .

Check sum: $12 + 64 + 144 = 220$ . Correct.

$E [X] = 1 \times \frac{12}{220} + 2 \times \frac{144}{220} + 3 \times \frac{64}{220}$ .

Numerator: $12 + 288 + 192 = 492$ .

$E [X] = \frac{492}{220}$ .

Divide by 4: $\frac{123}{55}$ .

Matches.

$123/55 = 2.23636...$

Rounding to 2 decimal places: 2.24.

More notes in this unit

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Probability and Random Events Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Probability and Random Events notes for XAT: 10 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

Let S be the set of all permutations of the digits {1, 2, 3, 4, 5, 6} . A permutation is chosen uniformly at random from S . Let A be the event that the digit 1 appears before the digit 2. Let B be the event that the digit 3 appears before the digit 4. Let C be the event that the digit 5 appears before the digit 6. What is the probability that exactly two of these events occur?

Question 2

A box contains 4 Red, 4 Green, and 4 Blue balls. Three balls are drawn simultaneously at random. Let X be the number of different colors among the drawn balls. What is the expected value of X ?

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Probability and Random Events Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Random Experiments and Sample Space

Random Experiments and Sample Space

Random Experiment

Sample Space (S)

Event (E)

Example: Tossing two coins

Classical Definition of Probability

Classical Definition of Probability

Basic Random Selection from a Group

Basic Random Selection from a Group

Complementary Events in Selection

Complementary Events in Selection

Probability and Random Events: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit

A question from this chapter

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

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Unlock the whole course

Sample Space ( $S$ )

Event ( $E$ )